Skip to main content

Mathematics AI · HL · Unit 1: Number and Algebra

Unit 1Topic 1.13Key Concepts99 min read

Polar and exponential forms of complex numbers

Reading progress0%
Back to course
On this page

Modulus, argument and Argand geometry

A non-zero complex number z=x+yiz=x+yi can be located by two Cartesian coordinates or by its distance and direction from the origin. The distance is its modulus r=∣z∣r=|z|, and the directed angle from the positive real axis is an argument θ\theta.

The modulus follows from Pythagoras: r=x2+y2r=\sqrt{x^2+y^2}. The argument is measured counterclockwise as positive and clockwise as negative, normally in radians unless a question explicitly uses degrees.

The horizontal and vertical components of the position vector are x=rcos⁡θx=r\cos\theta and y=rsin⁡θy=r\sin\theta. These equations connect Cartesian and polar descriptions.

A position vector showing modulus, argument and Cartesian projections on an Argand plane.The vector has modulus rr, argument θ\theta, and Cartesian projections rcos⁡θr\cos\theta and rsin⁡θr\sin\theta.

Open graph full size (opens in a new tab)

In the visual, the diagonal vector has length rr, the angle arc represents θ\theta, and the two perpendicular projections represent rcos⁡θr\cos\theta and rsin⁡θr\sin\theta. Exact labels remain in the surrounding mathematical text so the diagram stays visually clean.

Key terms

Modulus
The non-negative distance |z| from the origin to z.
Argument
A directed angle from the positive real axis to the position vector of a non-zero complex number.
Argand diagram
A coordinate plane whose horizontal and vertical axes represent real and imaginary parts.
Position vector
The directed segment from the origin to the point representing z.

Worked examples

Modulus from Cartesian coordinates

Find the modulus of z=3+4iz=3+4i.

  1. 1Identify x=3x=3 and y=4y=4.
  2. 2Use r=x2+y2r=\sqrt{x^2+y^2}.
  3. 3Calculate 9+16\sqrt{9+16}.

Answer: ∣z∣=5|z|=5.

Argument in the first quadrant

Find an argument of z=1+3iz=1+\sqrt3 i.

  1. 1The point lies in quadrant I.
  2. 2Use tan⁡θ=31\tan\theta=\frac{\sqrt3}{1}.
  3. 3Choose the quadrant-I angle.

Answer: θ=π3\theta=\frac{\pi}{3}.

Read Cartesian components from polar data

A complex number has modulus 6 and argument 2π3\frac{2\pi}{3}. Find its Cartesian coordinates.

  1. 1Use x=6cos⁡2π3x=6\cos\frac{2\pi}{3}.
  2. 2Use y=6sin⁡2π3y=6\sin\frac{2\pi}{3}.

Answer: The point is (−3,33)(-3,3\sqrt3), so z=−3+33iz=-3+3\sqrt3 i.

Rules and formulas

Modulus

r=∣z∣=x2+y2r=|z|=\sqrt{x^2+y^2}

Cartesian projections

x=rcos⁡θ,y=rsin⁡θx=r\cos\theta,\qquad y=r\sin\theta

Tangent relation

tan⁡θ=yx\tan\theta=\frac{y}{x}
xx
non-zero real component

Cartesian and polar forms

Cartesian form x+yix+yi displays horizontal and vertical components. Polar form r(cos⁡θ+isin⁡θ)r(\cos\theta+i\sin\theta) displays size and direction.

For a non-zero complex number, r>0r>0. Arguments that differ by a whole turn represent the same direction, so θ+2kπ\theta+2k\pi gives the same complex number for every integer kk.

Polar form is especially efficient for multiplication and division because moduli scale while arguments rotate. Cartesian form remains efficient for addition and subtraction.

Always state whether an angle is a principal argument, a particular argument, or the general family of arguments.

Key terms

Cartesian form
The representation x + yi.
Polar form
The representation r(cosθ+isinθ)r(cos θ + i sin θ), where r is the modulus.
Equivalent arguments
Angles differing by 2kπ that point in the same direction.

Worked examples

Cartesian to polar in quadrant I

Write 1+3i1+\sqrt3 i in polar form.

  1. 1Find r=1+3=2r=\sqrt{1+3}=2.
  2. 2The argument is π3\frac{\pi}{3}.
  3. 3Substitute r and θ into polar form.

Answer: 2(cos⁡π3+isin⁡π3)2\left(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}\right).

Polar to Cartesian

Convert 2(cos⁡2π3+isin⁡2π3)2\left(\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}\right) to Cartesian form.

  1. 1Use cos⁡2π3=−12\cos\frac{2\pi}{3}=-\frac12.
  2. 2Use sin⁡2π3=32\sin\frac{2\pi}{3}=\frac{\sqrt3}{2}.
  3. 3Multiply both components by 2.

Answer: −1+3i-1+\sqrt3 i.

A general argument

Give all arguments of z=1+iz=1+i.

  1. 1The principal direction is π4\frac{\pi}{4}.
  2. 2Add any integer multiple of 2π2\pi.

Answer: arg⁡(z)=π4+2kπ\arg(z)=\frac{\pi}{4}+2k\pi, where k∈Zk\in\mathbb Z.

Rules and formulas

Polar form

z=r(cos⁡θ+isin⁡θ)z=r(\cos\theta+i\sin\theta)

Equivalent arguments

θk=θ+2kπ,k∈Z\theta_k=\theta+2k\pi,\qquad k\in\mathbb Z

Cartesian reconstruction

z=rcos⁡θ+irsin⁡θz=r\cos\theta+i r\sin\theta

Cis notation

Cis notation abbreviates the trigonometric factor: cis⁡θ=cos⁡θ+isin⁡θ\operatorname{cis}\theta=\cos\theta+i\sin\theta. Therefore z=rcis⁡θz=r\operatorname{cis}\theta.

The notation is compact but carries exactly the same information as full polar form. It does not mean the product of three variables c, i, and s.

Because sine and cosine are both 2π2\pi-periodic, cis⁡(θ+2kπ)=cis⁡θ\operatorname{cis}(\theta+2k\pi)=\operatorname{cis}\theta.

On the unit circle, cis⁡θ\operatorname{cis}\theta has modulus 1 and argument θ\theta. Multiplication by r scales this unit vector to modulus r.

Key terms

Cis
The abbreviation cis θ = cos θ + i sin θ.
Unit complex number
A complex number of modulus 1.
Trigonometric factor
The unit-modulus directional part cos θ + i sin θ.

Worked examples

Expand cis notation

Write 4cis⁡π64\operatorname{cis}\frac{\pi}{6} in Cartesian form.

  1. 1Use cos⁡π6=32\cos\frac{\pi}{6}=\frac{\sqrt3}{2} and sin⁡π6=12\sin\frac{\pi}{6}=\frac12.
  2. 2Multiply both components by 4.

Answer: 23+2i2\sqrt3+2i.

Compress full polar form

Rewrite 7(cos⁡1.2+isin⁡1.2)7(\cos1.2+i\sin1.2) using cis notation.

  1. 1Replace the parenthesized trigonometric factor by cis⁡(1.2)\operatorname{cis}(1.2).

Answer: 7cis⁡(1.2)7\operatorname{cis}(1.2).

Equivalent cis angles

Explain why 3cis⁡9π4=3cis⁡π43\operatorname{cis}\frac{9\pi}{4}=3\operatorname{cis}\frac{\pi}{4}.

  1. 1Subtract the angles: 9π4−π4=2π\frac{9\pi}{4}-\frac{\pi}{4}=2\pi.
  2. 2A complete turn does not change the direction.

Answer: The two expressions represent the same complex number.

Rules and formulas

Cis definition

cis⁡θ=cos⁡θ+isin⁡θ\operatorname{cis}\theta=\cos\theta+i\sin\theta

Cis form

z=rcis⁡θz=r\operatorname{cis}\theta

Cis periodicity

cis⁡(θ+2kπ)=cis⁡θ\operatorname{cis}(\theta+2k\pi)=\operatorname{cis}\theta

Exponential form and Euler's relation

Euler's relation connects exponential and trigonometric notation: eiθ=cos⁡θ+isin⁡θe^{i\theta}=\cos\theta+i\sin\theta. Hence cis⁡θ=eiθ\operatorname{cis}\theta=e^{i\theta}.

The exponential form of a non-zero complex number is z=reiθz=re^{i\theta}, where r=∣z∣>0r=|z|>0 and θ is an argument.

The familiar exponential index laws explain polar multiplication and division: multiplying adds arguments, while dividing subtracts them.

Exponential form is a representation, not a numerical approximation. Keep angles exact in terms of π\pi when exact values are available.

Key terms

Euler's relation
The identity eiθ=cos⁡θ+isin⁡θe^{i\theta}=\cos\theta+i\sin\theta.
Exponential form
The representation reiθre^{i\theta} of a non-zero complex number.
Natural exponential base
The number e used in exponential functions.

Worked examples

Cis to exponential form

Write 5cis⁡(−π3)5\operatorname{cis}(-\frac{\pi}{3}) in exponential form.

  1. 1Use cis⁡θ=eiθ\operatorname{cis}\theta=e^{i\theta}.
  2. 2Substitute θ=−π3\theta=-\frac{\pi}{3}.

Answer: 5e−iπ35e^{\frac{-i\pi}{3}}.

Exponential to Cartesian form

Convert 5e−iπ65e^{\frac{-i\pi}{6}} to Cartesian form.

  1. 1Use Euler's relation.
  2. 2Evaluate cos⁡(−π6)=32\cos(-\frac{\pi}{6})=\frac{\sqrt3}{2} and sin⁡(−π6)=−12\sin(-\frac{\pi}{6})=-\frac12.
  3. 3Multiply by 5.

Answer: 532−52i\frac{5\sqrt3}{2}-\frac52 i.

Euler's famous special value

Evaluate eiπe^{i\pi}.

  1. 1Use eiπ=cos⁡π+isin⁡πe^{i\pi}=\cos\pi+i\sin\pi.
  2. 2Substitute cos⁡π=−1\cos\pi=-1 and sin⁡π=0\sin\pi=0.

Answer: eiπ=−1e^{i\pi}=-1, so eiπ+1=0e^{i\pi}+1=0.

Rules and formulas

Euler's relation

eiθ=cos⁡θ+isin⁡θe^{i\theta}=\cos\theta+i\sin\theta

Exponential form

z=reiθz=re^{i\theta}

Equivalent notation

reiθ=rcis⁡θre^{i\theta}=r\operatorname{cis}\theta

Principal and general arguments

A non-zero complex number has infinitely many arguments because adding 2π2\pi completes one full turn. The multivalued notation arg⁡z\arg z can describe this entire family.

The principal argument Arg⁡z\operatorname{Arg}z is the unique representative in a specified interval. In these notes the convention is −π<Arg⁡z≤π-\pi<\operatorname{Arg}z\le\pi.

Under this convention, a negative real number has principal argument π\pi, not −π-\pi. A negative imaginary number has principal argument −π2-\frac{\pi}{2}.

Some calculators or texts use 0≤θ<2π0\le\theta<2\pi. If so, convert the result to the convention requested by the question.

Key terms

General argument
The family θ + 2kπ for all integers k.
Principal argument
The unique argument in the adopted principal interval.
Principal interval
Here, the interval −π < Arg z ≤ π.

Worked examples

Convert to a principal argument

Find the principal argument equivalent to 4π3\frac{4\pi}{3}.

  1. 1The angle exceeds π\pi.
  2. 2Subtract 2π2\pi: 4π3−2π=−2π3\frac{4\pi}{3}-2\pi=-\frac{2\pi}{3}.

Answer: Arg⁡z=−2π3\operatorname{Arg}z=-\frac{2\pi}{3}.

Quadrant II argument

Find the principal argument of −3+i-\sqrt3+i.

  1. 1The point lies in quadrant II.
  2. 2Its reference angle is π6\frac{\pi}{6}.
  3. 3Use π−π6\pi-\frac{\pi}{6}.

Answer: Arg⁡z=5π6\operatorname{Arg}z=\frac{5\pi}{6}.

Quadrant III argument

Find the principal argument of −1−3i-1-\sqrt3 i.

  1. 1The point lies in quadrant III.
  2. 2The reference angle is π3\frac{\pi}{3}.
  3. 3Choose the clockwise representative inside the principal interval.

Answer: Arg⁡z=−2π3\operatorname{Arg}z=-\frac{2\pi}{3}.

Point on a negative axis

Find Arg⁡(−5)\operatorname{Arg}(-5) under −π<Arg⁡z≤π-\pi<\operatorname{Arg}z\le\pi.

  1. 1The point lies on the negative real axis.
  2. 2The included endpoint is π\pi.

Answer: Arg⁡(−5)=π\operatorname{Arg}(-5)=\pi.

Rules and formulas

Principal-argument convention

−π<Arg⁡z≤π-\pi<\operatorname{Arg}z\le\pi

General arguments

arg⁡z=Arg⁡z+2kπ,k∈Z\arg z=\operatorname{Arg}z+2k\pi,\qquad k\in\mathbb Z

Converting Cartesian form to polar, cis and exponential forms

Start with z=x+yiz=x+yi. First calculate r=x2+y2r=\sqrt{x^2+y^2}, then determine a quadrant-correct argument θ.

A reliable manual method is: sketch the point, find a positive reference angle from tan⁡−1(∣yx∣)\tan^{-1}(\frac{|y}{x|}), then adjust it for the quadrant.

Once r and θ are known, the three equivalent outputs are r(cos⁡θ+isin⁡θ)r(\cos\theta+i\sin\theta), rcis⁡θr\operatorname{cis}\theta, and reiθre^{i\theta}.

For points on an axis, read the direction directly rather than dividing by zero. For example, −3i-3i has modulus 3 and principal argument −π2-\frac{\pi}{2}.

Key terms

Reference angle
The acute positive angle between a position vector and the nearest real-axis direction.
Quadrant correction
Adjusting an inverse-tangent result so that it matches the signs of x and y.

Worked examples

First-quadrant exact conversion

Convert 1+3i1+\sqrt3 i to cis and exponential forms.

  1. 1Find r=2r=2.
  2. 2Find θ=π3\theta=\frac{\pi}{3}.
  3. 3Write both equivalent representations.

Answer: 2cis⁡π3=2eiπ32\operatorname{cis}\frac{\pi}{3}=2e^{\frac{i\pi}{3}}.

Second-quadrant exact conversion

Convert −3+i-\sqrt3+i to principal polar form.

  1. 1Find r=3+1=2r=\sqrt{3+1}=2.
  2. 2The reference angle is π6\frac{\pi}{6}.
  3. 3Quadrant II gives θ=5π6\theta=\frac{5\pi}{6}.

Answer: 2cis⁡5π62\operatorname{cis}\frac{5\pi}{6}.

Fourth-quadrant conversion

Convert (1-i) to principal exponential form.

  1. 1Find r=2r=\sqrt2.
  2. 2The point is in quadrant IV with reference angle π4\frac{\pi}{4}.
  3. 3Use the principal argument −π4-\frac{\pi}{4}.

Answer: 2e−iπ4\sqrt2 e^{\frac{-i\pi}{4}}.

Axis conversion

Convert −3i-3i to cis form using a principal argument.

  1. 1The modulus is 3.
  2. 2The direction is straight down, so θ is −π2-\frac{\pi}{2}.

Answer: 3cis⁡(−π2)3\operatorname{cis}(-\frac{\pi}{2}).

Rules and formulas

Cartesian-to-polar modulus

r=x2+y2r=\sqrt{x^2+y^2}

Cartesian-to-polar argument

θ=atan2⁡(y,x)\theta=\operatorname{atan2}(y,x)

Equivalent output forms

r(cos⁡θ+isin⁡θ)=rcis⁡θ=reiθr(\cos\theta+i\sin\theta)=r\operatorname{cis}\theta=re^{i\theta}

Converting polar, cis and exponential forms to Cartesian form

To convert from polar form, evaluate x=rcos⁡θx=r\cos\theta and y=rsin⁡θy=r\sin\theta, then write z=x+yiz=x+yi.

Use exact unit-circle values whenever the angle is a standard angle. Do not replace 32\frac{\sqrt3}{2} by a rounded decimal unless approximation is requested.

For a non-standard angle, use a calculator in radian mode and keep extra digits during the calculation. Round only the final real and imaginary parts as instructed.

Negative sine or cosine values automatically place the Cartesian result in the correct quadrant.

Key terms

Exact conversion
A conversion using exact trigonometric values, fractions, and radicals.
Approximate conversion
A decimal conversion rounded to a stated accuracy.

Worked examples

Cis to exact Cartesian form

Convert 4cis⁡5π64\operatorname{cis}\frac{5\pi}{6} to Cartesian form.

  1. 1Use cos⁡5π6=−32\cos\frac{5\pi}{6}=-\frac{\sqrt3}{2}.
  2. 2Use sin⁡5π6=12\sin\frac{5\pi}{6}=\frac12.
  3. 3Multiply by 4.

Answer: −23+2i-2\sqrt3+2i.

Exponential to exact Cartesian form

Convert 6e−iπ26e^{\frac{-i\pi}{2}} to Cartesian form.

  1. 1Replace the exponential factor by cos⁡(−π2)+isin⁡(−π2)\cos(-\frac{\pi}{2})+i\sin(-\frac{\pi}{2}).
  2. 2Use cosine 0 and sine −1-1.

Answer: −6i-6i.

Non-standard angle

Convert 3e1.1i3e^{1.1i} to Cartesian form, correct to three significant figures.

  1. 1Set the GDC to radian mode.
  2. 2Calculate x=3cos⁡1.1x=3\cos1.1 and y=3sin⁡1.1y=3\sin1.1.
  3. 3Round only the final components.

Answer: z≈1.36+2.67iz\approx1.36+2.67i.

Rules and formulas

Real component

x=rcos⁡θx=r\cos\theta

Imaginary component

y=rsin⁡θy=r\sin\theta

Polar-to-Cartesian conversion

rcis⁡θ=rcos⁡θ+irsin⁡θr\operatorname{cis}\theta=r\cos\theta+i r\sin\theta

Multiplication in polar form

If z1=r1cis⁡θ1z_1=r_1\operatorname{cis}\theta_1 and z2=r2cis⁡θ2z_2=r_2\operatorname{cis}\theta_2, then their product has modulus r1r2r_1r_2 and argument θ1+θ2\theta_1+\theta_2.

Geometrically, multiplication by r2cis⁡θ2r_2\operatorname{cis}\theta_2 scales every distance from the origin by r2r_2 and rotates every direction by θ2\theta_2.

After adding arguments, an answer may be normalized to the requested principal interval by adding or subtracting 2π2\pi.

The same rule follows immediately from exponential form: r1eiθ1r2eiθ2=r1r2ei(θ1+θ2)r_1e^{i\theta_1}r_2e^{i\theta_2}=r_1r_2e^{i(\theta_1+\theta_2)}.

Key terms

Complex multiplication
A combined scaling and rotation about the origin.
Argument addition
The rule that product directions add.

Worked examples

Multiply two cis forms

Calculate 2cis⁡π6×3cis⁡(−π4)2\operatorname{cis}\frac{\pi}{6}\times3\operatorname{cis}(-\frac{\pi}{4}).

  1. 1Multiply moduli: 2×3=62\times3=6.
  2. 2Add arguments: π6−π4=−π12\frac{\pi}{6}-\frac{\pi}{4}=-\frac{\pi}{12}.

Answer: 6cis⁡(−π12)6\operatorname{cis}(-\frac{\pi}{12}).

Normalize a product argument

Multiply 4cis⁡3π44\operatorname{cis}\frac{3\pi}{4} by 2cis⁡5π62\operatorname{cis}\frac{5\pi}{6}, using a principal argument.

  1. 1The modulus is 8.
  2. 2The angle sum is 19π12\frac{19\pi}{12}.
  3. 3Subtract 2π2\pi to obtain −5π12-\frac{5\pi}{12}.

Answer: 8cis⁡(−5π12)8\operatorname{cis}(-\frac{5\pi}{12}).

Cartesian product using polar structure

Find the polar form of (1+i)(3+i)(1+i)(\sqrt3+i).

  1. 1Write 1+i=2cis⁡π41+i=\sqrt2\operatorname{cis}\frac{\pi}{4}.
  2. 2Write 3+i=2cis⁡π6\sqrt3+i=2\operatorname{cis}\frac{\pi}{6}.
  3. 3Multiply moduli and add arguments.

Answer: 22cis⁡5π122\sqrt2\operatorname{cis}\frac{5\pi}{12}.

Rules and formulas

Product in cis form

z1z2=r1r2cis⁡(θ1+θ2)z_1z_2=r_1r_2\operatorname{cis}(\theta_1+\theta_2)

Product in exponential form

z1z2=r1r2ei(θ1+θ2)z_1z_2=r_1r_2e^{i(\theta_1+\theta_2)}

Product modulus

∣z1z2∣=∣z1∣∣z2∣|z_1z_2|=|z_1||z_2|

Powers in polar and exponential form

Polar form makes integer powers systematic. If z=rcis⁡θz=r\operatorname{cis}\theta, then each multiplication multiplies the modulus by rr and adds another copy of the argument θ\theta. Therefore zn=rncis⁡(nθ)z^n=r^n\operatorname{cis}(n\theta) for every integer nn, with z≠0z\ne0 when n<0n<0.

The same rule in exponential form is (reiθ)n=rneinθ(re^{i\theta})^n=r^n e^{in\theta}. This follows directly from the ordinary index laws and is a form of De Moivre's theorem.

After multiplying the argument by nn, normalize it to the requested interval only at the end. For an exact Cartesian answer, use exact sine and cosine values after the power has been simplified.

Geometrically, raising to the nth power raises the distance from the origin to the nth power and multiplies the direction angle by nn. This gives a fast reasonableness check for both magnitude and quadrant.

Key terms

De Moivre's theorem for powers
The rule that raises the modulus to n and multiplies the argument by n.
Integer power
A power whose exponent is an integer; negative powers require a non-zero base.
Argument normalization
Adding or subtracting whole turns so an angle lies in the required interval.

Worked examples

Raise an exact polar form to a power

Calculate (2cis⁡π6)4\left(2\operatorname{cis}\frac{\pi}{6}\right)^4 in exact Cartesian form.

  1. 1Raise the modulus: 24=162^4=16.
  2. 2Multiply the argument: 4(π6)=2π34\left(\frac{\pi}{6}\right)=\frac{2\pi}{3}.
  3. 3Convert 16cis⁡2π316\operatorname{cis}\frac{2\pi}{3} to Cartesian form.

Answer: −8+83i-8+8\sqrt3 i.

Power of a Cartesian complex number

Find (−1+i)6(-1+i)^6 using polar form.

  1. 1Write −1+i=2cis⁡3π4-1+i=\sqrt2\operatorname{cis}\frac{3\pi}{4}.
  2. 2Raise the modulus: (2)6=8(\sqrt2)^6=8.
  3. 3Multiply the argument and normalize: 6⋅3π4=9π2≡π2(mod2π)6\cdot\frac{3\pi}{4}=\frac{9\pi}{2}\equiv\frac{\pi}{2}\pmod{2\pi}.

Answer: (−1+i)6=8cis⁡π2=8i(-1+i)^6=8\operatorname{cis}\frac{\pi}{2}=8i.

Negative integer power

Write (2eiπ3)−2(2e^{\frac{i\pi}{3}})^{-2} in principal exponential form.

  1. 1Raise the modulus: 2−2=142^{-2}=\frac14.
  2. 2Multiply the argument by −2-2.

Answer: 14e−2πi3\frac14e^{\frac{-2\pi i}{3}}.

Rules and formulas

Power in cis form

(rcis⁡θ)n=rncis⁡(nθ)\left(r\operatorname{cis}\theta\right)^n=r^n\operatorname{cis}(n\theta)
nn
integer exponent

Power in exponential form

(reiθ)n=rneinθ\left(re^{i\theta}\right)^n=r^n e^{in\theta}
nn
integer exponent

Modulus of a power

∣zn∣=∣z∣n|z^n|=|z|^n

Division, reciprocals and conjugates in polar form

For z2≠0z_2\ne0, division divides the moduli and subtracts the arguments. This is usually shorter than rationalizing a complex denominator in Cartesian form.

The reciprocal of rcis⁡θr\operatorname{cis}\theta is 1rcis⁡(−θ)\frac1r\operatorname{cis}(-\theta): the scale is inverted and the rotation is reversed.

The conjugate of rcis⁡θr\operatorname{cis}\theta is rcis⁡(−θ)r\operatorname{cis}(-\theta). Conjugation preserves modulus and reflects the direction in the real axis.

Normalize the final argument only after subtracting. Preserve exact angles when possible.

Key terms

Polar quotient
A quotient found by dividing moduli and subtracting arguments.
Polar reciprocal
The inverse with reciprocal modulus and negated argument.
Polar conjugate
The reflection with unchanged modulus and negated argument.

Worked examples

Divide two cis forms

Calculate 8cis⁡(5π6)2cis⁡(π3)\frac{8\operatorname{cis}(\frac{5\pi}{6})}{2\operatorname{cis}(\frac{\pi}{3})}.

  1. 1Divide moduli: 82=4\frac{8}{2}=4.
  2. 2Subtract arguments: 5π6−π3=π2\frac{5\pi}{6}-\frac{\pi}{3}=\frac{\pi}{2}.
  3. 3Convert if useful.

Answer: 4cis⁡π2=4i4\operatorname{cis}\frac{\pi}{2}=4i.

Find a reciprocal

Write the reciprocal of 5cis⁡2π55\operatorname{cis}\frac{2\pi}{5} in cis form.

  1. 1Replace the modulus by 15\frac{1}{5}.
  2. 2Negate the argument.

Answer: 15cis⁡(−2π5)\frac15\operatorname{cis}(-\frac{2\pi}{5}).

Find a conjugate

Write the conjugate of 4e2πi34e^{\frac{2\pi i}{3}} in exponential form.

  1. 1Keep modulus 4.
  2. 2Negate the argument 2π3\frac{2\pi}{3}.

Answer: 4e−2πi34e^{\frac{-2\pi i}{3}}.

Solve a polar equation

Solve z(2cis⁡π3)=6cis⁡(−π6)z\left(2\operatorname{cis}\frac{\pi}{3}\right)=6\operatorname{cis}(-\frac{\pi}{6}).

  1. 1Divide moduli: 62=3\frac{6}{2}=3.
  2. 2Subtract arguments: −π6−π3=−π2-\frac{\pi}{6}-\frac{\pi}{3}=-\frac{\pi}{2}.

Answer: z=3cis⁡(−π2)=−3iz=3\operatorname{cis}(-\frac{\pi}{2})=-3i.

Rules and formulas

Quotient in cis form

z1z2=r1r2cis⁡(θ1−θ2)\frac{z_1}{z_2}=\frac{r_1}{r_2}\operatorname{cis}(\theta_1-\theta_2)
z2z_2
non-zero divisor

Reciprocal

1rcis⁡θ=1rcis⁡(−θ)\frac1{r\operatorname{cis}\theta}=\frac1r\operatorname{cis}(-\theta)

Conjugate in polar form

rcis⁡θ‾=rcis⁡(−θ)\overline{r\operatorname{cis}\theta}=r\operatorname{cis}(-\theta)

Quotient modulus

∣z1z2∣=∣z1∣∣z2∣\left|\frac{z_1}{z_2}\right|=\frac{|z_1|}{|z_2|}

Geometric interpretation: scaling and rotation

Multiplying z by a positive real number k stretches its position vector by factor k without changing its direction. Multiplying by a unit complex number cis⁡ϕ\operatorname{cis}\phi rotates it by ϕ\phi without changing its length.

Multiplication by kcis⁡ϕk\operatorname{cis}\phi combines both transformations: a dilation of scale factor k centered at the origin and a rotation of angle ϕ\phi.

Division by kcis⁡ϕk\operatorname{cis}\phi scales by 1k\frac{1}{k} and rotates by −ϕ-\phi.

This interpretation is a strong reasonableness check. A product modulus or direction inconsistent with the predicted transformation signals an arithmetic error.

Key terms

Dilation
A scaling from the origin by a positive factor.
Rotation
A turn about the origin through a directed angle.
Unit multiplier
A complex multiplier of modulus 1 that performs only a rotation.

Worked examples

Rotate without scaling

Describe multiplication by cis⁡π2\operatorname{cis}\frac{\pi}{2}.

  1. 1The multiplier modulus is 1.
  2. 2Its argument is π2\frac{\pi}{2}.

Answer: It rotates every point 90 degrees counterclockwise about the origin.

Scale and rotate

Describe multiplication by 2cis⁡(−π3)2\operatorname{cis}(-\frac{\pi}{3}).

  1. 1The modulus 2 gives scale factor 2.
  2. 2The argument gives a clockwise rotation of π3\frac{\pi}{3}.

Answer: A dilation by factor 2 followed by a 60-degree clockwise rotation.

Apply a transformation exactly

Find the image of z=1+iz=1+i under multiplication by 2cis⁡π62\operatorname{cis}\frac{\pi}{6}.

  1. 1Write the multiplier in Cartesian form: 3+i\sqrt3+i.
  2. 2Multiply (3+i)(1+i)(\sqrt3+i)(1+i).
  3. 3Use i2=−1i^2=-1.

Answer: The image is (3−1)+(3+1)i(\sqrt3-1)+(\sqrt3+1)i.

Rules and formulas

Scaling and rotation

z↦kcis⁡ϕ zz\mapsto k\operatorname{cis}\phi\,z

Modulus transformation

∣kcis⁡ϕ z∣=k∣z∣|k\operatorname{cis}\phi\,z|=k|z|
kk
positive scale factor

Argument transformation

arg⁡(kcis⁡ϕ z)=arg⁡z+ϕ(mod2π)\arg(k\operatorname{cis}\phi\,z)=\arg z+\phi\pmod{2\pi}

Applications: phasors, signals and transformations

Polar complex numbers are useful when a quantity has both magnitude and phase. In alternating-current models, a voltage, current, or impedance can be represented as a phasor reiθre^{i\theta}. The modulus gives amplitude and the argument gives phase angle.

Multiplication and division then describe physical scaling and phase shift. For example, the phasor model I=VZI=\frac{V}{Z} divides voltage magnitude by impedance magnitude and subtracts the impedance phase from the voltage phase.

Repeated rotations and dilations are modeled by powers. If each step multiplies a state by keiϕke^{i\phi}, then after nn steps the state is zn=z0kneinϕz_n=z_0k^n e^{in\phi}. This appears in iterative graphics, rotating mechanisms, signal processing and planar motion.

Phasors simplify operations only when the model's assumptions apply. State the meaning and units of the modulus and argument, and convert to Cartesian form before adding phasors with different directions.

Key terms

Phasor
A complex-number representation of a sinusoidal quantity using magnitude and phase.
Amplitude
The size of a modeled oscillation, represented by the phasor modulus.
Phase difference
The difference between two arguments, describing their relative angular shift.

Worked examples

AC phasor division

A voltage is modeled by V=230eiπ6V=230e^{\frac{i\pi}{6}} volts and an impedance by Z=10eiπ3Z=10e^{\frac{i\pi}{3}} ohms. Find I=VZI=\frac{V}{Z}.

  1. 1Divide the moduli: 23010=23\frac{230}{10}=23.
  2. 2Subtract the arguments: π6−π3=−π6\frac{\pi}{6}-\frac{\pi}{3}=-\frac{\pi}{6}.
  3. 3Interpret the modulus and phase with their units.

Answer: I=23e−iπ6I=23e^{\frac{-i\pi}{6}} amperes: magnitude 23 A and phase −π6-\frac{\pi}{6}.

Add two signal phasors

Combine 4cis⁡04\operatorname{cis}0 and 4cis⁡2π34\operatorname{cis}\frac{2\pi}{3}, giving the result in polar form.

  1. 1Convert to Cartesian form: 44 and −2+23i-2+2\sqrt3 i.
  2. 2Add components to obtain 2+23i2+2\sqrt3 i.
  3. 3Convert the sum back to polar form.

Answer: 4cis⁡π34\operatorname{cis}\frac{\pi}{3}.

Repeated rotation and dilation

A point starts at z0=2eiπ4z_0=2e^{\frac{i\pi}{4}}. Each step multiplies it by 1.5e−iπ61.5e^{\frac{-i\pi}{6}}. Find the state after three steps.

  1. 1Use z3=z0(1.5e−iπ6)3z_3=z_0(1.5e^{\frac{-i\pi}{6}})^3.
  2. 2Calculate the modulus: 2(1.5)3=6.752(1.5)^3=6.75.
  3. 3Calculate the argument: π4−3⋅π6=−π4\frac{\pi}{4}-3\cdot\frac{\pi}{6}=-\frac{\pi}{4}.

Answer: z3=6.75e−iπ4z_3=6.75e^{\frac{-i\pi}{4}}.

Rules and formulas

Phasor quotient model

I=VZI=\frac{V}{Z}
VV
voltage phasor
ZZ
non-zero impedance phasor

Repeated scale-and-rotate model

zn=z0(keiϕ)n=z0kneinϕz_n=z_0\left(ke^{i\phi}\right)^n=z_0k^n e^{in\phi}

Phase difference

Δθ=Arg⁡(z2)−Arg⁡(z1)\Delta\theta=\operatorname{Arg}(z_2)-\operatorname{Arg}(z_1)

Special cases, axes and boundary conventions

Positive real numbers have principal argument 0, negative real numbers have principal argument π\pi, positive pure imaginary numbers have principal argument π2\frac{\pi}{2}, and negative pure imaginary numbers have principal argument −π2-\frac{\pi}{2}.

The zero complex number may be written as 0(cos⁡θ+isin⁡θ)0(\cos\theta+i\sin\theta) for any θ, but this does not give zero a well-defined argument. For uniqueness, polar form is normally discussed for z≠0z\ne0.

A negative quantity should not be used as the modulus. If an expression begins with a negative radial coefficient, make the modulus positive and add π\pi to the argument.

Angles exactly at a principal-interval boundary must follow the stated convention. With −π<Arg⁡z≤π-\pi<\operatorname{Arg}z\le\pi, use π\pi instead of −π-\pi.

Key terms

Axis argument
A standard direction 0, π/2, π, or −π/2.
Positive modulus convention
For a non-zero number, polar radius r is strictly positive.
Boundary convention
The rule deciding which endpoint represents the negative real axis.

Worked examples

Repair a negative modulus

Rewrite −3cis⁡π4-3\operatorname{cis}\frac{\pi}{4} with positive modulus and a principal argument.

  1. 1Replace the negative factor by a rotation of π\pi.
  2. 2Add π\pi: π4+π=5π4\frac{\pi}{4}+\pi=\frac{5\pi}{4}.
  3. 3Convert to the principal interval by subtracting 2π2\pi.

Answer: 3cis⁡(−3π4)3\operatorname{cis}(-\frac{3\pi}{4}).

Pure imaginary direction

Write 7i7i in exponential form.

  1. 1The modulus is 7.
  2. 2The principal argument is π2\frac{\pi}{2}.

Answer: 7eiπ27e^{\frac{i\pi}{2}}.

Negative real direction

Write −4-4 in principal cis form.

  1. 1The modulus is 4.
  2. 2The principal argument is π\pi.

Answer: 4cis⁡π4\operatorname{cis}\pi.

Rules and formulas

Axis arguments

Arg⁡(a)=0 (a>0),Arg⁡(a)=π (a<0)\operatorname{Arg}(a)=0\ (a>0),\quad \operatorname{Arg}(a)=\pi\ (a<0)

Imaginary-axis arguments

Arg⁡(bi)=π2 (b>0),Arg⁡(bi)=−π2 (b<0)\operatorname{Arg}(bi)=\frac{\pi}{2}\ (b>0),\quad \operatorname{Arg}(bi)=-\frac{\pi}{2}\ (b<0)

Positive-modulus conversion

−rcis⁡θ=rcis⁡(θ+π)-r\operatorname{cis}\theta=r\operatorname{cis}(\theta+\pi)

Mixed conversion and operation problems

Mixed questions often present values in different forms. Convert them to the form that makes the requested operation simplest, then convert the final answer only if required.

For multiplication or division, identify each modulus and argument before operating. For addition or subtraction, convert each number to Cartesian form first.

Use exact standard-angle values throughout. If one angle is non-standard, retain full calculator precision until the final answer.

A complete AI HL solution communicates the chosen form, quadrant reasoning, operation rule, and final argument convention.

Key terms

Mixed representation
A problem using Cartesian, cis, polar, or exponential forms together.
Strategic conversion
Changing form specifically to simplify the requested operation.

Worked examples

Add numbers supplied in polar form

Find 2cis⁡0+2cis⁡π22\operatorname{cis}0+2\operatorname{cis}\frac{\pi}{2}.

  1. 1Addition is easiest in Cartesian form.
  2. 2Convert the terms to 2 and 2i2i.
  3. 3Add the components.

Answer: 2+2i=22cis⁡π42+2i=2\sqrt2\operatorname{cis}\frac{\pi}{4}.

Divide mixed forms

Find 4eiπ31+i\frac{4e^{\frac{i\pi}{3}}}{1+i} in principal cis form.

  1. 1Convert 1+i=2cis⁡π41+i=\sqrt2\operatorname{cis}\frac{\pi}{4}.
  2. 2Divide moduli: 42=22\frac{4}{\sqrt2}=2\sqrt2.
  3. 3Subtract arguments: π3−π4=π12\frac{\pi}{3}-\frac{\pi}{4}=\frac{\pi}{12}.

Answer: 22cis⁡π122\sqrt2\operatorname{cis}\frac{\pi}{12}.

Recover a missing factor

If zw=12eiπ2zw=12e^{\frac{i\pi}{2}} and w=3e−iπ6w=3e^{\frac{-i\pi}{6}}, find z.

  1. 1Divide moduli: 123=4\frac{12}{3}=4.
  2. 2Subtract arguments: π2−(−π6)=2π3\frac{\pi}{2}-(-\frac{\pi}{6})=\frac{2\pi}{3}.

Answer: z=4e2πi3z=4e^{\frac{2\pi i}{3}}.

Rules and formulas

Operation choice

Add or subtract → Cartesian; multiply or divide → polar

Calculator/GDC methods and exam strategy

Set the GDC to complex mode, radian angle mode, and the output form required by the question. Common output choices are rectangular a+bia+bi and polar reiθre^{i\theta}.

Use rectangular-to-polar and polar-to-rectangular commands where available. On other models, calculate r=x2+y2r=\sqrt{x^2+y^2} and use a two-argument angle function such as atan2y,xy,x.

Check whether the calculator reports an argument in (−π,π])or([0,2π)( -\pi,\pi ]) or ([0,2\pi), then convert to the required convention.

For exact standard angles, provide exact mathematical values even if the display gives decimals. Use the GDC to check, not replace, exact reasoning.

After multiplication or division, verify the modulus and the expected rotation independently. This catches incorrect angle signs quickly.

Key terms

Radian mode
The calculator angle setting required for standard complex exponential notation.
Rectangular mode
A display setting that returns a + bi.
Polar mode
A display setting that returns magnitude and angle.
atan2
A quadrant-aware angle function taking the imaginary and real coordinates separately.

Worked examples

GDC Cartesian-to-polar check

Check z=−2+23iz=-2+2\sqrt3 i on a GDC.

  1. 1Enter the complex number in rectangular mode.
  2. 2Use the polar-conversion command or calculate modulus and atan2.
  3. 3Convert the displayed angle to an exact standard angle.

Answer: ∣z∣=4|z|=4 and Arg⁡z=2π3\operatorname{Arg}z=\frac{2\pi}{3}.

GDC approximate conversion

Convert 3e1.1i3e^{1.1i} to Cartesian form on a GDC.

  1. 1Confirm radian mode.
  2. 2Enter the exponential expression with parentheses.
  3. 3Read both components before rounding.

Answer: z≈1.36+2.67iz\approx1.36+2.67i to three significant figures.

GDC operation check

Check 2e0.7i×5e−0.2i2e^{0.7i}\times5e^{-0.2i}.

  1. 1Predict modulus 2×5=102\times5=10.
  2. 2Predict argument 0.7−0.2=0.50.7-0.2=0.5.
  3. 3Use the GDC and compare.

Answer: 10e0.5i10e^{0.5i}.

Rules and formulas

Quadrant-aware argument

θ=atan2⁡(y,x)\theta=\operatorname{atan2}(y,x)

Post-operation check

∣z1z2∣=∣z1∣∣z2∣,arg⁡(z1z2)=arg⁡z1+arg⁡z2(mod2π)|z_1z_2|=|z_1||z_2|,\qquad \arg(z_1z_2)=\arg z_1+\arg z_2\pmod{2\pi}

Common mistakes and IB AI HL exam communication

Most errors come from an incorrect quadrant, an inconsistent principal interval, degree mode, a negative modulus, or applying polar rules to addition and subtraction.

A clear conversion solution should show the modulus, a reference-angle or atan2 calculation, the quadrant, and the selected argument.

A clear operation solution should identify the two moduli and arguments, state the multiplication or division rule, normalize the final argument if requested, and retain exact values.

Always distinguish arg⁡z\arg z, which may represent every coterminal argument, from Arg⁡z\operatorname{Arg}z, which denotes the chosen principal argument.

Key terms

Coterminal angles
Angles differing by complete turns and sharing the same terminal direction.
Normalization
Moving an angle into the required interval by adding or subtracting 2π.

Worked examples

Diagnose a quadrant error

A student claims Arg⁡(−1−i)=π4\operatorname{Arg}(-1-i)=\frac{\pi}{4}. Explain the error.

  1. 1The reference angle is π4\frac{\pi}{4}.
  2. 2But the point lies in quadrant III.
  3. 3Choose the representative in the principal interval.

Answer: Arg⁡(−1−i)=−3π4\operatorname{Arg}(-1-i)=-\frac{3\pi}{4}.

Diagnose the wrong operation rule

A student adds 2cis⁡0+3cis⁡π22\operatorname{cis}0+3\operatorname{cis}\frac{\pi}{2} by adding moduli and angles. Why is this invalid?

  1. 1Polar modulus-angle rules apply to products and quotients.
  2. 2Convert the addends to Cartesian form.

Answer: The correct sum is 2+3i2+3i, not 5cis⁡π25\operatorname{cis}\frac{\pi}{2}.

Normalize a final angle

Normalize −7π6-\frac{7\pi}{6} to (−π,π]-\pi,\pi].

  1. 1The angle is below −π-\pi.
  2. 2Add 2π2\pi.

Answer: 5π6\frac{5\pi}{6}.

Rules and formulas

Principal normalization

θ↦θ+2kπ∈(−π,π]\theta\mapsto\theta+2k\pi\in(-\pi,\pi]

IB AI HL-style practice with worked solutions

Attempt each problem before reading the steps. The sequence moves from basic conversion to mixed representation, geometric interpretation and multi-step operation problems.

Unless stated otherwise, give exact values and principal arguments in (−π,π]-\pi,\pi].

Key terms

Exact polar answer
A positive modulus and exact angle, normally expressed using π where appropriate.
Worked solution
A transparent sequence of mathematical steps leading to a verified answer.

Worked examples

Practice 1: first-quadrant conversion

Write 1+3i1+\sqrt3 i in cis and exponential forms.

  1. 1The modulus is 1+3=2\sqrt{1+3}=2.
  2. 2The argument is π3\frac{\pi}{3}.

Answer: 2cis⁡π3=2eiπ32\operatorname{cis}\frac{\pi}{3}=2e^{\frac{i\pi}{3}}.

Practice 2: quadrant-II argument

Find the modulus and principal argument of −3+i-\sqrt3+i.

  1. 1The modulus is 2.
  2. 2The reference angle is π6\frac{\pi}{6}.
  3. 3Quadrant II gives 5π6\frac{5\pi}{6}.

Answer: ∣z∣=2|z|=2 and Arg⁡z=5π6\operatorname{Arg}z=\frac{5\pi}{6}.

Practice 3: polar to Cartesian

Convert 4cis⁡5π64\operatorname{cis}\frac{5\pi}{6} to Cartesian form.

  1. 1Use cosine −32-\frac{\sqrt3}{2} and sine 12\frac12.
  2. 2Multiply both components by 4.

Answer: −23+2i-2\sqrt3+2i.

Practice 4: exponential to Cartesian

Convert 5e−iπ65e^{\frac{-i\pi}{6}} to Cartesian form.

  1. 1Use Euler's relation.
  2. 2Evaluate the exact cosine and sine values.
  3. 3Multiply by 5.

Answer: 532−52i\frac{5\sqrt3}{2}-\frac52 i.

Practice 5: principal-argument normalization

Write 3cis⁡4π33\operatorname{cis}\frac{4\pi}{3} using a principal argument.

  1. 1Subtract 2π2\pi from 4π3\frac{4\pi}{3}.
  2. 2Keep modulus 3.

Answer: 3cis⁡(−2π3)3\operatorname{cis}(-\frac{2\pi}{3}).

Practice 6: multiplication

Calculate 2cis⁡π6×3cis⁡(−π4)2\operatorname{cis}\frac{\pi}{6}\times3\operatorname{cis}(-\frac{\pi}{4}).

  1. 1Multiply moduli to obtain 6.
  2. 2Add arguments to obtain −π12-\frac{\pi}{12}.

Answer: 6cis⁡(−π12)6\operatorname{cis}(-\frac{\pi}{12}).

Practice 7: division

Calculate 8cis⁡(5π6)2cis⁡(π3)\frac{8\operatorname{cis}(\frac{5\pi}{6})}{2\operatorname{cis}(\frac{\pi}{3})}.

  1. 1Divide moduli to obtain 4.
  2. 2Subtract arguments to obtain π2\frac{\pi}{2}.

Answer: 4cis⁡π2=4i4\operatorname{cis}\frac{\pi}{2}=4i.

Practice 8: Cartesian product through polar form

Write (1+i)(3+i)(1+i)(\sqrt3+i) in polar form.

  1. 1Convert the factors to 2cis⁡π4\sqrt2\operatorname{cis}\frac{\pi}{4} and 2cis⁡π62\operatorname{cis}\frac{\pi}{6}.
  2. 2Multiply moduli and add arguments.

Answer: 22cis⁡5π122\sqrt2\operatorname{cis}\frac{5\pi}{12}.

Practice 9: solve for a polar factor

Solve z(2cis⁡π3)=6cis⁡(−π6)z(2\operatorname{cis}\frac{\pi}{3})=6\operatorname{cis}(-\frac{\pi}{6}).

  1. 1Divide the moduli: 62=3\frac{6}{2}=3.
  2. 2Subtract the arguments: −π6−π3=−π2-\frac{\pi}{6}-\frac{\pi}{3}=-\frac{\pi}{2}.

Answer: z=3cis⁡(−π2)=−3iz=3\operatorname{cis}(-\frac{\pi}{2})=-3i.

Practice 10: conjugate and reciprocal

For z=4cis⁡2π5z=4\operatorname{cis}\frac{2\pi}{5}, give z‾\overline z and z−1z^{-1} in cis form.

  1. 1Conjugation keeps modulus 4 and negates the argument.
  2. 2A reciprocal also replaces 4 by 14\frac{1}{4}.

Answer: z‾=4cis⁡(−2π5)\overline z=4\operatorname{cis}(-\frac{2\pi}{5}) and z−1=14cis⁡(−2π5)z^{-1}=\frac14\operatorname{cis}(-\frac{2\pi}{5}).

Practice 11: exact GDC-style conversion

Find the modulus and principal argument of −2+23i-2+2\sqrt3 i.

  1. 1The modulus is 4+12=4\sqrt{4+12}=4.
  2. 2The point is in quadrant II with reference angle π3\frac{\pi}{3}.
  3. 3Use π−π3\pi-\frac{\pi}{3}.

Answer: ∣z∣=4|z|=4 and Arg⁡z=2π3\operatorname{Arg}z=\frac{2\pi}{3}.

Practice 12: geometric transformation

The point z is multiplied by 2cis⁡π62\operatorname{cis}\frac{\pi}{6}. Describe the transformation and find the image of z=1+iz=1+i in Cartesian form.

  1. 1The multiplier gives a dilation by factor 2 and a counterclockwise rotation by π6\frac{\pi}{6}.
  2. 2Convert the multiplier to 3+i\sqrt3+i.
  3. 3Expand (3+i)(1+i)(\sqrt3+i)(1+i).

Answer: The image is (3−1)+(3+1)i(\sqrt3-1)+(\sqrt3+1)i.

Practice 13: integer power

Calculate (2cis⁡π6)4\left(2\operatorname{cis}\frac{\pi}{6}\right)^4 in exact Cartesian form.

  1. 1Raise the modulus to obtain 16.
  2. 2Multiply the argument by 4 to obtain 2π3\frac{2\pi}{3}.
  3. 3Convert the result to Cartesian form.

Answer: −8+83i-8+8\sqrt3 i.

Practice 14: applied phasor quotient

Given V=120eiπ4V=120e^{\frac{i\pi}{4}} and Z=8e−iπ12Z=8e^{\frac{-i\pi}{12}}, calculate I=VZI=\frac{V}{Z}.

  1. 1Divide the moduli: 1208=15\frac{120}{8}=15.
  2. 2Subtract the arguments: π4−(−π12)=π3\frac{\pi}{4}-(-\frac{\pi}{12})=\frac{\pi}{3}.

Answer: I=15eiπ3I=15e^{\frac{i\pi}{3}}.

Practice 15: repeated transformation

A state z0=3e−iπ6z_0=3e^{\frac{-i\pi}{6}} is multiplied twice by 2eiπ42e^{\frac{i\pi}{4}}. Find the final state in principal exponential form.

  1. 1Use z2=z0(2eiπ4)2z_2=z_0(2e^{\frac{i\pi}{4}})^2.
  2. 2The modulus is 3⋅22=123\cdot2^2=12.
  3. 3The argument is −π6+2⋅π4=π3-\frac{\pi}{6}+2\cdot\frac{\pi}{4}=\frac{\pi}{3}.

Answer: z2=12eiπ3z_2=12e^{\frac{i\pi}{3}}.

Rules and formulas

Conversion-operation workflow

identify form→find r,θ→operate→normalize→verify\text{identify form}\to\text{find }r,\theta\to\text{operate}\to\text{normalize}\to\text{verify}

Topic summary and cheat sheet

Polar, cis and exponential forms describe the same non-zero complex number using its modulus and direction. Conversion accuracy depends on a correct quadrant and a clear argument convention.

For products and quotients, operate directly on moduli and arguments; for sums and differences, return to Cartesian components.

Key terms

Polar fluency
Moving accurately between Cartesian, trigonometric, cis and exponential representations.

Worked examples

Final checklist

What should be checked before submitting a polar-form answer?

  1. 1Check first whether the result is zero. Zero has modulus 0 and no defined argument.
  2. 2For a nonzero result, the modulus is positive and the angle must be in the required interval and unit.
  3. 3The form matches the question.
  4. 4A conversion or geometric check confirms the result.

Answer: Submit an exact, convention-compliant form unless an approximation is requested.

Rules and formulas

Modulus

r=x2+y2r=\sqrt{x^2+y^2}

Polar form

z=r(cos⁡θ+isin⁡θ)z=r(\cos\theta+i\sin\theta)

Cis form

z=rcis⁡θz=r\operatorname{cis}\theta

Exponential form

z=reiθz=re^{i\theta}

Product

z1z2=r1r2cis⁡(θ1+θ2)z_1z_2=r_1r_2\operatorname{cis}(\theta_1+\theta_2)

Quotient

z1z2=r1r2cis⁡(θ1−θ2)\frac{z_1}{z_2}=\frac{r_1}{r_2}\operatorname{cis}(\theta_1-\theta_2)

Topic summary

  • For z=x+yiz=x+yi, the modulus is r=∣z∣=x2+y2r=|z|=\sqrt{x^2+y^2}.
  • An argument is a directed angle from the positive real axis to the position vector of z.
  • The argument of zero is undefined.
  • Cartesian and polar components satisfy x=rcos⁡θx=r\cos\theta and y=rsin⁡θy=r\sin\theta.
  • Polar form is z=r(cos⁡θ+isin⁡θ)z=r(\cos\theta+i\sin\theta) with r>0r>0 for z≠0z\ne0.
  • Cis notation is defined by cis⁡θ=cos⁡θ+isin⁡θ\operatorname{cis}\theta=\cos\theta+i\sin\theta.
  • Euler's relation is eiθ=cos⁡θ+isin⁡θe^{i\theta}=\cos\theta+i\sin\theta.
  • The exponential form is z=reiθz=re^{i\theta}.
  • Arguments differing by 2kπ2k\pi represent the same direction.
  • These notes use the principal interval −π<Arg⁡z≤π-\pi<\operatorname{Arg}z\le\pi.
  • A negative real number has principal argument π\pi under this convention.
  • Use the signs of x and y to choose the correct quadrant.
  • A two-argument atan2 calculation is safer than an unadjusted inverse tangent.
  • To convert to Cartesian form, calculate rcos⁡θ+irsin⁡θr\cos\theta+i r\sin\theta.
  • Use exact unit-circle values whenever possible.
  • To multiply in polar form, multiply moduli and add arguments.
  • To divide in polar form, divide moduli and subtract arguments.
  • Multiplication by kcis⁡ϕk\operatorname{cis}\phi scales by k and rotates by φ.
  • The conjugate of rcis⁡θr\operatorname{cis}\theta is rcis⁡(−θ)r\operatorname{cis}(-\theta).
  • The reciprocal of rcis⁡θr\operatorname{cis}\theta is 1rcis⁡(−θ)\frac1r\operatorname{cis}(-\theta).
  • Use Cartesian form for addition and subtraction.
  • Normalize a final angle only after completing the operation.
  • Check calculator complex mode, rectangular or polar output, and radian mode.
  • Verify results using modulus, argument, Cartesian conversion, or geometric scaling and rotation.
  • For integer nn, (rcis⁡θ)n=rncis⁡(nθ)\left(r\operatorname{cis}\theta\right)^n=r^n\operatorname{cis}(n\theta).
  • A negative integer power requires a non-zero base and inverts the modulus.
  • After a power, normalize the multiplied argument only when presenting the final answer.
  • Phasor modulus represents magnitude or amplitude, while its argument represents phase.
  • Repeated scaling and rotation can be modeled by zn=z0(keiϕ)nz_n=z_0(ke^{i\phi})^n.
  • For applied answers, state the meanings and units of both modulus and phase.
  • Complex numbers are removed from the revised AI HL syllabus first assessed in May 2029; check which specification applies.