Precalculus Polar Form and De Moivre’s Theorem Worksheet Solutions (PDF)

Precalculus math worksheet solutions: Polar Form and De Moivre's Theorem – Maths PDF

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This polar complex answers for Precalculus covers polar form and de moivre's theorem (Common Core N-CN.B.4, N-CN.B.5, N-CN.B.6), with a free PDF to print.

Use this key to check each polar computation. Solutions show the modulus, the argument and the rule used, with final answers in bold.

Precalculus · Common Core State Standards: N-CN.B.4, N-CN.B.5, N-CN.B.6

Polar complex answers: exercises 1 to 10

Exercise 1 – Modulus and argument

r = 2 and angle 60 degrees.

\(r=\sqrt{1+3}=2\). \(\tan\theta=\sqrt3\) in quadrant I, so \(\theta=60^{\circ}\).

\(z=2(\cos60^{\circ}+i\sin60^{\circ})\)

Exercise 2 – Quadrant II

\(r=\sqrt{9+9}=3\sqrt2\). The point is in quadrant II with reference angle \(45^{\circ}\), so \(\theta=135^{\circ}\).

\(3\sqrt2(\cos135^{\circ}+i\sin135^{\circ})\)

Exercise 3 – On an axis

\(r=4\). The point \((0,-4)\) is on the negative imaginary axis, at \(270^{\circ}\).

\(4(\cos270^{\circ}+i\sin270^{\circ})\)

Exercise 4 – Polar to rectangular

\(6\cos120^{\circ}=-3\). \(6\sin120^{\circ}=3\sqrt3\).

\(-3+3\sqrt3\,i\)

Exercise 5 – Polar to rectangular in radians

\(4\cdot\frac{\sqrt3}{2}=2\sqrt3\). \(4\cdot\frac12=2\).

\(2\sqrt3+2i\)

Exercise 6 – Quadrant IV

\(r=5\sqrt2\). The point is in quadrant IV with reference angle \(45^{\circ}\), so \(\theta=315^{\circ}\) (or \(-45^{\circ}\)).

\(5\sqrt2(\cos315^{\circ}+i\sin315^{\circ})\)

Exercise 7 – Multiply in polar form

6 cis 75 degrees.

Moduli: \(2\cdot3=6\). Angles: \(30^{\circ}+45^{\circ}=75^{\circ}\). Rectangular: \(6\cos75^{\circ}\approx 1.553\), \(6\sin75^{\circ}\approx 5.796\).

\(6(\cos75^{\circ}+i\sin75^{\circ})\approx 1.553+5.796i\)

Exercise 8 – Divide in polar form

\(\frac{12}{4}=3\) and \(200^{\circ}-50^{\circ}=150^{\circ}\). Rectangular: \(3\cos150^{\circ}\approx -2.598\) and \(3\sin150^{\circ}=1.5\).

\(3\,\text{cis}\,150^{\circ}\approx -2.598+1.500i\)

Exercise 9 – Check with rectangular multiplication

\(1+i=\sqrt2\,\text{cis}\,45^{\circ}\) and \(\sqrt3+i=2\,\text{cis}\,30^{\circ}\). Product: \(2\sqrt2\,\text{cis}\,75^{\circ}\approx 0.732+2.732i\). Rectangular: \(\sqrt3+i+\sqrt3\,i+i^{2}=(\sqrt3-1)+(\sqrt3+1)i\approx 0.732+2.732i\).

\(2\sqrt2\,\text{cis}\,75^{\circ}\approx 0.732+2.732i\)

Exercise 10 – De Moivre with a unit modulus

\(\cos(9\cdot20^{\circ})+i\sin(9\cdot20^{\circ})=\cos180^{\circ}+i\sin180^{\circ}\).

\(-1\)

Exercises 11 and up

Exercise 11 – Power of 1 + i

\(1+i=\sqrt2\,\text{cis}\,45^{\circ}\). \((\sqrt2)^{8}=16\) and \(8\cdot45^{\circ}=360^{\circ}\).

Exercise 12 – Power with a negative angle

\(r=2\), \(\theta=-30^{\circ}\). \(2^{6}=64\) and \(6\cdot(-30^{\circ})=-180^{\circ}\): \(64\,\text{cis}(-180^{\circ})\).

\(-64\)

Exercise 13 – Power in rectangular form

\(2^{4}=16\), \(4\cdot15^{\circ}=60^{\circ}\): \(16\cdot\frac12+16\cdot\frac{\sqrt3}{2}i\).

\(8+8\sqrt3\,i\)

Exercise 14 – Cube roots of 8

z0, z1, z2 at 0, 120 and 240 degrees.

\(8=8\,\text{cis}\,0^{\circ}\). The roots have modulus \(\sqrt[3]{8}=2\) and arguments \(0^{\circ},120^{\circ},240^{\circ}\): \(2\), \(-1+\sqrt3\,i\), \(-1-\sqrt3\,i\).

\(2,\ -1+\sqrt3\,i,\ -1-\sqrt3\,i\)

Exercise 15 – Sixth roots of unity

Six points spaced 60 degrees apart.

Since \(1=1\,\text{cis}\,0^{\circ}\), \(\theta=\frac{360^{\circ}k}{6}=60^{\circ}k\) for \(k=0,\text{ to },5\).

\(\text{cis}\,0^{\circ},\ 60^{\circ},\ 120^{\circ},\ 180^{\circ},\ 240^{\circ},\ 300^{\circ}\)

Exercise 16 – Fourth roots of -16

\(-16=16\,\text{cis}\,180^{\circ}\). Modulus \(\sqrt[4]{16}=2\). Arguments \(\frac{180^{\circ}+360^{\circ}k}{4}=45^{\circ},135^{\circ},225^{\circ},315^{\circ}\). Each has \(2\cos45^{\circ}=\sqrt2\).

\(\sqrt2+\sqrt2\,i,\ -\sqrt2+\sqrt2\,i,\ -\sqrt2-\sqrt2\,i,\ \sqrt2-\sqrt2\,i\)

Exercise 17 – Solve z cubed = 27i

\(27i=27\,\text{cis}\,90^{\circ}\). Roots: \(3\,\text{cis}\,30^{\circ}\), \(3\,\text{cis}\,150^{\circ}\), \(3\,\text{cis}\,270^{\circ}\). That is \(\frac{3\sqrt3}{2}+\frac32i\), \(-\frac{3\sqrt3}{2}+\frac32i\) and \(-3i\).

\(z=\frac{3\sqrt3}{2}+\frac32i,\ -\frac{3\sqrt3}{2}+\frac32i,\ -3i\)

Exercise 18 – AC circuit

\(|Z|=5\) and \(\text{arg}\,Z=\arctan\frac43\approx53.13^{\circ}\). Multiply: \(10\cdot5=50\), \(0^{\circ}+53.13^{\circ}\).

\(Z=5\,\text{cis}\,53.13^{\circ}\); \(V=50\,\text{cis}\,53.13^{\circ}\) volts

Exercise 19 – Rotate a point

The point moves along a circle of radius 5.

\((4+3i)\left(\frac12+\frac{\sqrt3}{2}i\right)=2+2\sqrt3\,i+\frac32i+\frac{3\sqrt3}{2}i^{2}=\left(2-\frac{3\sqrt3}{2}\right)+\left(2\sqrt3+\frac32\right)i\approx -0.60+4.96i\).

\(\approx -0.60+4.96i\)

Exercise 20 – Cube root of unity

\(\omega=\text{cis}\,120^{\circ}\), so \(\omega^{3}=\text{cis}\,360^{\circ}=1\). \(\omega^{2}=\text{cis}\,240^{\circ}=-\frac12-\frac{\sqrt3}{2}i\). Then \(1+\left(-\frac12+\frac{\sqrt3}{2}i\right)+\left(-\frac12-\frac{\sqrt3}{2}i\right)=0\).

Both statements hold

Key facts behind this polar complex answers

Key facts

z = a + bi = r(cos theta + i sin theta) with r = sqrt(a^2 + b^2) and tan theta = b/a.
Product: multiply moduli, add arguments. Quotient: divide moduli, subtract arguments.
De Moivre: [r cis theta]^n = r^n cis(n theta).
nth roots: (r)^(1/n) cis((theta + 360 k)/n), k = 0, …, n – 1.
The n roots form a regular n-gon on a circle of radius r^(1/n).

How to use this polar complex answers

Cover each solution, solve the exercise again, then uncover it and compare every line. If your answer differs, find the first step that changed and redo the exercise from there.

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