Complex Numbers in Polar Form: Precalculus Math Worksheet Answer Key – PDF Download

Precalculus Complex Polar Worksheet Answer Key: problems 1–5

Problem 1: Plot and find the modulus

1. \(3+4i\), \(|z|=\sqrt{9+16}=\)\(5\)
2. \(-6+8i\), \(|z|=\sqrt{36+64}=\)\(10\)
3. \(8-6i\), \(|z|=\sqrt{64+36}=\)\(10\)
4. \(2-2i\), \(|z|=\sqrt{4+4}=\)\(2\sqrt{2}\approx 2.83\)

Problem 2: Conjugates and products

1. \(\overline{z}=3+2i\); \((3-2i)(3+2i)=9+4=\)13
2. \(\overline{z}=5-i\); \(25+1=\)26
3. \(\overline{z}=4i\); \((-4i)(4i)=-16i^{2}=\)16
4. \(\overline{z}=-2+6i\); \(4+36=\)40

Each product equals \(|z|^{2}\), a real number.

Problem 3: Dividing complex numbers

1. \(\frac{(2+3i)(1+i)}{(1-i)(1+i)}=\frac{2+5i+3i^{2}}{2}=\frac{-1+5i}{2}=\)\(-\frac{1}{2}+\frac{5}{2}i\)
2. \(\frac{5(2-i)}{(2+i)(2-i)}=\frac{5(2-i)}{5}=\)\(2-i\)
3. \(\frac{(3-i)(1-2i)}{5}=\frac{3-7i+2i^{2}}{5}=\frac{1-7i}{5}=\)\(\frac{1}{5}-\frac{7}{5}i\)
4. \(\frac{(4+2i)(i)}{(-i)(i)}=\frac{4i+2i^{2}}{1}=\)\(-2+4i\)

Problem 4: Finding the argument

1. \(\tan\theta=1\), quadrant I: \(45^{\circ}\)
2. Reference angle \(45^{\circ}\), quadrant II: \(135^{\circ}\)
3. \(\tan\theta=\frac{-1}{-\sqrt{3}}=\frac{1}{\sqrt{3}}\), reference \(30^{\circ}\), quadrant III: \(180^{\circ}+30^{\circ}=\)\(210^{\circ}\)
4. \(\tan\theta=-\sqrt{3}\), reference \(60^{\circ}\), quadrant IV: \(360^{\circ}-60^{\circ}=\)\(300^{\circ}\)

Problem 5: Rectangular to polar form

1. \(r=\sqrt{27+9}=6\), \(\tan\theta=\frac{3}{3\sqrt{3}}=\frac{1}{\sqrt{3}}\), \(\theta=30^{\circ}\): \(6(\cos 30^{\circ}+i\sin 30^{\circ})\)
2. \(r=\sqrt{8}=2\sqrt{2}\), quadrant II, \(\theta=135^{\circ}\): \(2\sqrt{2}(\cos 135^{\circ}+i\sin 135^{\circ})\)
3. \(r=5\), on the negative imaginary axis, \(\theta=270^{\circ}\): \(5(\cos 270^{\circ}+i\sin 270^{\circ})\)
4. \(r=\sqrt{1+3}=2\), quadrant IV, \(\theta=300^{\circ}\): \(2(\cos 300^{\circ}+i\sin 300^{\circ})\)

Precalculus Complex Polar Worksheet Answer Key: problems 6–10

Problem 6: Polar to rectangular form

1. \(4\cdot\frac{1}{2}+4\cdot\frac{\sqrt{3}}{2}i=\)\(2+2\sqrt{3}\,i\)
2. \(10(-1)+10(0)i=\)\(-10\)
3. \(6(-\frac{\sqrt{2}}{2})+6(-\frac{\sqrt{2}}{2})i=\)\(-3\sqrt{2}-3\sqrt{2}\,i\)
4. \(2\sqrt{2}\cdot\frac{\sqrt{2}}{2}+2\sqrt{2}(-\frac{\sqrt{2}}{2})i=\)\(2-2i\)

Problem 7: Radian measure

1. \(2\cdot\frac{\sqrt{3}}{2}+2\cdot\frac{1}{2}i=\)\(\sqrt{3}+i\)
2. \(5(0)+5(1)i=\)\(5i\)
3. \(8\cdot\frac{1}{2}+8(-\frac{\sqrt{3}}{2})i=\)\(4-4\sqrt{3}\,i\)
4. \(6(-\frac{\sqrt{2}}{2})+6\cdot\frac{\sqrt{2}}{2}i=\)\(-3\sqrt{2}+3\sqrt{2}\,i\)

Problem 8: Conjugate in polar form

1. Conjugation reflects across the real axis, so the modulus stays 6 and the angle changes sign: \(-50^{\circ}\), which is the same as \(310^{\circ}\). \(6(\cos 310^{\circ}+i\sin 310^{\circ})\)
2. Multiply moduli, add angles: \(6\cdot 6=36\) and \(50^{\circ}+310^{\circ}=360^{\circ}\), so \(36(\cos 360^{\circ}+i\sin 360^{\circ})=\)\(36\), which equals \(|z|^{2}\).

Problem 9: Multiplying in polar form

1. \(14(\cos 60^{\circ}+i\sin 60^{\circ})=\)\(7+7\sqrt{3}\,i\)
2. \(12(\cos 180^{\circ}+i\sin 180^{\circ})=\)\(-12\)
3. \(2(\cos 360^{\circ}+i\sin 360^{\circ})=\)\(2\)

Problem 10: Dividing in polar form

1. \(4(\cos 120^{\circ}+i\sin 120^{\circ})=\)\(-2+2\sqrt{3}\,i\)
2. \(2(\cos(-45^{\circ})+i\sin(-45^{\circ}))=2(\cos 315^{\circ}+i\sin 315^{\circ})=\)\(\sqrt{2}-\sqrt{2}\,i\)
3. \(3(\cos\frac{\pi}{2}+i\sin\frac{\pi}{2})=\)\(3i\)

Precalculus Complex Polar Worksheet Answer Key: problems 11–15

Problem 11: Check a product two ways

1. \((1+i)(\sqrt{3}+i)=\sqrt{3}+i+\sqrt{3}\,i+i^{2}=\)\((\sqrt{3}-1)+(\sqrt{3}+1)i\)
2. \(z_{1}=\sqrt{2}(\cos 45^{\circ}+i\sin 45^{\circ})\), \(z_{2}=2(\cos 30^{\circ}+i\sin 30^{\circ})\), so \(z_{1}z_{2}=\)\(2\sqrt{2}(\cos 75^{\circ}+i\sin 75^{\circ})\)
3. \(2\sqrt{2}\cos 75^{\circ}\approx 2.828\times 0.2588\approx 0.73\) and \(\sqrt{3}-1\approx 0.73\). They agree.

Problem 12: Rotation and scaling

\(2i(4+2i)=8i+4i^{2}=\)\(-4+8i\).

The modulus of \(z\) is \(\sqrt{20}=2\sqrt{5}\) and the modulus of the result is \(\sqrt{16+64}=4\sqrt{5}\): it doubled. The argument grew by \(90^{\circ}\) because \(2i\) has modulus 2 and argument \(90^{\circ}\).

Problem 13: Modulus of products and quotients

1. \((3+4i)(5-12i)=15-36i+20i-48i^{2}=\)\(63-16i\), and \(|63-16i|=\sqrt{3969+256}=\sqrt{4225}=65\).
2. \(|z_{1}|=5\) and \(|z_{2}|=13\), and \(5\times 13=65\). Verified.
3. \(\frac{|z_{1}|}{|z_{2}|}=\)\(\frac{5}{13}\)

Problem 14: De Moivre's theorem: basics

1. \(\cos 180^{\circ}+i\sin 180^{\circ}=\)\(-1\)
2. \(2^{4}(\cos 60^{\circ}+i\sin 60^{\circ})=16(\frac{1}{2}+\frac{\sqrt{3}}{2}i)=\)\(8+8\sqrt{3}\,i\)
3. \(\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}=\)\(i\)

Problem 15: Powers of complex numbers

1. \(r=\sqrt{2}\), \(\theta=45^{\circ}\): \((\sqrt{2})^{8}(\cos 360^{\circ}+i\sin 360^{\circ})=\)\(16\)
2. \(r=\sqrt{2}\), \(\theta=-45^{\circ}\): \((\sqrt{2})^{6}[\cos(-270^{\circ})+i\sin(-270^{\circ})]=8(0+i)=\)\(8i\)
3. \(r=2\), \(\theta=120^{\circ}\): \(2^{3}(\cos 360^{\circ}+i\sin 360^{\circ})=\)\(8\)
4. \(r=2\), \(\theta=30^{\circ}\): \(2^{5}(\cos 150^{\circ}+i\sin 150^{\circ})=32(-\frac{\sqrt{3}}{2}+\frac{1}{2}i)=\)\(-16\sqrt{3}+16i\)

Precalculus Complex Polar Worksheet Answer Key: problems 16–20

Problem 16: Distance in the complex plane

1. \(|(1+2i)-(5+5i)|=|-4-3i|=\sqrt{16+9}=\)5
2. \(|(-3+4i)-(2-8i)|=|-5+12i|=\sqrt{25+144}=\)13
3. \(|(7-i)-(-1+5i)|=|8-6i|=\sqrt{64+36}=\)10

Problem 17: Midpoints

1. \(\frac{(1+2i)+(5+5i)}{2}=\frac{6+7i}{2}=\)\(3+3.5i\)
2. \(\frac{-2+4i}{2}=\)\(-1+2i\)
3. The midpoint is the average, so \(w=2(1+4i)-(3-i)=2+8i-3+i=\)\(-1+9i\). Check: \(\frac{(3-i)+(-1+9i)}{2}=1+4i\).

Problem 18: Triangle in the complex plane

\(|3+4i-0|=5\) and \(|(3+4i)-6|=|-3+4i|=5\), so two sides have length 5: isosceles. The base \(|6-0|=6\).

Perimeter \(5+5+6=\)16. The height is the imaginary part, 4, so the area is \(\frac{1}{2}(6)(4)=\)12 square units.

Problem 19: Circle in the complex plane

The distance from \(z\) to \(2+i\) is always 3, so the center is \(2+i\) and the radius is 3.

1. \(|3|=3\): on the circle
2. \(|-3|=3\): on the circle
3. \(|3i|=3\): on the circle
4. \(|2+2i|=\sqrt{8}\approx 2.83 < 3\): inside the circle

Problem 20: Robot vacuum rotation

Multiplying by \(i\) rotates by \(90^{\circ}\): \(i(2+3i)=2i+3i^{2}=\)\(-3+2i\).

For \(30^{\circ}\) multiply by \(\cos 30^{\circ}+i\sin 30^{\circ}=\frac{\sqrt{3}}{2}+\frac{1}{2}i\): real part \(\sqrt{3}-\frac{3}{2}\approx 0.23\), imaginary part \(1+\frac{3\sqrt{3}}{2}\approx 3.60\). The position is about \(0.23+3.60i\).

Check: the distance from the dock stays \(\sqrt{13}\approx 3.61\) ft.

Precalculus Complex Polar Worksheet Answer Key: problems 21–22

Problem 21: AC circuit impedance

\(|Z|=\sqrt{64+36}=\)10 ohms. Phase angle \(\tan^{-1}(\frac{6}{8})\approx\) \(36.87^{\circ}\).

\(I=\frac{120(8-6i)}{(8+6i)(8-6i)}=\frac{960-720i}{100}=\)\(9.6-7.2i\) amperes.

\(|I|=\sqrt{92.16+51.84}=\sqrt{144}=\)12 amperes, which also equals \(\frac{120}{10}\).

Problem 22: Powers: multi-step

1. \(r=2\), \(\theta=-30^{\circ}\): \(2^{6}(\cos(-180^{\circ})+i\sin(-180^{\circ}))=64(-1+0i)=\)\(-64\)
2. We need \(40n\) to be a multiple of 360, so the smallest positive solution of \(40n=360\) is \(n=9\) since \(40\times 9=360\). \(n=9\)
3. The argument \(30n^{\circ}\) must equal \(90^{\circ}\) (or \(270^{\circ}\), and so on). \(n=1,2\) give \(30^{\circ}\) and \(60^{\circ}\), which are not on the imaginary axis; \(n=3\) gives \(90^{\circ}\): \((\sqrt{3}+i)^{3}=8i\). \(n=3\)

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