Algebra 2 Quiz: Complex Numbers In Rectangular Polar Form

20 questions · exam conditions

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Complex Numbers In Rectangular Polar FormQuestion 1 of 20

Find the modulus $r$ and argument $\theta$ (in degrees) of $3-3i$ , then write it in polar form $r(\cos\theta+i\sin\theta)$ . Use $r=\sqrt{a^2+b^2}$ and $\theta=\arctan\left(\frac{b}{a}\right)$ with quadrant adjustment.

3\sqrt{2}\left(\cos 45^\circ+i\sin 45^\circ\right)

3\sqrt{2}\left(\cos 315^\circ+i\sin 315^\circ\right)

6\left(\cos 315^\circ+i\sin 315^\circ\right)

3\sqrt{2}\left(\cos 135^\circ+i\sin 135^\circ\right)

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Algebra 2 Quiz

Algebra 2 Quiz: Complex Numbers In Rectangular Polar Form

Practice Complex Numbers In Rectangular Polar Form in Algebra 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Complex Numbers In Rectangular Polar Form, giving you a quick way to practice the rules, question types, and explanations that matter most for Algebra 2.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

$3\sqrt{2}\left(\cos 45^\circ+i\sin 45^\circ\right)$
$3\sqrt{2}\left(\cos 315^\circ+i\sin 315^\circ\right)$ (correct answer)
$6\left(\cos 315^\circ+i\sin 315^\circ\right)$
$3\sqrt{2}\left(\cos 135^\circ+i\sin 135^\circ\right)$

Explanation: This question tests your ability to convert complex numbers between rectangular form a + bi and polar form r(cos θ + i sin θ), which represent the same number using Cartesian coordinates versus magnitude and direction. Rectangular form a + bi uses horizontal (real) and vertical (imaginary) components, while polar form r(cos θ + i sin θ) uses distance from origin (modulus r) and angle from positive real axis (argument θ, measured counterclockwise). To convert FROM rectangular TO polar: find r = √(a² + b²) using Pythagorean theorem, then find θ = arctan(b/a) BUT adjust for quadrant (arctan only gives reference angle!). For 3 - 3i: (1) Find modulus: r = √(3² + (-3)²) = √(9 + 9) = √18 = 3√2. (2) Find argument: reference angle = arctan(|-3|/|3|) = arctan(1) = 45°. Since a = 3 > 0 and b = -3 < 0, we're in Quadrant 4, so θ = 360° - 45° = 315°. (3) Write polar form: 3√2(cos 315° + i sin 315°). Choice B correctly calculates modulus using Pythagorean theorem and determines argument with proper quadrant adjustment for Quadrant 4. Choice A would be correct if the complex number were 3 + 3i (Quadrant 1), but fails to recognize that negative imaginary part places us in Quadrant 4, requiring the 360° - reference angle adjustment. Rectangular to polar recipe: (1) Calculate r = √(a² + b²) (always positive). (2) Calculate reference angle = arctan(|b|/|a|). (3) Determine quadrant from signs of a and b: Q1 (++), Q2 (-+), Q3 (--), Q4 (+-). (4) Adjust angle: Q1: θ = reference; Q2: θ = 180° - reference; Q3: θ = 180° + reference; Q4: θ = 360° - reference. For 3 - 3i, we get r = 3√2 and θ = 315°, giving 3√2(cos 315° + i sin 315°).

Question 2

Convert the complex number $-2+2i$ from rectangular form $a+bi$ to polar form $r(\cos\theta+i\sin\theta)$ . Use $r=\sqrt{a^2+b^2}$ and $\theta=\arctan\!\left(\frac{b}{a}\right)$ with the correct quadrant adjustment. (Angle in degrees; $\theta$ is the argument measured counterclockwise from the positive real axis.)

$2\sqrt{2}\left(\cos 45^\circ+i\sin 45^\circ\right)$
$2\sqrt{2}\left(\cos 135^\circ+i\sin 135^\circ\right)$ (correct answer)
$4\left(\cos 135^\circ+i\sin 135^\circ\right)$
$2\sqrt{2}\left(\cos (-45^\circ)+i\sin (-45^\circ)\right)$

Explanation: This question tests your ability to convert complex numbers between rectangular form a + bi and polar form r(cos θ + i sin θ), which represent the same number using Cartesian coordinates versus magnitude and direction. Rectangular form a + bi uses horizontal (real) and vertical (imaginary) components, while polar form r(cos θ + i sin θ) uses distance from origin (modulus r) and angle from positive real axis (argument θ, measured counterclockwise). To convert -2 + 2i to polar form: (1) Find modulus: r = √((-2)² + 2²) = √(4 + 4) = √8 = 2√2. (2) Find argument: reference angle = arctan(|2|/|-2|) = arctan(1) = 45°. Since a = -2 < 0 and b = 2 > 0, we're in Quadrant 2, so θ = 180° - 45° = 135°. (3) Write polar form: 2√2(cos 135° + i sin 135°). Choice B correctly calculates the modulus as 2√2 and determines the argument as 135° with proper quadrant adjustment for Quadrant 2. Choice A incorrectly places the complex number in Quadrant 1 with θ = 45°, forgetting that negative real part means we're in Quadrant 2 or 3; Choice D uses -45° which would be for 2 - 2i in Quadrant 4. Rectangular to polar recipe: Calculate r = √(a² + b²), find reference angle, determine quadrant from signs (Q2 for -+), adjust angle (Q2: θ = 180° - reference), then write r(cos θ + i sin θ). Always check your quadrant—the signs of a and b tell you where your complex number lives on the complex plane!

Question 3

Convert the complex number $3+4i$ from rectangular form $a+bi$ to polar form $r(\cos\theta+i\sin\theta)$ . Use $r=\sqrt{a^2+b^2}$ and $\theta=\arctan\left(\frac{b}{a}\right)$ (with quadrant adjustment). The argument $\theta$ is measured counterclockwise from the positive real axis.

$7\left(\cos 53.13^\circ+i\sin 53.13^\circ\right)$
$5\left(\cos 233.13^\circ+i\sin 233.13^\circ\right)$
$5\left(\cos 53.13^\circ+i\sin 53.13^\circ\right)$ (correct answer)
$5\left(\cos 36.87^\circ+i\sin 36.87^\circ\right)$

Explanation: This question tests your ability to convert complex numbers between rectangular form a + bi and polar form r(cos θ + i sin θ), which represent the same number using Cartesian coordinates versus magnitude and direction. Rectangular form a + bi uses horizontal (real) and vertical (imaginary) components, while polar form r(cos θ + i sin θ) uses distance from origin (modulus r) and angle from positive real axis (argument θ, measured counterclockwise). To convert 3 + 4i to polar form: (1) Find modulus: r = √(3² + 4²) = √(9 + 16) = √25 = 5. (2) Find argument: reference angle = arctan(4/3) ≈ 53.13°. Since a = 3 > 0 and b = 4 > 0, we're in Quadrant 1, so θ = 53.13° (no adjustment needed for Q1). (3) Write polar form: 5(cos 53.13° + i sin 53.13°). Choice A correctly calculates modulus using Pythagorean theorem and determines argument with proper quadrant consideration. Choice B incorrectly calculates r = 3 + 4 = 7 instead of using the Pythagorean theorem—remember, modulus is the distance from origin, not the sum of components! Rectangular to polar recipe: (1) Calculate r = √(a² + b²) (always positive). (2) Calculate reference angle = arctan(|b|/|a|). (3) Determine quadrant from signs of a and b, then adjust angle accordingly. For 3 + 4i in Q1, no adjustment needed, giving 5(cos 53.13° + i sin 53.13°).

Question 4

Convert the rectangular complex number $1-\sqrt{3}i$ to polar form $r(\cos\theta+i\sin\theta)$ . Use $r=\sqrt{a^2+b^2}$ and $\theta=\arctan\!\left(\frac{b}{a}\right)$ with quadrant adjustment. (Angle in degrees.)

$2\left(\cos 300^\circ+i\sin 300^\circ\right)$ (correct answer)
$2\left(\cos 60^\circ+i\sin 60^\circ\right)$
$\sqrt{3}\left(\cos 300^\circ+i\sin 300^\circ\right)$
$2\left(\cos (-60^\circ)+i\sin (-60^\circ)\right)$

Explanation: This question tests your ability to convert complex numbers between rectangular form a + bi and polar form r(cos θ + i sin θ), which represent the same number using Cartesian coordinates versus magnitude and direction. Rectangular form a + bi uses horizontal (real) and vertical (imaginary) components, while polar form r(cos θ + i sin θ) uses distance from origin (modulus r) and angle from positive real axis (argument θ, measured counterclockwise). To convert 1 - √3i to polar form: (1) Find modulus: r = √(1² + (-√3)²) = √(1 + 3) = √4 = 2. (2) Find argument: reference angle = arctan(|-√3|/|1|) = arctan(√3) = 60°. Since a = 1 > 0 and b = -√3 < 0, we're in Quadrant 4, so θ = 360° - 60° = 300° (or equivalently -60°). (3) Write polar form: 2(cos 300° + i sin 300°). Choice A correctly calculates the modulus as 2 and determines the argument as 300° with proper quadrant adjustment for Quadrant 4. Choice D also correctly represents the same angle as -60°, which is coterminal with 300°; both are valid! Choice B incorrectly uses 60°, which would be for 1 + √3i in Quadrant 1. Rectangular to polar recipe: Calculate r = √(a² + b²), find reference angle, determine quadrant from signs (Q4 for +-), adjust angle (Q4: θ = 360° - reference or use negative angle), then write r(cos θ + i sin θ). In Quadrant 4, you can use either positive angles (300° to 360°) or negative angles (-60° to 0°)—they represent the same direction!

Question 5

Convert the complex number $-3-3i$ from rectangular form $a+bi$ to polar form $r(\cos\theta+i\sin\theta)$ . Use $r=\sqrt{a^2+b^2}$ and $\theta=\arctan\!\left(\frac{b}{a}\right)$ with quadrant adjustment. (Angle in degrees.)

$3\sqrt{2}\left(\cos 45^\circ+i\sin 45^\circ\right)$
$6\left(\cos 225^\circ+i\sin 225^\circ\right)$
$3\sqrt{2}\left(\cos 225^\circ+i\sin 225^\circ\right)$ (correct answer)
$3\sqrt{2}\left(\cos 135^\circ+i\sin 135^\circ\right)$

Explanation: This question tests your ability to convert complex numbers between rectangular form a + bi and polar form r(cos θ + i sin θ), which represent the same number using Cartesian coordinates versus magnitude and direction. Rectangular form a + bi uses horizontal (real) and vertical (imaginary) components, while polar form r(cos θ + i sin θ) uses distance from origin (modulus r) and angle from positive real axis (argument θ, measured counterclockwise). To convert -3 - 3i to polar form: (1) Find modulus: r = √((-3)² + (-3)²) = √(9 + 9) = √18 = 3√2. (2) Find argument: reference angle = arctan(|-3|/|-3|) = arctan(1) = 45°. Since a = -3 < 0 and b = -3 < 0, we're in Quadrant 3, so θ = 180° + 45° = 225°. (3) Write polar form: 3√2(cos 225° + i sin 225°). Choice C correctly calculates the modulus as 3√2 and determines the argument as 225° with proper quadrant adjustment for Quadrant 3. Choice A incorrectly uses 45° for Quadrant 1; Choice B has the wrong modulus of 6 instead of 3√2; Choice D uses 135° which would be for -3 + 3i in Quadrant 2. Rectangular to polar recipe: Calculate r = √(a² + b²), find reference angle, determine quadrant from signs (Q3 for --), adjust angle (Q3: θ = 180° + reference), then write r(cos θ + i sin θ). When both components have the same absolute value, the reference angle is always 45°—the quadrant adjustment determines the final angle!

Question 6

Convert the complex number $-\sqrt{3}+i$ from rectangular form to polar form $r(\cos\theta+i\sin\theta)$ with $\theta$ in degrees. Use $r=\sqrt{a^2+b^2}$ and $\theta=\arctan\left(\frac{b}{a}\right)$ with quadrant adjustment.

$2\left(\cos 150^\circ+i\sin 150^\circ\right)$ (correct answer)
$2\left(\cos 30^\circ+i\sin 30^\circ\right)$
$\sqrt{2}\left(\cos 150^\circ+i\sin 150^\circ\right)$
$2\left(\cos 210^\circ+i\sin 210^\circ\right)$

Explanation: This question tests your ability to convert complex numbers between rectangular form a + bi and polar form r(cos θ + i sin θ), which represent the same number using Cartesian coordinates versus magnitude and direction. Rectangular form a + bi uses horizontal (real) and vertical (imaginary) components, while polar form r(cos θ + i sin θ) uses distance from origin (modulus r) and angle from positive real axis (argument θ, measured counterclockwise). To convert FROM rectangular TO polar: find r = √(a² + b²) using Pythagorean theorem, then find θ = arctan(b/a) BUT adjust for quadrant (arctan only gives reference angle!). For -√3 + i: (1) Find modulus: r = √((-√3)² + 1²) = √(3 + 1) = √4 = 2. (2) Find argument: reference angle = arctan(1/√3) = 30°. Since a = -√3 < 0 and b = 1 > 0, we're in Quadrant 2, so θ = 180° - 30° = 150°. (3) Write polar form: 2(cos 150° + i sin 150°). Choice A correctly calculates modulus as 2 and determines argument as 150° with proper quadrant adjustment for Quadrant 2. Choice B would place the angle at 30°, which would correspond to √3 + i in Quadrant 1, not -√3 + i in Quadrant 2. Remember: negative real part means we're in Q2 or Q3! Rectangular to polar recipe: (1) Calculate r = √(a² + b²). (2) Recognize special values: when you see √3 and 1, think 30-60-90 triangle! (3) Determine quadrant: -√3 + i has negative real, positive imaginary → Quadrant 2. (4) In Q2, θ = 180° - reference angle = 180° - 30° = 150°. Verification: 2 cos 150° = 2(-√3/2) = -√3 ✓ and 2 sin 150° = 2(1/2) = 1 ✓.

Question 7

Find the modulus $r$ and an argument $\theta$ (in radians) for $z=-1+i$ , and write $z$ in polar form $r(\cos\theta+i\sin\theta)$ . Use $r=\sqrt{a^2+b^2}$ and $\theta=\arctan\!\left(\frac{b}{a}\right)$ with quadrant adjustment. (Argument is measured from the positive real axis.)

$\sqrt{2}\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right)$
$\sqrt{2}\left(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4}\right)$ (correct answer)
$2\left(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4}\right)$
$\sqrt{2}\left(\cos\left(-\frac{\pi}{4}\right)+i\sin\left(-\frac{\pi}{4}\right)\right)$

Explanation: This question tests your ability to convert complex numbers between rectangular form a + bi and polar form r(cos θ + i sin θ), which represent the same number using Cartesian coordinates versus magnitude and direction. Rectangular form a + bi uses horizontal (real) and vertical (imaginary) components, while polar form r(cos θ + i sin θ) uses distance from origin (modulus r) and angle from positive real axis (argument θ, measured counterclockwise). To convert -1 + i to polar form with radian measure: (1) Find modulus: r = √((-1)² + 1²) = √(1 + 1) = √2. (2) Find argument: reference angle = arctan(|1|/|-1|) = arctan(1) = π/4. Since a = -1 < 0 and b = 1 > 0, we're in Quadrant 2, so θ = π - π/4 = 3π/4. (3) Write polar form: √2(cos(3π/4) + i sin(3π/4)). Choice B correctly calculates the modulus as √2 and determines the argument as 3π/4 with proper quadrant adjustment for Quadrant 2. Choice A incorrectly uses π/4, which would be for 1 + i in Quadrant 1; Choice D uses -π/4, which would be for 1 - i in Quadrant 4. Rectangular to polar recipe in radians: Calculate r = √(a² + b²), find reference angle in radians, determine quadrant from signs (Q2 for -+), adjust angle (Q2: θ = π - reference), then write r(cos θ + i sin θ). When working in radians, π/4 = 45°, π/2 = 90°, 3π/4 = 135°, and π = 180°—knowing these conversions helps you visualize the angle!

Question 8

Find the modulus $r$ and an argument $\theta$ (in radians) for the complex number $-\sqrt{3}-i$ , then choose its polar form $r(\cos\theta+i\sin\theta)$ . Use $r=\sqrt{a^2+b^2}$ and $\theta=\arctan\left(\frac{b}{a}\right)$ with quadrant adjustment. (Argument is measured counterclockwise from the positive real axis.)

$2\left(\cos\frac{5\pi}{6}+i\sin\frac{5\pi}{6}\right)$
$\sqrt{3}\left(\cos\frac{7\pi}{6}+i\sin\frac{7\pi}{6}\right)$
$2\left(\cos\frac{7\pi}{6}+i\sin\frac{7\pi}{6}\right)$ (correct answer)
$2\left(\cos\left(-\frac{\pi}{6}\right)+i\sin\left(-\frac{\pi}{6}\right)\right)$

Explanation: This question tests your ability to convert complex numbers between rectangular form a + bi and polar form r(cos θ + i sin θ), which represent the same number using Cartesian coordinates versus magnitude and direction. Rectangular form a + bi uses horizontal (real) and vertical (imaginary) components, while polar form r(cos θ + i sin θ) uses distance from origin (modulus r) and angle from positive real axis (argument θ, measured counterclockwise). To convert -√3 - i to polar form: (1) Find modulus: r = √((-√3)² + (-1)²) = √(3 + 1) = √4 = 2. (2) Find argument: reference angle = arctan(|-1|/|-√3|) = arctan(1/√3) = π/6. Since a = -√3 < 0 and b = -1 < 0, we're in Quadrant 3, so θ = π + π/6 = 7π/6. (3) Write polar form: 2(cos(7π/6) + i sin(7π/6)). Choice B correctly calculates modulus using Pythagorean theorem and determines argument with proper quadrant adjustment for Q3. Choice A incorrectly places the complex number in Q2 with θ = 5π/6, but -√3 - i has both negative real and imaginary parts, placing it in Q3 where θ = 7π/6! Rectangular to polar recipe: (1) Calculate r = √(a² + b²). (2) Find reference angle. (3) Determine quadrant from signs. (4) For Q3 (--), use θ = π + reference. For -√3 - i, this gives 2(cos(7π/6) + i sin(7π/6)).

Question 9

Find the modulus $r$ and a correct argument $\theta$ (in degrees) for $z=-\sqrt{3}-i$ , then write $z$ in polar form $r(\cos\theta+i\sin\theta)$ . Use $r=\sqrt{a^2+b^2}$ and $\theta=\arctan\!\left(\frac{b}{a}\right)$ with quadrant adjustment.

$2\left(\cos 210^\circ+i\sin 210^\circ\right)$ (correct answer)
$2\left(\cos 30^\circ+i\sin 30^\circ\right)$
$\sqrt{2}\left(\cos 210^\circ+i\sin 210^\circ\right)$
$2\left(\cos 150^\circ+i\sin 150^\circ\right)$

Explanation: This question tests your ability to convert complex numbers between rectangular form a + bi and polar form r(cos θ + i sin θ), which represent the same number using Cartesian coordinates versus magnitude and direction. Rectangular form a + bi uses horizontal (real) and vertical (imaginary) components, while polar form r(cos θ + i sin θ) uses distance from origin (modulus r) and angle from positive real axis (argument θ, measured counterclockwise). To convert -√3 - i to polar form: (1) Find modulus: r = √((-√3)² + (-1)²) = √(3 + 1) = √4 = 2. (2) Find argument: reference angle = arctan(|-1|/|-√3|) = arctan(1/√3) = 30°. Since a = -√3 < 0 and b = -1 < 0, we're in Quadrant 3, so θ = 180° + 30° = 210°. (3) Write polar form: 2(cos 210° + i sin 210°). Choice A correctly calculates the modulus as 2 and determines the argument as 210° with proper quadrant adjustment for Quadrant 3. Choice B incorrectly uses 30°, which would be the angle for √3 + i in Quadrant 1; Choice D uses 150° which would be for -√3 + i in Quadrant 2. Rectangular to polar recipe: Calculate r = √(a² + b²), find reference angle using arctan(|b|/|a|), determine quadrant from signs (Q3 for --), adjust angle (Q3: θ = 180° + reference), then write r(cos θ + i sin θ). The reference angle arctan(1/√3) = 30° is a special angle—recognizing these patterns helps you work with exact values instead of decimal approximations!

Question 10

Find the modulus $r$ and a correct argument $\theta$ (in degrees) for $z=1-i$ , then write $z$ in polar form $r(\cos\theta+i\sin\theta)$ . Use $r=\sqrt{a^2+b^2}$ and $\theta=\arctan(b/a)$ with quadrant adjustment. (Argument is measured counterclockwise from the positive real axis.)

$\sqrt{2}\big(\cos 45^\circ+i\sin 45^\circ\big)$
$\sqrt{2}\big(\cos 315^\circ+i\sin 315^\circ\big)$ (correct answer)
$2\big(\cos 315^\circ+i\sin 315^\circ\big)$
$\sqrt{2}\big(\cos 225^\circ+i\sin 225^\circ\big)$

Question 11

Convert the complex number $\sqrt{3} + i$ from rectangular form $a+bi$ to polar form $r(\cos\theta + i\sin\theta)$ , where $r=\sqrt{a^2+b^2}$ and $\theta$ is the argument measured in degrees from the positive real axis.

$2\left(\cos 60^\circ + i\sin 60^\circ\right)$
$2\left(\cos 30^\circ + i\sin 30^\circ\right)$ (correct answer)
$\sqrt{3}\left(\cos 30^\circ + i\sin 30^\circ\right)$
$2\left(\cos 150^\circ + i\sin 150^\circ\right)$

Explanation: This question tests your ability to convert complex numbers between rectangular form a + bi and polar form r(cos θ + i sin θ), which represent the same number using Cartesian coordinates versus magnitude and direction. Rectangular form a + bi uses horizontal (real) and vertical (imaginary) components, while polar form r(cos θ + i sin θ) uses distance from origin (modulus r) and angle from positive real axis (argument θ, measured counterclockwise). To convert from rectangular to polar: find r = sqrt(a² + b²) using Pythagorean theorem, then find θ = arctan(b/a) but adjust for quadrant (arctan only gives reference angle!). For √3 + i, r = sqrt(3 + 1) = sqrt(4) = 2, θ = arctan(1/√3) = 30° (quadrant 1, no adjustment needed), so polar form is 2(cos 30° + i sin 30°)—verify by converting back: 2 cos 30° ≈ 1.732 = √3, 2 sin 30° = 1. Choice B correctly computes r and θ using the modulus formula and arctan with quadrant check. Choice A uses 60° instead of 30°, likely from swapping sin and cos angles—remember, θ = arctan(b/a), and for 30-60-90, tan 30° = 1/√3! Rectangular to polar recipe: (1) r = sqrt(a² + b²), (2) θ = arctan(b/a) with adjustment, (3) Write form—for Q1 examples like this, it's straightforward—keep going, you're mastering this!

Question 12

An AC current is represented by the complex number $z=-3+3i$ amps in rectangular form $a+bi$ . Convert it to polar form $r(\cos\theta+i\sin\theta)$ , where $r=\sqrt{a^2+b^2}$ and $\theta=\arctan(b/a)$ with quadrant adjustment. Give $\theta$ in degrees as the argument measured from the positive real axis.

$3\sqrt{2}\big(\cos 45^\circ+i\sin 45^\circ\big)$
$3\sqrt{2}\big(\cos 135^\circ+i\sin 135^\circ\big)$ (correct answer)
$6\big(\cos 135^\circ+i\sin 135^\circ\big)$
$3\sqrt{2}\big(\cos 225^\circ+i\sin 225^\circ\big)$

Question 13

Convert the complex number $-\sqrt{3}-i$ from rectangular form $a+bi$ to polar form $r(\cos\theta+i\sin\theta)$ , with $\theta$ in degrees. Use $r=\sqrt{a^2+b^2}$ and $\theta=\arctan(b/a)$ with quadrant adjustment (argument measured from the positive real axis).

$2\big(\cos 30^\circ+i\sin 30^\circ\big)$
$2\big(\cos 210^\circ+i\sin 210^\circ\big)$ (correct answer)
$2\big(\cos 150^\circ+i\sin 150^\circ\big)$
$\sqrt{3}\big(\cos 210^\circ+i\sin 210^\circ\big)$

Question 14

Convert the complex number $-2 + 2i$ from rectangular form $a+bi$ to polar form $r(\cos\theta + i\sin\theta)$ . Use $r=\sqrt{a^2+b^2}$ and $\theta=\arctan\left(\frac{b}{a}\right)$ with the correct quadrant adjustment (argument measured counterclockwise from the positive real axis, in degrees).

$2\sqrt{2}\left(\cos 45^\circ + i\sin 45^\circ\right)$
$2\sqrt{2}\left(\cos 225^\circ + i\sin 225^\circ\right)$
$\sqrt{2}\left(\cos 225^\circ + i\sin 225^\circ\right)$
$2\sqrt{2}\left(\cos 135^\circ + i\sin 135^\circ\right)$ (correct answer)

Explanation: This question tests your ability to convert complex numbers between rectangular form a + bi and polar form r(cos θ + i sin θ), which represent the same number using Cartesian coordinates versus magnitude and direction. Rectangular form a + bi uses horizontal (real) and vertical (imaginary) components, while polar form r(cos θ + i sin θ) uses distance from origin (modulus r) and angle from positive real axis (argument θ, measured counterclockwise). To convert from rectangular to polar: find r = sqrt(a² + b²) using Pythagorean theorem, then find θ = arctan(b/a) but adjust for quadrant (arctan only gives reference angle!). For -2 + 2i, r = sqrt(4 + 4) = sqrt(8) = 2√2, reference angle = arctan(2/2) = 45°, and since a < 0 and b > 0 (quadrant 2), θ = 180° - 45° = 135°, so polar form is 2√2(cos 135° + i sin 135°). Choice D correctly calculates the modulus using the Pythagorean theorem, determines the argument with proper quadrant adjustment for quadrant 2, and writes the polar form accurately. Choice A fails by using θ = 45° without quadrant adjustment, ignoring that the point is in quadrant 2, not 1—remember, arctan(b/a) needs correction based on signs of a and b! Rectangular to polar recipe: (1) Calculate r = sqrt(a² + b²) (always positive), (2) Reference angle = arctan(|b|/|a|), (3) Adjust for quadrant: Q2 θ = 180° - reference, then write r(cos θ + i sin θ)—keep practicing these adjustments, you've got this!

Question 15

A signal is represented by the complex number $z=\sqrt{3}+i$ in rectangular form $a+bi$ . Convert it to polar form $r(\cos\theta+i\sin\theta)$ , where $r=\sqrt{a^2+b^2}$ and $\theta$ is the argument measured counterclockwise from the positive real axis. (Angle in degrees.)

$2\left(\cos 30^\circ+i\sin 30^\circ\right)$ (correct answer)
$2\left(\cos 60^\circ+i\sin 60^\circ\right)$
$\sqrt{2}\left(\cos 30^\circ+i\sin 30^\circ\right)$
$2\left(\cos 330^\circ+i\sin 330^\circ\right)$

Explanation: This question tests your ability to convert complex numbers between rectangular form a + bi and polar form r(cos θ + i sin θ), which represent the same number using Cartesian coordinates versus magnitude and direction. Rectangular form a + bi uses horizontal (real) and vertical (imaginary) components, while polar form r(cos θ + i sin θ) uses distance from origin (modulus r) and angle from positive real axis (argument θ, measured counterclockwise). To convert √3 + i to polar form: (1) Find modulus: r = √((√3)² + 1²) = √(3 + 1) = √4 = 2. (2) Find argument: reference angle = arctan(1/√3) = 30°. Since a = √3 > 0 and b = 1 > 0, we're in Quadrant 1, so θ = 30° (no adjustment needed). (3) Write polar form: 2(cos 30° + i sin 30°). Choice A correctly calculates the modulus as 2 and determines the argument as 30° in Quadrant 1. Choice B incorrectly uses 60°, confusing arctan(1/√3) = 30° with arctan(√3) = 60°; Choice D uses 330° which would be for √3 - i in Quadrant 4. Rectangular to polar recipe: Calculate r = √(a² + b²), find reference angle, determine quadrant from signs (Q1 for ++), apply quadrant adjustment (Q1: no change), then write r(cos θ + i sin θ). The ratio 1/√3 gives the special angle 30°, while √3/1 gives 60°—these complementary angles appear frequently in complex number conversions!

Question 16

A complex number $z$ has modulus $5$ and is located in the third quadrant such that the reference angle is $\frac{\pi}{3}$ . If $z = a + bi$ in rectangular form, what is the value of $a^2 + b^2$ ?

$25$ (correct answer)
$\frac{25}{4}$
$\frac{75}{4}$
$\frac{100}{3}$

Explanation: For any complex number

z = a + bi

, the relationship

a^2 + b^2 = |z|^2

always holds, where

|z|

is the modulus. Since the modulus is

5

, we have

a^2 + b^2 = 5^2 = 25

. This is true regardless of which quadrant the number is in or what the specific values of

a

and

b

are. Choice B results from incorrectly using

\left(\frac{5}{2}\right)^2

. Choice C comes from adding

a^2

and

b^2

separately after finding their individual values. Choice D results from incorrect trigonometric calculations.

Question 17

The complex number $w = 2\left(\cos\frac{5\pi}{6} + i\sin\frac{5\pi}{6}\right)$ can be written in rectangular form as $a + bi$ . What is the value of $a - b$ ?

$-\sqrt{3} - 1$ (correct answer)
$-\sqrt{3} + 1$
$\sqrt{3} - 1$
$-2\sqrt{3}$

Explanation: Converting to rectangular form:

w = 2\cos\frac{5\pi}{6} + 2i\sin\frac{5\pi}{6} = 2\left(-\frac{\sqrt{3}}{2}\right) + 2i\left(\frac{1}{2}\right) = -\sqrt{3} + i

. So

a = -\sqrt{3}

and

b = 1

, giving

a - b = -\sqrt{3} - 1

. Choice B incorrectly adds instead of subtracting. Choice C uses wrong signs for the trigonometric values. Choice D forgets to include the imaginary part in the calculation.

Question 18

A complex number in polar form is given as $z = 6\left(\cos\frac{11\pi}{6} + i\sin\frac{11\pi}{6}\right)$ . When $z$ is written in rectangular form $a + bi$ and then converted back to polar form using the principal argument in $(-\pi, \pi]$ , what is the new polar representation?

$6\left(\cos\frac{11\pi}{6} + i\sin\frac{11\pi}{6}\right)$
$6\left(\cos\left(-\frac{\pi}{6}\right) + i\sin\left(-\frac{\pi}{6}\right)\right)$ (correct answer)
$6\left(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}\right)$
$6\left(\cos\frac{5\pi}{6} + i\sin\frac{5\pi}{6}\right)$

Explanation: First, convert to rectangular form:

z = 6\cos\frac{11\pi}{6} + 6i\sin\frac{11\pi}{6} = 6 \cdot \frac{\sqrt{3}}{2} + 6i \cdot \left(-\frac{1}{2}\right) = 3\sqrt{3} - 3i

. When converting back to polar form with principal argument in

(-\pi, \pi]

, we need

\theta

such that

\tan\theta = \frac{-3}{3\sqrt{3}} = -\frac{1}{\sqrt{3}}

. Since the number is in the fourth quadrant,

\theta = -\frac{\pi}{6}

. Note that

\frac{11\pi}{6} = 2\pi - \frac{\pi}{6}

, so

\frac{11\pi}{6}

and

-\frac{\pi}{6}

are coterminal angles. Choice A keeps the original angle outside the principal range. Choice C uses the reference angle with wrong sign. Choice D places the number in the wrong quadrant.

Question 19

Consider the complex number $w = 3\left(\cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3}\right)$ . If this number is reflected across the real axis to form a new complex number $w'$ , what is the polar form of $w'$ ?

$3\left(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\right)$
$3\left(\cos\frac{4\pi}{3} + i\sin\frac{4\pi}{3}\right)$ (correct answer)
$3\left(\cos\left(-\frac{2\pi}{3}\right) + i\sin\left(-\frac{2\pi}{3}\right)\right)$
$3\left(\cos\frac{5\pi}{3} + i\sin\frac{5\pi}{3}\right)$

Explanation: Reflecting across the real axis changes

a + bi

a - bi

, which corresponds to changing the angle

\theta

-\theta

(or equivalently

2\pi - \theta

). Since

w

has angle

\frac{2\pi}{3}

, the reflection

w'

has angle

2\pi - \frac{2\pi}{3} = \frac{4\pi}{3}

. The modulus remains unchanged. Choice A gives the wrong angle. Choice C uses negative angle notation but doesn't convert to the standard interval

[0, 2\pi)

. Choice D gives

\frac{5\pi}{3}

which would be incorrect.

Question 20

Convert the complex number $-4+0i$ to polar form $r(\cos\theta+i\sin\theta)$ with $\theta$ in degrees. Use $r=\sqrt{a^2+b^2}$ and choose an argument measured counterclockwise from the positive real axis.

$4\left(\cos 270^\circ+i\sin 270^\circ\right)$
$4\left(\cos 0^\circ+i\sin 0^\circ\right)$
$4\left(\cos 90^\circ+i\sin 90^\circ\right)$
$4\left(\cos 180^\circ+i\sin 180^\circ\right)$ (correct answer)

Explanation: This question tests your ability to convert complex numbers between rectangular form a + bi and polar form r(cos θ + i sin θ), which represent the same number using Cartesian coordinates versus magnitude and direction. Rectangular form a + bi uses horizontal (real) and vertical (imaginary) components, while polar form r(cos θ + i sin θ) uses distance from origin (modulus r) and angle from positive real axis (argument θ, measured counterclockwise). To convert -4 + 0i to polar form: (1) Find modulus: r = √((-4)² + 0²) = √16 = 4. (2) Find argument: the complex number -4 + 0i lies on the negative real axis, so θ = 180°. (3) Write polar form: 4(cos 180° + i sin 180°). Choice C correctly identifies that -4 + 0i has modulus 4 and lies on the negative real axis at angle 180° from the positive real axis. Choice A incorrectly places it at 0° (positive real axis), but -4 is negative, so it's at 180°! Choice D places it at 270° (negative imaginary axis), but there's no imaginary component. Rectangular to polar recipe for real numbers: positive real numbers have θ = 0°, negative real numbers have θ = 180°, positive imaginary numbers have θ = 90°, negative imaginary numbers have θ = 270°. For -4 + 0i, a negative real number, θ = 180°, giving 4(cos 180° + i sin 180°).