Algebra 2 Quiz: Complex Numbers In Rectangular Polar Form

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Complex Numbers In Rectangular Polar FormQuestion 1 of 20

The complex number $z = -2 - 2\sqrt{3}i$ is written in polar form as $r(\cos\theta + i\sin\theta)$ where $\theta \in [0, 2\pi)$ . What is the exact value of $r + \theta$ ?

2\sqrt{3} + \frac{5\pi}{3}

4 + \frac{5\pi}{3}

2\sqrt{3} + \frac{4\pi}{3}

4 + \frac{4\pi}{3}

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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

The complex number $z = -2 - 2\sqrt{3}i$ is written in polar form as $r(\cos\theta + i\sin\theta)$ where $\theta \in [0, 2\pi)$ . What is the exact value of $r + \theta$ ?

$2\sqrt{3} + \frac{5\pi}{3}$
$4 + \frac{5\pi}{3}$
$2\sqrt{3} + \frac{4\pi}{3}$
$4 + \frac{4\pi}{3}$ (correct answer)

Explanation: Converting complex numbers from rectangular to polar form requires finding both the modulus (r) and argument (θ). When you see this type of problem, visualize the complex number as a point in the coordinate plane to determine which quadrant it's in. For

z = -2 - 2\sqrt{3}i

, first find the modulus:

r = \sqrt{(-2)^2 + (-2\sqrt{3})^2} = \sqrt{4 + 12} = \sqrt{16} = 4

. Next, find the argument. Since both real and imaginary parts are negative, the complex number is in the third quadrant. The reference angle is

\tan^{-1}\left(\frac{|-2\sqrt{3}|}{|-2|}\right) = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3}

. In the third quadrant,

\theta = \pi + \frac{\pi}{3} = \frac{4\pi}{3}

. Therefore,

r + \theta = 4 + \frac{4\pi}{3}

, which is answer choice D. Choice A uses

r = 2\sqrt{3}

instead of 4 - this comes from incorrectly calculating

\sqrt{4 + 12} = \sqrt{16}

. Choice B correctly finds

r = 4

but uses

\theta = \frac{5\pi}{3}

, which would place the number in the fourth quadrant where the imaginary part is positive. Choice C makes both errors: using the incorrect modulus

2\sqrt{3}

and the correct argument. Remember: always check which quadrant your complex number belongs to based on the signs of its real and imaginary parts. The argument must reflect the correct quadrant, and the modulus is always the distance from the origin.

Question 2

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.

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

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 4

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 5

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 6

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 7

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 8

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 9

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 10

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 11

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 12

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 13

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 14

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 15

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 16

Find the polar form of $-5i$ as $r(\cos\theta + i\sin\theta)$ with $\theta$ in degrees, where $r=\sqrt{a^2+b^2}$ and $\theta$ is the argument measured from the positive real axis.

$5\left(\cos 90^\circ + i\sin 90^\circ\right)$
$-5\left(\cos 270^\circ + i\sin 270^\circ\right)$
$5\left(\cos 270^\circ + i\sin 270^\circ\right)$ (correct answer)
$5\left(\cos 180^\circ + i\sin 180^\circ\right)$

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 -5i (0 - 5i), r = sqrt(0 + 25) = 5, and since it's on the negative imaginary axis (a=0, b<0), θ = 270° (or -90°), so polar form is 5(cos 270° + i sin 270°)—verify: cos 270°=0, sin 270°=-1, giving 0 -5i. Choice C correctly identifies r positive and θ=270° for the negative imaginary axis. Choice D uses negative r, but modulus is always positive—r represents distance, so it's non-negative! Rectangular to polar recipe: For axis cases, like negative imaginary: θ=270°, r=|b|—you're handling special cases well, stay encouraged!

Question 17

Convert $3\left(\cos 60^\circ+i\sin 60^\circ\right)$ to rectangular form $a+bi$ using $a=r\cos\theta$ and $b=r\sin\theta$ .

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

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(cos 60° + i sin 60°) to rectangular form: (1) Identify r = 3 and θ = 60°. (2) Calculate real part: a = r cos θ = 3 cos 60° = 3(1/2) = 3/2. (3) Calculate imaginary part: b = r sin θ = 3 sin 60° = 3(√3/2) = 3√3/2. (4) Write rectangular form: 3/2 + (3√3/2)i. Choice A correctly applies the conversion formulas a = r cos θ and b = r sin θ, using the special angle values cos 60° = 1/2 and sin 60° = √3/2. Choice B swaps the values, incorrectly using a = r sin θ and b = r cos θ—remember cosine goes with the real part (horizontal) and sine with the imaginary part (vertical), just like x = r cos θ and y = r sin θ in parametric equations! Polar to rectangular recipe: From r(cos θ + i sin θ), calculate a = r cos θ for the real part and b = r sin θ for the imaginary part, then write a + bi. For special angles like 30°, 45°, 60°, memorize the exact values from the unit circle—they appear frequently and give cleaner answers than decimal approximations!

Question 18

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 19

Convert $4\,\text{cis}\,150^\circ$ to rectangular form $a+bi$ . (Recall $r\,\text{cis}\,\theta=r(\cos\theta+i\sin\theta)$ and use $a=r\cos\theta$ , $b=r\sin\theta$ .)

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

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. The notation "cis" is shorthand: r cis θ means r(cos θ + i sin θ), so 4 cis 150° = 4(cos 150° + i sin 150°). To convert to rectangular form: (1) Identify r = 4 and θ = 150°. (2) Calculate real part: a = r cos θ = 4 cos 150° = 4(-√3/2) = -2√3. (3) Calculate imaginary part: b = r sin θ = 4 sin 150° = 4(1/2) = 2. (4) Write rectangular form: -2√3 + 2i. Choice A correctly applies the conversion formulas using cos 150° = -√3/2 (negative in Quadrant 2) and sin 150° = 1/2 (positive in Quadrant 2). Choice B incorrectly makes the imaginary part negative; Choice D swaps the coefficients, confusing which trig value goes where—remember cos 150° = -cos 30° = -√3/2 and sin 150° = sin 30° = 1/2. Polar to rectangular recipe: From r cis θ or r(cos θ + i sin θ), calculate a = r cos θ and b = r sin θ, then write a + bi. For angles like 150° = 180° - 30°, use reference angle relationships: cos 150° = -cos 30° and sin 150° = sin 30°, making calculations easier with special angle values!

Question 20

Convert from polar to rectangular form using $a=r\cos\theta$ and $b=r\sin\theta$ : $5\left(\cos 45^\circ + i\sin 45^\circ\right).$

$\frac{5\sqrt{2}}{2} + \frac{5\sqrt{2}}{2}i$ (correct answer)
$\frac{5}{2} + \frac{5\sqrt{2}}{2}i$
$\frac{5\sqrt{2}}{2} - \frac{5\sqrt{2}}{2}i$
$5\sqrt{2} + 5\sqrt{2}\,i$

Explanation: This question tests your ability to convert complex numbers between rectangular form

a + bi

and polar form

r(\cos \theta + i \sin \theta)

, 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 \theta + i \sin \theta)

uses distance from origin (modulus r) and angle from positive real axis (argument

\theta

, measured counterclockwise). To convert from polar to rectangular: use

a = r \cos \theta

and

b = r \sin \theta

—both forms describe the same point, just different perspectives! For

5(\cos 45^\circ + i \sin 45^\circ)

a = 5 \cos 45^\circ = 5(\sqrt{2}/2) = (5\sqrt{2})/2

b = 5 \sin 45^\circ = (5\sqrt{2})/2

, giving

(5\sqrt{2})/2 + (5\sqrt{2})/2 i

—note

\cos 45^\circ = \sin 45^\circ = \sqrt{2}/2

from 45-45-90 triangle. Choice A correctly evaluates the trig functions and applies the formulas. Choice D forgets to multiply by 1/2, using

\sqrt{2}

instead of

\sqrt{2}/2

—remember to use exact values for special angles! Polar to rectangular recipe: (1) Identify r and

\theta

, (2)

a = r \cos \theta

, (3)

b = r \sin \theta

, (4) Simplify— for 45°, it's symmetric, great for practice—you're doing fantastically!