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Precalculus : Polar Coordinates and Complex Numbers

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Example Questions

Example Question #3 : Evaluate Powers Of Complex Numbers Using De Moivre's Theorem

Evaluate $(-3 + 3i ) ^{ 12}$

Possible Answers:

34,012,224

-531,441

-34,012,224

$17,006,112 \sqrt2 + 17,006,112 \sqrt 2 i$

$-17,006, 112 \sqrt2 + 17,006,112 \sqrt 2 i$

Correct answer:

-34,012,224

Explanation:

First convert the complex number into polar form:

$r = \sqrt{3^2 + (-3)^2 } = \sqrt{9+9} = \sqrt{2 \cdot 9 } = 3 \sqrt2$

$\sin \theta = \frac{3}{3 \sqrt2} = \frac{ 1}{\sqrt2}$

$\theta = \sin ^{-1 } (\frac{1}{\sqrt2}) = \frac{\pi}{4} \enspace or \enspace \frac{ 3 \pi }{4}$

Since the real part is negative but the imaginary part is positive, the angle should be in quadrant II, so it is $\frac{ 3 \pi }{4}$

We are evaluating $(3\sqrt2 (\cos \frac{ 3 \pi }{4} + i \sin \frac{ 3 \pi }{4} ) ) ^ {12}$

Using DeMoivre's Theorem:

DeMoivre's Theorem is

$(\cos(x)+i\sin(x))^n=\cos(nx)+i\sin(nx)$

We apply it to our situation to get.

$(3\sqrt2)^{12 } (\cos \frac{ 3 \pi }{4} \cdot 12 + i \sin \frac{3 \pi }{4} \cdot 12 )$ simplify and take the exponent

$34,012,224 (\cos 9\pi + i \sin 9 \pi )$

$9\pi$ is coterminal with $\pi$ since it is an odd multiple of pi

$34, 012, 224 ( \cos \pi + i \sin \pi ) = 34, 012, 224 ( -1 + 0i) = -34,012, 224$

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Example Question #6 : Powers And Roots Of Complex Numbers

Use DeMoivre's Theorem to evaluate the expression $(\sqrt3 - 3i ) ^ 6$ .

Possible Answers:

Correct answer:

Explanation:

First convert this complex number to polar form:

$r = \sqrt{ (\sqrt3)^2 + 1^2 } = \sqrt{ 3 + 1 } = \sqrt4 = 2$

$\sin \theta = \frac{ 1}{ 2}$ so $\theta = \frac{\pi }{6} \enspace or \enspace \frac{5 \pi }{6}$

Since this number has positive real and imaginary parts, it is in quadrant I, so the angle is $\frac{ \pi }{6}$

So we are evaluating $(2(\cos \frac{ \pi }{6} + i \sin \frac{ \pi }{6} ))^3$

Using DeMoivre's Theorem:

DeMoivre's Theorem is

$(\cos(x)+i\sin(x))^n=\cos(nx)+i\sin(nx)$

We apply it to our situation to get.

$2^3 (\cos \frac{3\pi}{6} + i \sin \frac{ 3\pi }{6} ) = 8 ( \cos \frac{\pi}{2} + i \sin \frac{ \pi }{2} ) = 8(0 + i ) = 8i$

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Example Question #2 : Powers And Roots Of Complex Numbers

Evaluate: (1-i)^8

Possible Answers:

256

128

16i

Correct answer:

Explanation:

First, convert this complex number to polar form.

$r= \sqrt{ a^2 + b^2 } = \sqrt{1^2 +(-1)^2 }=\sqrt{2}$

$\sin \theta = \frac{-1}{\sqrt2}$

$\theta = \frac{5 \pi }{4} \enspace or \enspace \frac{7 \pi }{4}$

Since the point has a positive real part and a negative imaginary part, it is located in quadrant IV, so the angle is $\frac{7\pi}{4}$ .

This gives us $(\sqrt{2} ( \cos \frac{7 \pi }{4} + i \sin \frac{ 7 \pi }{4} )) ^ 8$

To evaluate, use DeMoivre's Theorem:

DeMoivre's Theorem is

$(\cos(x)+i\sin(x))^n=\cos(nx)+i\sin(nx)$

We apply it to our situation to get.

$(\sqrt2 )^8 ( \cos \frac{ 7 \pi }{4} \cdot 8 + i \sin \frac{ 7 \pi }{4} \cdot 8 )$ simplifying

$2 ^ 4 ( \cos 14 \pi + i \sin 14 \pi )$ , $14 \pi$ is coterminal with since it is an even multiple of $\pi$

$2^ 4 (\cos 0 + i \sin 0 ) = 16(1 + 0i) = 16$

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Example Question #1 : Evaluate Powers Of Complex Numbers Using De Moivre's Theorem

(2 + 2i )^9

Possible Answers:

11,585.238 i

512

11,585.238

8,192 + 8,192 i

Correct answer:

8,192 + 8,192 i

Explanation:

First, convert the complex number to polar form:

$r = \sqrt{ 2^2 + 2^2 } = \sqrt{4 + 4 } = \sqrt{2 \cdot 4 } = 2 \sqrt 2$

$\sin \theta = \frac{ 2 }{ 2 \sqrt 2 } = \frac{ 1 }{ \sqrt2 }$

$\theta = \sin ^{-1 } (\frac{ 1}{ \sqrt 2 }) = \frac{ \pi }{4} \enspace or \enspace \frac{ 3 \pi }{4}$

Since both the real and the imaginary parts are positive, the angle is in quadrant I, so it is $\frac{ \pi }{4}$

This means we're evaluating

$(2 \sqrt2 (\cos \frac{ \pi }{4} + i \sin \frac{ \pi }{4} ))^ 9$

Using DeMoivre's Theorem:

DeMoivre's Theorem is

$(\cos(x)+i\sin(x))^n=\cos(nx)+i\sin(nx)$

We apply it to our situation to get.

$(2 \sqrt2 )^9 (\cos \frac{\pi }{4} \cdot 9 + i \sin \frac{ \pi }{4} \cdot 9 )$

First, evaluate $(2\sqrt2)^9$ . We can split this into $2^9 \cdot (\sqrt2)^9$ which is equivalent to $2^9 \cdot (\sqrt2)^8 \cdot \sqrt2 = 2^9 \cdot 2^4 \cdot \sqrt 2$

[We can re-write the middle exponent since $\sqrt 2$ is equivalent to $2^{\frac{1}{2}}$ ]

This comes to $8,192 \sqrt 2$

Evaluating sine and cosine at $\frac{\pi }{4} \cdot 9 = \frac{ 9 \pi }{4}$ is equivalent to evaluating them at $\frac{ \pi }{4}$ since $\frac{ 9 \pi }{4} - 2 \pi = \frac{9 \pi }{4 } - \frac{ 8 \pi }{4} = \frac{ \pi }{4}$

This means our expression can be written as:

$8,192 \sqrt2 ( \cos \frac{\pi }{4} + i \sin \frac{ \pi }{4} ) = 8,192 \sqrt2 (\frac{1}{\sqrt2} + i \frac{ 1}{\sqrt2}) = 8,192 + 8,192i$

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Example Question #472 : Pre Calculus

Evaluate ${}(1+i)^{2n}$ , where is a natural number and is the complex number ${}i^{2}=-1$ .

Possible Answers:

${}2^{n}i^{n}$

${}2^{n+1}i^{n}$

${}-2^{n}i^{n}$

${}-2^{n+1}i^{n}$

${}2^{n}i^{n+1}$

Correct answer:

${}2^{n}i^{n}$

Explanation:

Note that,

${}(1+i)^{2n}=((1+i)^2)^{n}}= (2i)^{n}= 2^{n}i^{n}$

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Example Question #1 : Find The Roots Of Complex Numbers

What is the length of

$({1-i})^{3}$ ?

Possible Answers:

$2\sqrt{2}$

$\sqrt{2}$

$2-\sqrt{2}$

$2+\sqrt{2}$

Correct answer:

$2\sqrt{2}$

Explanation:

We have

$|1-i|=\sqrt{2}$ .

Hence,

$|(1-i)^3|=|1-i|^{3}=(\sqrt2)^3=2\sqrt{2}$ .

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Example Question #2 : Find The Roots Of Complex Numbers

Solve for $\small z^3=1$ (there may be more than one solution).

Possible Answers:

$\small z=1, \frac{-1+\sqrt{3}}{2}, \frac{-1-\sqrt{3}}{2}$

$\small z=1, \frac{-1+i\sqrt{3}}{2}, \frac{-1-i\sqrt{3}}{2}$

$\small z=1$

$\small z=-1$

Correct answer:

$\small z=1, \frac{-1+i\sqrt{3}}{2}, \frac{-1-i\sqrt{3}}{2}$

Explanation:

Solving that equation is equivalent to solving the roots of the polynomial $\small f(z)=z^3-1$ .

Clearly, one of roots is 1.

Thus, we can factor the polynomial as $\small f(z)=(z-1)(z^2+z+1)$

so that we solve for the roots of $\small z^2+z+1$ .

Using the quadratic equation, we solve for roots, which are $\small z=\frac{-1+i\sqrt{3}}{2}, \frac{-1-i\sqrt{3}}{2}$ .

This means the solutions to $\small z^3=1$ are

$z=1, \frac{-1+i\sqrt{3}}{2}, \frac{-1-i\sqrt{3}}{2}$

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Example Question #4 : Find The Roots Of Complex Numbers

Recall that $z=rcis(\theta)$ is just shorthand for $z=r(cos(\theta)+isin(\theta))$ when dealing with complex numbers in polar form.

Express $\frac{1}{(\sqrt{3}-i)^4}$ in polar form.

Possible Answers:

$2cis\left(\frac{\pi}{3}\right)$

$2cis\left(\frac{2\pi}{3}\right)$

$\frac{1}{16}cis\left(\frac{-2\pi}{3}\right)$

$\frac{1}{16}cis\left(\frac{2\pi}{3}\right)$

$2cis\left(\frac{2\pi}{12}\right)$

Correct answer:

$\frac{1}{16}cis\left(\frac{2\pi}{3}\right)$

Explanation:

First we recognize that we are trying to solve $z^{-4}$ where $z=\sqrt{3}-i$ .

Then we want to convert into polar form using,

$arg(z)=\theta =arctan(\frac{b}{a})$ and $\left | z\right |=r=\sqrt{a^2+b^2}$ .

Then since De Moivre's theorem states,

$z^n=r^ncis(n\theta)$ if is an integer, we can say

$z^{-4}=2^{-4}cis(-4*\frac{-\pi}{6})=\frac{1}{16}cis\left(\frac{2\pi}{3}\right)$ .

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Example Question #1 : Find The Roots Of Complex Numbers

Solve for z^3-z^2+z=0 (there may be more than one solution).

Possible Answers:

$0, 1, \frac{1}{2}$

$0, \frac{1}{2}\pm\frac{\sqrt{3}}{2}$

$0, \frac{1}{2}\pm\frac{\sqrt{3i}}{2}$

$0, \frac{1}{2}\pm\sqrt{3i}$

Correct answer:

$0, \frac{1}{2}\pm\frac{\sqrt{3i}}{2}$

Explanation:

To solve for the roots, just set equal to zero and solve for z using the quadratic formula ( $\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ ) : $z^3-z^2+z\rightarrow z(z^2-z+1)$ and now setting both and z^2-z+1 equal to zero we end up with the answers z=0 and $z=\frac{1\pm 3i}{2}$

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Example Question #5 : Find The Roots Of Complex Numbers

Compute (3+3i)^5

Possible Answers:

972i

972

-972-972i

972-972i

972+972i

Correct answer:

-972-972i

Explanation:

To solve this question, you must first derive a few values and convert the equation into exponential form: $re^{\pi*z}$ : $r=\sqrt{9+9}=3\sqrt{2}\rightarrow tan(\theta)=\frac{3}{3}=1\rightarrow Arg(z)=\frac{\pi}{4}$

Now plug back into the original equation and solve: $(3+3i)^5=(3\sqrt{2})^5\cdot e^{\frac{i5\pi}{4}=972\sqrt{2}(cos(\frac{5\pi}{4})+isin(\frac{5\pi}{4}))=972\sqrt{2}(\frac {-\sqrt{2}{2}}-\frac ({\sqrt{2}\cdot i}{2})=-972-972i$

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Precalculus : Polar Coordinates and Complex Numbers

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #3 : Evaluate Powers Of Complex Numbers Using De Moivre's Theorem

Example Question #6 : Powers And Roots Of Complex Numbers

Example Question #2 : Powers And Roots Of Complex Numbers

Example Question #1 : Evaluate Powers Of Complex Numbers Using De Moivre's Theorem

Example Question #472 : Pre Calculus

Example Question #1 : Find The Roots Of Complex Numbers

Example Question #2 : Find The Roots Of Complex Numbers

Example Question #4 : Find The Roots Of Complex Numbers

Express $\frac{1}{(\sqrt{3}-i)^4}$ in polar form.

Example Question #1 : Find The Roots Of Complex Numbers

Example Question #5 : Find The Roots Of Complex Numbers

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Precalculus : Polar Coordinates and Complex Numbers

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Example Questions

Example Question #3 : Evaluate Powers Of Complex Numbers Using De Moivre's Theorem

Example Question #6 : Powers And Roots Of Complex Numbers

Example Question #2 : Powers And Roots Of Complex Numbers

Example Question #1 : Evaluate Powers Of Complex Numbers Using De Moivre's Theorem

Example Question #472 : Pre Calculus

Example Question #1 : Find The Roots Of Complex Numbers

Example Question #2 : Find The Roots Of Complex Numbers

Example Question #4 : Find The Roots Of Complex Numbers

Express in polar form.

Example Question #1 : Find The Roots Of Complex Numbers

Example Question #5 : Find The Roots Of Complex Numbers

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Express $\frac{1}{(\sqrt{3}-i)^4}$ in polar form.