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Precalculus : Polar Coordinates and Complex Numbers

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Example Questions

Example Question #463 : Pre Calculus

Divide.

$\frac{4-i^3}{5+i}$

Possible Answers:

$\frac{7}{13}-\frac{21i}{6}$

$\frac{26}{21}+\frac{26}{i}$

$\frac{21}{26}+\frac{i}{26}$

$\frac{21}{26}-\frac{i}{26}$

Correct answer:

$\frac{21}{26}+\frac{i}{26}$

Explanation:

To divide complex numbers, multiply both the numerator and denominator by the conjugate of the denominator.

To find the conjugate, just change the sign in the denominator. The conjugate used will be 5-i .

$\left(\frac{4-i^3}{5+i}\right)\left(\frac{5-i}{5-i}\right)$

Now, distribute and simplify.

$\left(\frac{4-i^3}{5+i}\right)\left(\frac{5-i}{5-i}\right)=\frac{20-4i-5i^3+i^4}{25-i^2}$

Recall that $i^2=-1, i^3=-i,\text{ and }i^4=1$

$\frac{20-4i-5i^3+i^4}{25-i^2}=\frac{20-4i-5(-i)+1}{25-(-1)}=\frac{21+i}{26}=\frac{21}{26}+\frac{i}{26}$

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Example Question #464 : Pre Calculus

Divide

$\frac{5-8i^2}{3+i}$

Possible Answers:

$\frac{39}{10}+\frac{13i}{10}$

$\frac{39}{10}-\frac{13i}{10}$

$\frac{1}{2}+\frac{13i}{10}$

$\frac{10}{39}-\frac{10}{13i}$

Correct answer:

$\frac{39}{10}-\frac{13i}{10}$

Explanation:

Start by simplifying the fraction.

Recall that i^2=-1

$\frac{5-8i^2}{3+i}=\frac{5-8(-1)}{3+i}=\frac{13}{3+i}$

To divide complex numbers, multiply both the numerator and denominator by the conjugate of the denominator.

To find the conjugate, just change the sign in the denominator. The conjugate used will be 3-i .

$\left(\frac{13}{3+i}\right)\left(\frac{3-i}{3-i}\right)$

Now, distribute and simplify.

$\left(\frac{13}{3+i}\right)\left(\frac{3-i}{3-i}\right)=\frac{39-13i}{9-i^2}$

Recall that i^2=-1

$\frac{39-13i}{9-(-1)}=\frac{39-13i}{10}=\frac{39}{10}-\frac{13i}{10}$

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Example Question #465 : Pre Calculus

Simplify $\frac{3+5i}{2-4i}$ into a number of the form a+bi .

Possible Answers:

3+i

None of the other answers.

$\frac{1}{2}-\frac{3}{2}i$

-1+5i

$-\frac{7}{10}+\frac{11}{10}i$

Correct answer:

$-\frac{7}{10}+\frac{11}{10}i$

Explanation:

We have

$\frac{3+5i}{2-4i}$

Multiply by the complex conjugate of the denominator.

The complex conjugate is the denominator with the sign changed:

$\frac{3+5i}{2-4i}\times \frac{2+4i}{2+4i}$

Multiply fractions

$\frac{(3+5i)(2+4i)}{(2-4i)(2+4i)}$

FOIL the numerator and denominator

$\frac{6+12i+10i+20i^2}{4+8i-8i-16i^2}$

Apply the rule of i^2 = -1 :

$\frac{6+12i+10i-20}{4+8i-8i+16}$

Simplify.

$\frac{-14+22i}{20}$

Simplify further using the addition of fractions rule, then factor the i out of the 2nd fraction.

$-\frac{7}{10}+\frac{11}{10}i$

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Example Question #21 : Polar Coordinates And Complex Numbers

Divide:

$\frac{5-i}{3+6i}$

Possible Answers:

$\frac{3-11i}{5}$

$\frac{4}{5}-\frac{5i}{8}$

$15+\frac{15}{11}i$

$\frac{1}{15}+\frac{11i}{15}$

Correct answer:

$\frac{3-11i}{5}$

Explanation:

To divide complex numbers, multiply both the numerator and denominator by the conjugate of the denominator.

To find the conjugate, just change the sign in the denominator. The conjugate used will be 3-6i .

$\left ( \frac{5-i}{3+6i} \right )\left ( \frac{3-6i}{3-6i} \right )$

Now, distribute and simplify.

$\left ( \frac{5-i}{3+6i} \right )\left ( \frac{3-6i}{3-6i} \right )=\frac{15-30i-3i+6i^{2}}{9-6i^{2}}$

Recall that i^2=-1

$\frac{15-33i-6}{9+6}$

Then combine like terms:

$\frac{9-33i}{15}$

Then since each term is a multiple of you can simplify:

$\frac{3-11i}{5}$

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Example Question #21 : Polar Coordinates And Complex Numbers

Find the magnitude of the complex number described by {}(1+3i) .

Possible Answers:

$\sqrt{4}$

$\sqrt{10}$

$\sqrt8$

${}\sqrt{2}$

$\sqrt{3}$

Correct answer:

$\sqrt{10}$

Explanation:

To find the magnitude of a complex number we use the formula:

$|z|=\sqrt{a^2+b^2}$ ,

where our complex number is in the form (a+bi) .

Therefore,

$|z|=\sqrt{1^2+3^2}=\sqrt{10}$

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Example Question #22 : Polar Coordinates And Complex Numbers

Find the magnitude of :

${}(1+3i)^{7}$ , where the complex number satisfies ${}i^{2}=-1$ .

Possible Answers:

${}10^3\sqrt{10}$

${}10^7\sqrt{10}$

${}10^2\sqrt{10}$

${}10^{14}\sqrt{10}$

${}10\sqrt{10}$

Correct answer:

${}10^3\sqrt{10}$

Explanation:

Note for any complex number z, we have:

${}|z^n|=|z|^n$ .

Let {}z=1+i3 . Hence ${}|z|=\sqrt{10}$

Therefore:

${}|z^{7}|=|z|^{7}=(\sqrt{10})^{7}=(\sqrt{10})^{6}(\sqrt{10})$

This gives the result.

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

What is the magnitude of (2+3i)^2 ?

Possible Answers:

$\sqrt13$

Correct answer:

Explanation:

To find the magnitude of a complex number we use the following formula:

$|z|=\sqrt{a^2+b^2}$ , where (a+bi) .

Therefore we get,

$|z|=\sqrt{2^2+3^2}=\sqrt{13}$ .

Now to find

$|z|^2=(\sqrt{13})^2$

z=13 .

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Example Question #1 : Powers And Roots Of Complex Numbers

Simplify

$\small (\cos x+i\sin x)^3$

Possible Answers:

$\small \small \sin 3x+i\cos 3x$

$\small \small 3\cos x+3i\sin x$

$\small \small \cos 3x+i\sin3x$

$\small \small \cos 3x+i\sin x$

Correct answer:

$\small \small \cos 3x+i\sin3x$

Explanation:

We can use DeMoivre's formula which states:

$z^n=r^n(\cos n\theta + i \sin n \theta)$

Now plugging in our values of $\small n=3$ and r=1 we get the desired result.

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

$\cos 3x +i \sin 3x$

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

$(3 \sqrt3 -3i) ^ 6$

Possible Answers:

15,625

-46,656

23,328 - 23,328 i

23,328 + 23,328 i

46,656

Correct answer:

-46,656

Explanation:

First convert this point to polar form:

$r = \sqrt{(3\sqrt3)^2 + (-3)^2 } = \sqrt{9 \cdot 3 + 9 } = \sqrt{36} = 6$

$\cos \theta = \frac{ 3\sqrt3}{6 } = \frac{\sqrt3}{2}$

$\theta = \cos ^{-1} (\frac{\sqrt3}{2} ) = \frac{ \pi }{6} \enspace \or \enspace \frac{ 11 \pi }{6}$

Since this number has a negative imaginary part and a positive real part, it is in quadrant IV, so the angle is $\frac{ 11 \pi }{6}$

We are evaluating $(6 \cos \frac{11 \pi }{6} + i \sin \frac{ 11 \pi }{6} ) ^ 6$

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.

$6^6 (\cos \frac{ 11 \pi }{6} \cdot 6 + i \sin \frac{ 11 \pi }{6} \cdot 6)$

$\frac{11 \pi }{6} \cdot 6 = 11 \pi$ which is coterminal with $\pi$ since it is an odd multiplie

$46,656(\cos \pi + i \sin \pi ) = 46,656 (-1 + 0i) = -46,656$

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Example Question #4 : Powers And Roots Of Complex Numbers

Evaluate $(2 -2 \sqrt3)^6$

Possible Answers:

-4,096

4,096i

4,096

Correct answer:

4,096

Explanation:

First, convert this complex number to polar form:

$r = \sqrt{2^2 + (2\sqrt3)^2 } = \sqrt{4 +12} = \sqrt{16} = 4$

$\sin \theta = \frac{ -2\sqrt3 }{4} = -\frac{\sqrt3}{2}$

$\theta = \sin ^{-1} (-\frac{\sqrt3}{2}) = \frac{4 \pi }{3} \enspace or \enspace \frac{ 5 \pi }{3}$

Since the real part is positive and the imaginary part is negative, this is in quadrant IV, so the angle is $\frac{5 \pi }{3}$

So we are evaluating $(4(\cos \frac{5 \pi }{3} + i \sin \frac{5 \pi }{3} ) )^6$

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.

$4^6 (\cos \frac{ 5 \pi }{3} \cdot 6 + i \sin \frac{ 5 \pi }{3} \cdot 6 ) = 4,096 ( \cos 10 \pi + i \sin 10 \pi)$

$10 \pi$ is coterminal with since it is an even multiple of $\pi$

$4,096( \cos 0 + i \sin 0 ) = 4,096 ( 1 + 0i) = 4,096$

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Precalculus : Polar Coordinates and Complex Numbers

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #463 : Pre Calculus

Example Question #464 : Pre Calculus

Example Question #465 : Pre Calculus

Example Question #21 : Polar Coordinates And Complex Numbers

Example Question #21 : Polar Coordinates And Complex Numbers

Example Question #22 : Polar Coordinates And Complex Numbers

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

Example Question #1 : Powers And Roots Of Complex Numbers

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

Example Question #4 : Powers And Roots Of Complex Numbers

All Precalculus Resources

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