Question

Determine which of the following equations are true by converting the complex numbers from standard form $z = x + yi$ to polar form $z = re^{i\theta}$. Check all true statements. $-2 + 2i = \sqrt{8}e^{i\frac{3\pi}{4}}$ $2 - 2i = \sqrt{8}e^{i\frac{7\pi}{4}}$ $2 + 2i = \sqrt{8}e^{i\frac{\pi}{4}}$ $-2 - 2i = \sqrt{8}e^{i\frac{5\pi}{4}}$

          Determine which of the following equations are true by converting the complex numbers from standard form $z = x + yi$ to polar form $z = re^{i\theta}$. Check all true statements.

$-2 + 2i = \sqrt{8}e^{i\frac{3\pi}{4}}$
$2 - 2i = \sqrt{8}e^{i\frac{7\pi}{4}}$
$2 + 2i = \sqrt{8}e^{i\frac{\pi}{4}}$
$-2 - 2i = \sqrt{8}e^{i\frac{5\pi}{4}}$

Determine which of the following equations are true by converting the complex numbers from standard form z = x + yi to polar form z = re^iθ. Check all true statements.

-2 + 2i = √(8)e^i(3π)/(4)
2 - 2i = √(8)e^i(7π)/(4)
2 + 2i = √(8)e^i(π)/(4)
-2 - 2i = √(8)e^i(5π)/(4)

Added by Jon F.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Adi S

07/02/2023

Step 1

For -2+2i: r = sqrt((-2)^2 + 2^2) = sqrt(8) theta = arctan(2/-2) = -pi/4 Therefore, -2+2i = sqrt(8)e^((-pi/4)i) For 2-2i: r = sqrt(2^2 + (-2)^2) = sqrt(8) theta = arctan(-2/2) = -pi/4 + pi = 3pi/4 Therefore, 2-2i = sqrt(8)e^((3pi/4)i) For 2+2i: r = sqrt(2^2 + Show more…

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Determine which of the following equations are true by converting the complex numbers from standard form z=x+yi to polar form z=re^(i heta ). Check all true statements. ◻ -2+2i=sqrt(8)e^((3pi )/(4)i) 2-2i=sqrt(8)e^((7pi )/(4)i) 2+2i=sqrt(8)e^((pi )/(4)i) -2-2i=sqrt(8)e^((5pi )/(4)i) Determine which of the following equations are true by converting the complex numbers from standard form z = + yi to polar form z=rei. Check all true statements. -2+2i=8e 2-2i=8e 2+2i=v8e7 -2-2i=8e