I have to remember to convert (-2 + 2i) in to polar form.

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I have to remember to convert (-2 + 2i) in to polar form.

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Polar form

Magnitude = sqrt(2^2 + 2^2) = sqrt(8) = 2sqrt(2)

Angle = tan-1 (2/-2)CALCULATOR FEEDS BACK -PI/4, BUT KNOW BETTER BECAUSE IT IS A 2ND QUADRANT ANGLE = 3PI/4.

This means problems translates to (2sqrt(2))^8 * e^i(3PI/4)^8

(2sqrt(2))^8 = 4096

e^i(3PI/4)^8 = e^i(24PI/4) = e^i(6PI) = +1

this means that the answer is entirely real and = 4096.

(-2 + 2i)^{8}

= [2sqrt(2)cis(3pi/4)]^{8}

= 2^{12}cis(6pi)

= 4096

------

Attn: (-2 + 2i ) is in the second quadrant, and cis(x) = cos(x) + i sin(x)

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