What is the value of (-2 + 2i)^8 ?

Question

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

 

Answer 1

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.

Answer 2

(-2 + 2i)⁸

= [2sqrt(2)cis(3pi/4)]⁸

= 2¹²cis(6pi)

= 4096

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