I have to remember to convert (-2 + 2i) in to polar form.
I have to remember to convert (-2 + 2i) in to polar form.
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
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Attn: (-2 + 2i ) is in the second quadrant, and cis(x) = cos(x) + i sin(x)