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\[-i = e ^{-i \frac{\pi}{2}}\]

\[ z^5 = e ^{-i \frac{\pi}{2} +2\pi ki}\]

then factor out the i\[z^5 = e ^{i \pi [ 2k-\frac{1}{2} ] } \]

take 5 root both sides\[ z= e ^{\frac{i \pi}{5} [2k -\frac{1}{2} ]}\]

then sub in k=-2,-1,0,1,2
to get the five solutions

\[e ^{\frac{-i \pi}{2} }\]

prob cant even even see that , ahh you can do the rest

just subing in numbers

thanks man, EE for life

how did you know to use k=-2,-1,0,1,2?

so if you had a power of 5 you would do as I did

if you had power of 3 you use k= -1,0,1

oh ok, thanks :D

you have to keep going until you find n different solutions ( for the general eqn z^n = x+iy )

also, you should go through and make sure that the arguments of the solutions are in the range -pi

is \[-i=e^\pi/2\] an identity or something?

-i=e^(pi/2)

-i = e^(-pi/2) , probably cant see it well above.