Solve
z^5 =-1
with the solutions in polar form.
I've solved the equation z^5 =1, which I have
Letz=r(cosθ+isinθ). Also,1=cos0+isin0. Sotheequationis (r(cosθ+isinθ))5 =cos0+isin0. how would i go about it with the negative 1 though?
e^iT = cos T + i sin T =
for z^5 = -1
z^5 = 1cos pi + i sin pi
that pi angle gives you real pat of -1 and imaginary part of 0
|z|= r = 1^(1/5) = 1
T = pi
pi + 2pi/5
pi + 4pi/5
pi + 6 pi/5
pi + 8 pi/5
subtract 2 pi if they are over to get angles
the first one is at 180 deg
the second is at 180 + 72 degrees etc