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Question
Find all complex numbers z which satisfy:
(2-2i)z^4=(sqrt(3)-i)^5

Thank you so much for your help!

Answer
This is most easily solved by first expressing the complex number in polar form, which you should be familiar with if you have been introduced to complex numbers already. That is, for z an arbitrary complex number

z = a + ib = Zexp(iθ)

where

Z = (a^2 +b^2)^1/2 = modulus

θ = tan^-1(b/a) = angle in the complex plane

and of course  i = sqrt(-1).

So first, rewrite your equation

q・z^4 = p^5

as

z =  { (p^5)/q }^1/4

where q  and p are the complex numbers

q = 2-2i = sqrt(8)exp(-iπ/4)

p = 2exp(-iπ/6).

so that

z = { [2exp(-iπ/6)^5]/sqrt(8)exp(-iπ/4) }^1/4

Going thru the algebra I get

z = { [16/sqrt(2)]exp(-i7π/12) }^1/4

We're almost there except for the fact that the fourth root of complex z actually has 4 roots corresponding to distinct angles, as follows:

let θ = 7π/12 for now and recall that [exp(θ)]^1/4 = exp(θ/4)

so that we want to find the 4 roots αj = θ/4, such that (αj)^4 = θ for j = 1,2,3,4 (j is a subscript here). To this end, let

α1 = θ/4  so that α1^4 = θ   (for this problem, α1 = (7π/12)/4 = 7π/48)
α2 = θ/4 + π/2      --> α2^4 = θ + 2π
α3 = θ/4 +π          --> α3^4 = θ + 4π
α4 = θ/4 + 3π/2  --> α4^4 = θ + 6π.

Note that we have just added π/2 increments to θ, where π/2 is just a quarter of the way around the unit circle, giving us the 4 roots since (with n = integer)

exp(i(θ+n2π)) = exp(iθ)・exp(in2π) = exp(iθ)     because exp(in2π) = 1.

So now you can calculate the 4th root of the modulus, i.e, [16/sqrt(2)]^1/4 = 2^7/8 and combine this with the 4 angles above to get your 4 solutions for z.

I hope I didn't make any mistakes in the algebra, but at any rate, this approach is how I would solve it.

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