Find all the complex numbers $z$ such that $\left(\frac{z-1}{1+i}\right)^{4} = -1+i$

Question

The title corresponds to the statement of the exercise. So far what I did was:

$(\frac{z-1}{1+i})^{4} = -1+i$

$\frac{(z-1)^4}{(1+i)^4} = -1+i$

$(z-1)^{4} = (-1+i)(1+i) = -2$

$(z-1) = (-2)^{1/4}$

it's okay?

How do I continue it?

Answer 1

No, it's not okay. From $\left(\frac{z-1}{1+i}\right)^4=1-i$, the conclusion is that$$(z-1)^4=(1+i)^4(1-i)=-4+4i.$$So, compute the $4$ fourth roots of $-4+4i$ and add $1$ to each of them.

Answer 2

Being w = z +1. I used the formula w^n = r^n [cos (an) + i sin(an)];

a = a'/n + k*2π/n, with 0 ≤ k < n and r = r'^(1/n)

then w_k = r (cos(a) + i sin (a)

In this case r' = 4*(2^(1/2)) then r = (4*(2^(1/2)))^(1/4) and a' = 2π - arctan(4/4) = 7*π/4

w_0 = r ( cos (7*π/16) + i sin (7*π/16) )

w_1 = r ( cos (15*π/16) + i sin (15*π/16) )

w_2 = r ( cos (23*π/16) + i sin (23*π/16) )

w_3 = r (cos (31*π/16) + i sin(31*π/16) )

And as you said the 4 fourth roots are: z_k = w_k + 1, 0 ≤ k < n

So I got the roots but is there any geometric way to get them or to be more intuitive? Because what I did was through the formula. I would like to understand the concept geometrically.