I have this complex equation: $$\left(\frac{z-1}{1+i}\right)^{4} = -1+i$$
I solved this problem in this answer using the approach I duplicate below. Is there any geometric way to get the roots? Or is there a more intuitive way to get the roots? Because what I did was through the formula. I would like to understand the concept geometrically.
Here's my work from the previous answer:
Working on that equation I get to: $$(z-1)^4=(1+i)^4(1-i)=-4+4i.$$
To get the four roots of this equation I do the following:
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 $r = r'(\frac 1 n)$
then $w_k = r (\cos(a) + i \sin (a)$
In this case $r' = 4\sqrt 2$ then $r = (4\sqrt 2)^{1/4}$ and $a' = 2π - \arctan(4/4) = 7*π/4$
$w_0 = r ( \cos (7*π/16) + i \sin (7*π/16) )$
$w_1 = r ( \cos (15*π/16) + i \sin (15*π/16) )$
$w_2 = r ( \cos (23*π/16) + i \sin (23*π/16) )$
$w_3 = r (\cos (31*π/16) + i \sin(31*π/16) )$
Finally the $4$ fourth roots are: $z_k = w_k + 1, 0 ≤ k < n$
Complex roots have a very nice geometric intepretation, the complex $n$ roots of unity indeed form a regular polygon with $n$ sides and each vertex lies on the unit circle.
Then in your case we are looking for $w^4=-4+4i=4\sqrt 2e^{i\frac34\pi}$ then the roots you have found form the vertices of a square which lie on a circle with radius $(4\sqrt 2)^\frac14$.