Find all numbers $z \in \mathbb{C}$ such that $(z−i)^5 = \sqrt{3} +i$ This is a follow-up question to finding all solutions for $z \in \mathbb{C}$ such that $z^5 = \sqrt{3} +i$ but I have no idea how to approach this question (might just be having a brain fart)
The only way I can think of solving it would be to expand it into a polynomial and then solve for z but that seems like a lot more work than necessary, and I would think this answer would have something to do with my answer to the previous question.
 A: If $w^5 = x$ then $z = w+i$ satisfies $(z-i)^5 = x$, and conversely. In other words, add $i$ to your solutions to the previous problem. 
A: .If you can make $\sqrt{3}+i$ into polar form, then it will be easy for you to remove the exponent.
The way to convert it is very simple: $\sqrt{\sqrt{3}^2+1^2} e^{i(\tan^{-1}(3^{-0.5}))} = 2 e^{\frac{i\pi}{6}}$.
So if we write $(z-i)^5 = 2e^{\frac{i\pi}{6}}$, then taking fifth root on both sides and then adding $i$, we get:
$z = i + \sqrt[5]{2}e^{\frac{(12n+1)i\pi}{30}}$, $n=0,1,2,3,4$. These are the solutions, you can now expand out the $z$ and get the roots more precisely.
A: Generalize the problem, solve $\text{z}$ when $\text{a}\in\mathbb{R}^+$, $\text{b}\in\mathbb{C}$ and $\text{n}\in\mathbb{N}^+$:
$$\left(\text{z}-\text{a}i\right)^{\text{n}}=\text{b}\Longleftrightarrow\text{z}=\left|\text{b}\right|^{\frac{1}{\text{n}}}e^{\frac{1}{\text{n}}\left(\arg\left(\text{b}\right)+2\pi k\right)i}+\text{a}i$$
Where $k\in\mathbb{Z}$ and $k:0-(\text{n}-1)$.
A: Hint:
Write $\sqrt 3+i$ in exponential form., and you'll find out it's a problem of finding the $5$th roots of a complex number with modulus $1$.
