How to show that $\sqrt{1+ni}=\sqrt[4]{n^2+1}\cos\left(\frac{1}{2}\tan^{-1}(n)\right)+i\sqrt[4]{n^2+1}\sin\left(\frac{1}{2}\tan^{-1}(n)\right)$

Question

Wolfram Alpha gives me that $$\begin{align} \sqrt{1+i}&=\sqrt[4]{2}\cos\left(\frac{1}{2}\tan^{-1}(1)\right)+i\sqrt[4]{2}\sin\left(\frac{1}{2}\tan^{-1}(1)\right)\\ &=\sqrt[4]{2}e^{\frac{1}{2}i\tan^{-1}(1)}\\ \sqrt{1+2i}&=\sqrt[4]{5}\cos\left(\frac{1}{2}\tan^{-1}(2)\right)+i\sqrt[4]{5}\sin\left(\frac{1}{2}\tan^{-1}(2)\right)\\ &=\sqrt[4]{5}e^{\frac{1}{2}i\tan^{-1}(2)}\\ \sqrt{1+3i}&=\sqrt[4]{10}\cos\left(\frac{1}{2}\tan^{-1}(3)\right)+i\sqrt[4]{10}\sin\left(\frac{1}{2}\tan^{-1}(3)\right)\\ &=\sqrt[4]{10}e^{\frac{1}{2}i\tan^{-1}(3)} \end{align}$$ In general, how can we show that $$\begin{align} \sqrt{1+ni}&=\sqrt[4]{n^2+1}\cos\left(\frac{1}{2}\tan^{-1}(n)\right)+i\sqrt[4]{n^2+1}\sin\left(\frac{1}{2}\tan^{-1}(n)\right)\\ &=\sqrt[4]{n^2+1}e^{\frac{1}{2}i\tan^{-1}(n)} \end{align}$$

I tried to draw a complex plane, but it did not help significantly. Perhaps I forgot some fundamental concepts. Any hints?

Answer 1

From the polar form $$z=\sqrt{n^2+1}\,e^{i\arctan n}$$ you draw

$$\sqrt z=\pm\sqrt[4]{n^2+1}\,e^{i\arctan n\,/2}.$$

Then back to Cartesian coordinates,

$$\pm\sqrt[4]{n^2+1}\left(\cos\frac{\arctan n}2+i\sin\frac{\arctan n}2\right).$$

Answer 2

$$\begin{align} z&=1+ni,~r=\sqrt{n^2+1}\\ 1+ni &=\sqrt{n^2+1}\cos(\tan^{-1}(n))+i\sqrt{n^2+1}\sin(\tan^{-1}(n))\\ &= \sqrt{n^2+1}e^{i\tan^{-1}(n)}\\ \sqrt{1+ni}&=\sqrt[4]{n^2+1}e^{\frac{1}{2}\tan^{-1}(n)}\\ &=\sqrt[4]{n^2+1}\cos\left(\frac{1}{2}\tan^{-1}(n)\right)+i\sqrt[4]{n^2+1}\sin\left(\frac{1}{2}\tan^{-1}(n)\right) \end{align}$$