Simplify $(1+i^\frac{1}{2})^\frac{1}{2}$ How can I simplify $\sqrt{1+\sqrt{i}}$?

I thought about making $z^2=1+\sqrt{i}$ and then $w=z^2$

But I'm not really sure
$$(1+i^\frac{1}{2})^\frac{1}{2}=(1+(e^{i\frac\pi2+i2\pi n})^\frac12 )^\frac12
=(1+e^{i\frac\pi4+i\pi n} )^\frac12$$

$$=\left(1+\cos(\frac\pi4 +\pi n)+ i \sin(\frac\pi4 +\pi n) \right)^\frac12
=(re^{i\theta})^\frac12\tag 1$$

where,
$$r=\sqrt{\left(1+\cos(\frac\pi4 +\pi n)\right)^2+\sin^2(\frac\pi4 +\pi n) }
=\sqrt{2+2\cos(\frac\pi4 +\pi n) }=2| \cos(\frac\pi8 +\frac{\pi n}2) |$$

$$ \tan\theta = \frac{\sin(\frac\pi4 +\pi n)}{1+\cos(\frac\pi4 +\pi n)} 
= \tan (\frac\pi8 +\frac{\pi n}2)\implies \theta =\frac\pi8 +\frac{\pi n}2+k\pi $$

Substitute $r$ and $\theta$ into (1),

$$(1+i^\frac{1}{2})^\frac{1}{2}
=r^\frac12 e^{i\frac{\theta}2} = \sqrt{2| \cos(\frac\pi8 +\frac{\pi n}2) |}
\&gt;e^{i\pi(\frac1{16} +\frac{ n}4+\frac{k}2)}$$

which assumes multiple values. For the special case $n=k=0$, 

$$(1+i^\frac{1}{2})^\frac{1}{2}
= \sqrt{2\cos\frac\pi8}\&gt;e^{\frac{i\pi}{16}}
=\sqrt{2\cos\frac\pi8}\&gt;(\cos\frac{\pi}{16} + i\sin\frac{\pi}{16})$$
Using polar coordinates $\displaystyle \sqrt{1+\sqrt{i}} = \sqrt{1+\sqrt{e^{i \pi/2}}} = \sqrt{1 + e^{i \pi/4}} = \sqrt{1 + \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2}} \, $ from this we have

$\displaystyle r^2 = \left( 1 + \frac{\sqrt{2}}{2} \right)^2 + \left( \frac{\sqrt{2}}{2} \right)^2 = 1 + \sqrt{2} + \frac{1}{2} + \frac{1}{2} = 2 + \sqrt{2} \implies r = \sqrt{2 + \sqrt{2}} $

and

$\displaystyle \tan \theta  = \frac{\frac{\sqrt{2}}{2}}{1 + \frac{\sqrt{2}}{2}} = \frac{1}{\sqrt{2} + 1} = \sqrt{2} - 1 \implies \theta = \arctan(\sqrt{2} - 1)$

So, $\displaystyle \sqrt{1 + \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2}} = \sqrt{re^{i\theta}} = \sqrt{r} \, e^{i\theta/2}$

Finally, $\displaystyle \sqrt{1+\sqrt{i}} = \sqrt[4]{2 + \sqrt{2}}  \, \left( \cos\left( \frac{\arctan(\sqrt{2} - 1)}{2} \right) + i\sin\left( \frac{\arctan(\sqrt{2} - 1)}{2} \right) \right)$  

We can also obtain cartesian form:

$\displaystyle a + ib = \sqrt{1+\sqrt{i}}  = \sqrt{1 + \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2}}$ yields

$\displaystyle a^2 + b^2 +i(-2ab) = 1 + \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2} \implies a^2 + b^2 = 1 + \frac{\sqrt{2}}{2}$ and $\displaystyle -2ab = \frac{\sqrt{2}}{2}$. So we have

$\displaystyle (a-b)^2 = 1 + \sqrt{2} \,$ and $\displaystyle (a+b)^2 = 1 $ which gives $\displaystyle a = \frac{1 + \sqrt{1 + \sqrt{2}}}{2} $ and $\displaystyle  b = \frac{-1 + \sqrt{1 + \sqrt{2}}}{2}$