<s>[INST] 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
 [/INST] 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 $\displaystyle  b = \frac{-1 + \sqrt{1 + \sqrt{2}}}{2}$
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