Show that the exact values of the square roots of $z=1+i$ are... Show that the exact values of the square roots of $z=1+i$ are...
$w_0=\sqrt{\frac{1+\sqrt{2}}{2}}+i\sqrt{\frac{-1+\sqrt{2}}{2}}$ 
$w_1=-\sqrt{\frac{1+\sqrt{2}}{2}}-\sqrt{\frac{-1+\sqrt{2}}{2}}$
My attempt 
Let $z=1+i\in \mathbb{C}$. Then
$r=|z|=\sqrt{2}$
Moreover,
$\theta=\tan^{-1}(1)=\frac{\pi}{4}$
Then, the polar form of $z$ is
$z=\sqrt{2}(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4})$
Let $w\in\mathbb{C}$ such that $w^2=z$
Then
$w_k=\sqrt{\sqrt{2}}(\cos(\frac{\frac{\pi}{4}+2k\pi}{2})+i\sin(\frac{\frac{\pi}{4}+2k\pi}{2})$ for $k=0,1$.
Then the square roots, are:
$w_0=\sqrt[4]2(\cos\frac{\pi}{8}+i\sin\frac{\pi}{8})=1,84+0,76i.$
$w_1=\sqrt[4]2(\cos\frac{9\pi}{8}+i\sin\frac{9\pi}{8})=-1,84-0,76i.$
But, here i'm stuck because the $w_0=\sqrt{\frac{1+\sqrt{2}}{2}}+i\sqrt{\frac{-1+\sqrt{2}}{2}}\not = 1,84+0,76i$

Setting $$\sqrt{1+i}=A+Bi$$ then $$1+i=A^2-B^2+2ABi$$ and we have to solve
$$A^2-B^2=1$$
$$2AB=1$$
