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 $w_0=\sqrt{\frac{1+\sqrt{2}}{2}}+i\sqrt{\frac{-1+\sqrt{2}}{2}}\not = 1,84+0,76i$
 A: Setting $$\sqrt{1+i}=A+Bi$$ then $$1+i=A^2-B^2+2ABi$$ and we have to solve
$$A^2-B^2=1$$
$$2AB=1$$
A: Well, you could cheat and just do
$$
w_0^2=\frac{1+\sqrt{2}}{2}-\frac{-1+\sqrt{2}}{2}+2i\sqrt{\frac{1+\sqrt{2}}{2}\cdot\frac{-1+\sqrt{2}}{2}}=1+i
$$
and note that $w_1=-w_0$.
Without cheating, you did quite well: $|1+i|=\sqrt{2}$, so
$$
1+i=\sqrt{2}\left(\frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}}\right)=
\sqrt{2}\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right)
$$
Hence we have
$$
w_0=\sqrt[4]{2}\left(\cos\frac{\pi}{8}+i\sin\frac{\pi}{8}\right)
$$
and you can also compute $w_1$ (but it must be $w_1=-w_0$).
It remains to compute $\cos(\pi/8)$ and $\sin(\pi/8)$:
$$
\cos\frac{\pi}{8}=\sqrt{\frac{1+\cos(\pi/4)}{2}}=
\sqrt{\frac{\sqrt{2}+1}{2\sqrt{2}}}
$$
and
$$
\sqrt[4]{2}\sqrt{\frac{\sqrt{2}+1}{2\sqrt{2}}}=
\sqrt{\sqrt{2}\frac{\sqrt{2}+1}{2\sqrt{2}}}=
\sqrt{\frac{1+\sqrt{2}}{2}}
$$
You can do similarly, with
$$
\sin\frac{\pi}{8}=\sqrt{\frac{1-\cos(\pi/4)}{2}}
$$
A: Just to expand on Dr. Sonnhard Graubner's answer, square-rooting the sum of the squares of these equations gives $A^2+B^2=\sqrt{2}$. (The right-hand side won't be $-\sqrt{2}$, for obvious reasons.) So $A^2=\frac{\sqrt{2}+1}{2},\,B^2=\frac{\sqrt{2}-1}{2}$. Although square-rooting these equations introduces $\pm$ signs, they're not independent; $AB>0$ so $A,\,B$ have the same sign. This gives the two desired results.
