Let $f(z) = \sqrt z$. Using the branch that is defined everywhere except where $z = x + iy$ with $y = 0$ and $x < 0$. What are the formulas for real valued functions $u(x, y)$ and $v(x, y)$ such that $f(x + iy) = u(x, y) + iv(x, y)$?
So far I know that I'm gonna have to convert $z$ into exponential form such that $z=r^{1/2}\exp(i\pi/2)$ but I am confused where to proceed from there.
P.s. Where can I get some formatting help for this site? I feel bad that my posts are so ugly.
The goal is to transform $$(x,y)=(r\cos t,r\sin t),\qquad r\gt0,\qquad |t|\lt\pi,$$ into $$(u,v)=(\sqrt{r}\cos(t/2),\sqrt{r}\sin(t/2)).$$ Recall that $$\cos t=2\cos^2(t/2)-1,\quad\sin t=2\sin(t/2)\cos(t/2),$$ and that $\cos(t/2)\gt0$ when $|t|\lt\pi$ hence $$\cos(t/2)=\sqrt{\frac{1+\cos t}2},\qquad\sin(t/2)=\frac{\sin t}{2\cos(t/2)}=\frac{\sin t}{2\sqrt{\frac{1+\cos t}2}}.$$ Adding to the pot the obvious identity $r=\sqrt{x^2+y^2}$, all this yields $$u=\sqrt{r}\sqrt{\frac{1+\cos t}2}=\frac1{\sqrt2}\sqrt{r+r\cos t}=\frac1{\sqrt2}\sqrt{\sqrt{x^2+y^2}+x},$$ for every $(x,y)$, and $$v=\sqrt{r}\frac{\sin t}{2\sqrt{\frac{1+\cos t}2}}=\frac{r\sin t}{\sqrt2\sqrt{r+r\cos t}}=\frac{y}{\sqrt2\sqrt{\sqrt{x^2+y^2}+x}},$$ for every $(x,y)$ except if $x\leqslant0$ and $y=0$ (if $(x,y)=(0,0)$, use $v=0$).