Problem: sketch the curve $w=\sqrt z$ in principal branch, and THE other branch in another color:
$z=x+i$; that is, $y=1$
Work done so far:
I let $w=u(x,y)+iv(x,y)$, then
$$f(z)=\sqrt z= u+vi$$
$u$ and $v$ of $x$ and $y$ dropped from hereon due to redundancy.
Then, $$x+i=u^2-v^2+2uvi$$ Then, $$x=u^2-v^2 \land 1=2uv$$
I don't know how to go ahead. I want to eliminate $x$, so that I can get $w=g(u,v)$; that is, get $w$ as a function of $u$ and $v$ alone. If some one can give me an algebraic solution to this I will really appreciate it. Plus, i want to know how can I restrict my domain in $w$ to get only the values for the principle square root; that is, +\sqrt (z).
$(u^{2}-v^{2})^{2}=x^{2}$ and $4 u^{2}v^{2}=1$. Add these two to get $(u^{2}+v^{2})^{2}=x^{2}+1$. Hence $u^{2}+v^{2}=\sqrt {1+x^{2}}$. Combined with the equation $u^{2}-v^{2}=x$ we get $u^{2}=\frac {x+\sqrt {1+x^{2}}} 2$ and $v^{2}=\frac {\sqrt {1+x^{2}} -x} 2$. Can you take it from here?