Little sidenote: Im new to function theory, so I'm inexperienced and don't have many tools regarding this topic.
$\mathbb{C^{*}} =$ { $z \in \mathbb{C} | z \neq 0$ } with the function $ f(z) = z^2 $.
Determine the equations regarding the image curves of the lines $ x=a $ and $ y=b $ with $a,b \neq 0 $ and $ z = x+iy$.
Show that these image curves are perpendicular to each other.
My thoughts:
So we consider $ t \rightarrow (a + it)^2 $ and $ t \rightarrow (t +ib)^2 $
So we get $(a + it)^2 = a^2 - t^2 + i2at =: c_1 $ and $(t +ib)^2 = t^2 - b^2 + i2tb =: c_2$ .
So I thought knowing the fact that $f$ is holomorphic could help. I mean this allows us to use the Cauchy-Riemann equations. The real part of $c_1$ is $a^2 -t^2$ and imaginary part $2at$. For $c_2$ we have real part $t^2-b^2$ and imaginary part $ 2tb$. Maybe we could with their partial derivates to show they are perpendicular on each other? Thinking about scalar product here.
The curves $v(t)=a^2-t^2+2ati$ and $h(u)=u^2-b^2+2bui$ meet when $t=b$ and $u=a$. Besides, $v'(b)=-2b+2ai$ and $h'(a)=2a+2bi$. And\begin{align}\langle v'(b),h'(a)\rangle&=\operatorname{Re}\left((-2b+2ai)\times\overline{(2a+2bi)}\right)\\&=\operatorname{Re}\left((-2b+2ai)\times(2a-2bi)\right)\\&=0.\end{align}So, yes, $v'(b)$ and $h'(a)$ are orthogonal.
Note: If $z=a+bi$ and $w=c+di$, then\begin{align}\langle(a,b),(c,d)\rangle&=ac+bd\\&=\operatorname{Re}\left((a+bi)\times\overline{(c+di)}\right)\\&=\operatorname{Re}\left(z\times\overline w\right).\end{align}