Consider the map
$$f(z):=\frac{1}{2}(z+\frac{1}{z})$$
and show that $f(B)$ is a hyperbola where $B=\{re^{i \theta}: \text{$r>1$ is variable and $\theta$ fixed}\}$
So I know I have to substitute in for $z$ thus I have
$$f(re^{i\theta})=\frac{1}{2}(re^{i \theta} + \frac{1}{re^{i \theta}})$$
So do I combine the terms inside the parenthesis and set it equal to say $X+iY$ then set real parts equal and imaginary parts equal?
I.e., $X=\frac{1}{2}(r+\frac{1}{r})\cos \theta, Y = \frac{1}{2}(r-\frac{1}{r})\sin \theta$ then do I use $\cos^2 \theta + \sin^2 \theta=1$?
Sorry if this is a duplicate, I couldn't find one like it. Also, at which points is the map angle preserving? is it all the $z$ such that
$$f'(z) \neq 0$$
also, how do I show $X^2-Y^2=1$?
I keep getting
$$X^2-Y^2=\frac{r^2}{4}(\cos^2 \theta - \sin^2 \theta)+ \frac{1}{4r^2}(\cos^2 \theta - \sin^2 \theta)+ \frac{1}{2}.$$
any help??