Let's say we have a following function:
$$f(z)=\frac{1}{2}(z+\frac{1}{z})$$ Find the image of a $B(\sqrt{2}i,1)$, when mapped by given function.
My results:
$f(z) = \frac{1}{2}(z + \frac{1}{z}) = \frac{1}{2} \frac{(z|z|^{2}+\overline z)}{|z|^{2}} = \frac{1}{2} (z + \frac{\overline z}{|z|^{2}}) $
So I prepared simple, when $B(0,1)$ I computed the result $[-1,1]$
Because we have: $z \in B(0,1) $ is represented by $z = e^{i\alpha}$
so using the equation $$w = \frac{1}{2}(e^{i\alpha} + e^{- i\alpha})$$ it is $\cos(\alpha)$ so the value of $\cos$ is in $[-1,1]$
How to prove that the image of $B(\sqrt{2}i,1)$ is like lower?
edit:
parametric plot for $2 (x^{2} + y^{2})^{2} == 2 y^{2} + x^{2}$
edit2: $$u+vi = f(z) $$ $(u,v)$ in ${\mathbb C} - [-1,1]$ I prepered $$z=re^{i\alpha}$$ $r < 1$ such that $${\displaystyle \bigg({u \over (r + {1 \over r})}\bigg)^2 + \bigg({v \over (r - {1 \over r})}\bigg)^2 = {1 \over 4}}$$ so it's an elipse.
Why my image of $B(\sqrt2i,1)$ looks like the half eclipse?
I show that the foci are the points $(-1,0) , (1,0)$
$B(0,1)$ means $z=e^{i\phi}$. Changing the center to $\sqrt 2 i$ means you just need to add this value to $z$, so $$z=e^{i\phi}+\sqrt 2 i$$ Then $$\begin{align}f(z)&=\frac 12(e^{i\phi}+\sqrt 2 i+\frac 1{e^{i\phi}+\sqrt 2 i})\\ &=\frac 12(e^{i\phi}+\sqrt 2 i+\frac {e^{-i\phi}-\sqrt 2 i}{(e^{i\phi}+\sqrt 2 i)({e^{-i\phi}-\sqrt 2 i})})\\&=\frac 12(e^{i\phi}+\sqrt 2 i+\frac {e^{-i\phi}-\sqrt 2 i}{3+2\sqrt2\sin\phi})\\&=\frac{\cos\phi}{2}(1+\frac 1{3+2\sqrt2\sin\phi})+\frac{\sin\phi+\sqrt 2}{2}(1-\frac 1{3+2\sqrt 2\sin\phi})i\end{align}$$