Let $\Omega = \{z=x+iy\in \mathbb{C}:|y|\leq1 \}.$
If $f(z)=z^2+2,$ then draw a sketch of
$f(\Omega) = \{f(z) : z \in \Omega \}.$ Justify your answer.
My Approach:
Substituted $z = x+iy$ in $f(z)=z^2+2.$
After simplifying, $f(z) = (x^2-y^2+2) + i(2xy)$
Problem: How do I find the locus of this complex number?
I took $h=x^2-y^2+2$ and $k = 2xy.$
I can't simplify further to get the locus in $h$ and $k.$
The boundary of $f(\Omega)$ is the transformed of the boundary of $\Omega$: inserting $y=\pm1$ in the formulas for $h$ and $k$ we then find $$ k=\pm2x,\quad h=x^2+1={1\over4}k^2+1. $$ Hence the boundary of $f(\Omega)$ is the parabola of equation $x={1\over4}y^2+1$.