Sketching a complex function $f(\Omega) = \{ f(z) : z \in \Omega\}$

Question

I've been working on this question for some time now, and have not had any significant progress towards a solution:

Let $\Omega = \{z = x +iy \in \mathbb{C} : |y| \leq 1\}.$ If $f(z) = z^2 +2$, then draw a sketch of $$f(\Omega) = \{ f(z) : z \in \Omega\}.$$ Justify your answer.

Naturally I first determined that the set $\Omega$ essentially refers to the strip of complex numbers in the plane which have their imaginary part between $i$ and $-i$, so I then set about attempting to plot the particualar set. However, I'm not sure how to go about drawing this — I tried 'mapping' random points on the curve by drawing arrows from $z$ to $f(z)$, and found some success:

• The purely imaginary numbers 'map' to the region $[1, 2)$ on the real line.
• The purely real numbers 'map' to the region $[0, \infty)$ on the real line.
• The complex numbers of the form $x+i$ trace out the path of the curve $y=2\sqrt{x+1}$, as would be expected.

After this, though, I'm having a really hard time trying to find any sort of pattern between these observations, and I'm nowhere near trying to find the drawing of the curve as I'd like to be. The closest I've come is this crude sketch on Desmos.

                       

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Just as an add-on, this question is expected to be solved on a normal sheet of paper. It appeared in this test last year which was expected to be solved by high-schoolers.

Answer 1

The boundary,$|y|=1$, maps to the parabola $\displaystyle x= \left(\frac{y}{2}\right)^2+1$. The rest of the image $|y|<1$ lies to the right of the parabola with $\displaystyle x > \left(\frac{y}{2}\right)^2+1$.

Answer 2

Continuing my comment (as I cannot add images there). Dividing the function in two functions that map to the reals, which in this case is: $$f(z) = f(x+iy)=(x+iy)^2+2 = \underbrace{x^2-y^2+2}_{v(x,y)} + i\underbrace{2xy}_{u(x,y)}$$ yields the following plot (I used MATLAB here). The z-axes are the real part and imaginary part of $f$, which is maybe hard to see as the font in Matlab sucks sometimes :P

I think it is not too hard to sketch $v$ and $u$ by hand, it is just much easier using a computer :P. I hope this helps you out!