<s>[INST] Transformation of $|z-1|=1$ and $\mathrm{Re}(z)^2 = \mathrm{Im }(z)^2-1$ under $f(z) = \sqrt{z}$ I have two regions in the complex plane, defined by
$|z-1|=1$ and $(\,\mathrm{Im}(z))^2 = (\,\mathrm{Re}(z))^2-1$, $\mathrm{Re}(z)>0$. 
I am being asked to find and sketch the image of those regions  under the mapping $f(z) = \sqrt{z}$ . 
My attempt
We have $f(z) = \sqrt{r}e^{i\frac{\theta}{2}}$ and $|z-1|=1 \iff r = 2\cos(\theta)$
If $|z-1|=1 $, then:
 $$f(z) =  \sqrt{2|\cos(\theta)|}\cos\left(\frac{\theta}{2}\right) + 
i\sqrt{2|\cos(\theta)|}\sin\left(\frac{\theta}{2}\right)$$
It means that $f(z)$ lies on the trace of the curve $\left(\sqrt{2|\cos(\theta)|}\cos\left(\frac{\theta}{2}\right); 
\sqrt{2|\cos(\theta)|}\sin\left(\frac{\theta}{2}\right) \right)$ if $|z-1| = 1.$
Am I right so far?
I couldn't recognize this plane curve.
And I don't know how to proceed in the region  $(\,\mathrm{Im}(z))^2 = (\,\mathrm{Re}(z))^2-1$.
How can I do it?
 [/INST] Possible suggestion:
Let $z = x + yi$. We will emulate this problem in the Cartesian plane.
By transforming $Re(z)^2 = Im(z)^2 - 1$ under $f$, we are taking the root of each point. Conversely, $z^2 = (x^2 - y^2) + (2xy)i$ satisfies $Re(z^2)^2 = Im(z^2)^2 - 1$ instead for every point $z$ on the transformed curve.
This is the Cartesian graph $x^4 - 2x^2y^2 + y^4 = 4x^2y^2 - 1$ 
I.E. $x^4 - 6x^2y^2 + y^4 =  - 1$
The first problem can be done similarly. The transformed curve is $|z^2 - 1| = 1$
This translates to the Cartesian plane as $4x^2y^2 + x^4 + y^4 + 1 - 2x^2y^2 + 2y^2 - 2x^2 = 1$
I.E. $0 = 2x^2y^2 + x^4 + y^4 + 2y^2 - 2x^2$
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