<s>[INST] Find the image of the parabola $x = \frac{9}{4} − \frac{y^2}{9}$ under the principal square root mapping $w = z^{1/2}$ with $z \in \mathbb{C}$. Find the image of  the parabola $x = \frac{9}{4} − \frac{y^2}{9}$ under the principal square root
mapping $w = z^{1/2}$ with $z \in \mathbb{C}$.
Definition of principal square root mapping: $z^{1/2} = \sqrt{|z|}e^{i\operatorname{Arg}(z)/2}$.
Arg is the principal argument of $z$, so $-\pi<\operatorname{Arg}(z) \le \pi$
I've tried plugging in the expression $\frac{9}{4} − \frac{y^2}{9}+yi$ into the principal square root mapping, but can't seem to get anywhere.
 [/INST] Suppose $w=(u,v)$ is in the image of the parabola under the map principal branch of square root of $z$.
Then $\exists z=(x, y) $ on the parabola such that $√z=w$
Then $z=w^2$ implies
$x+iy=u^2-v^2+2iuv$
Hence $x=u^2-v^2 , y=2uv$
$(x, y) $ lies on the parabola $x=\frac{9}{4}-\frac{y^2}{9}$ implies
$u^2-v^2=\frac{9}{4}-\frac{4u^2v^2}{9}$
On simplification, we get
$36u^2-36v^2+16u^2v^2=81$
Hence image of the parabola is the set $\{(u,v):36u^2-36v^2+16u^2v^2=81\}$

$$\text{ Parabola} \space  x=\frac{9}{4}-\frac{y^2}{9}$$
After solving, we get a pair of straight lines $u=\pm\frac{3}{2}$

$$\text{ Pair of straight lines}  \space u=\pm\frac{3}{2}$$
For any point $w$ in  the image, we have $-\frac{\pi}{2}<\operatorname{Arg} (w) \le\frac{\pi}{2}$.
Hence image of the parabola under the principal branch of $z^{\frac{1}{2}} $ is the line $u=\frac{3}{2}$

Image credit: GeoGebra
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