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 $z \in \mathbb{C}$.

Question

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.

@AnotherUser the textbook I am using says that that the answer is x=3/2, but I can't see how this works. Can you help me see this? — , Jul 05 '22 at 10:31
Start with $w=u+iv$, compute $w^2 = x+iy$ and substitute that into the equation of f the parabola — Martin R, Jul 05 '22 at 10:55

Sourav Ghosh · Accepted Answer · 2022-07-05T12:34:39.320

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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