Show that the mapping $w (z) = z^2 $ takes the upper-half complex plane $\text{Im} (z) \ge 0$ into the whole plane.

Question

I am supposed to Show that the mapping $w (z) = z^2 $ takes the upper-half complex plane $\text{Im} (z) \ge 0$ into the whole plane. $w = u+iv$ and $z=x+iy$

I tried doing it as follows:

$$w = u+iv = (x+iy)(x+iy) = x^2-y^2 + i2yx $$

This implies that $$u = x^2-y^2 \space\space\space\space\space \text{and} \space\space\space\space\space v=2yx$$

The condition $\text{Im}(z) \geq 0$ would mean that $v \geq0$ and that $x^2 \geq u$. Since x can take any value from the condition we see that u can be any positive value.

This would mean that we have transferred the initial complex number into the upper right quadrant but not the entire quadrant. I probably did something wrong but am very confused.

Any help would be appreciated.

Answer 1

You can find a parametric equation of the upper half plane, namely $$\text{upper-half plane}= \{ \alpha +i\beta:-\infty\lt\alpha\lt+\infty ,0\leq\beta\lt+\infty\}$$ Then the image of the upper half plane under $f(z)={z^2}$ is ${(\alpha+i\beta)^2}={\alpha^2}-{\beta^2}+i\cdot2\alpha\beta $. Since $-\infty\lt\alpha\lt+\infty$ and $0\leq\beta\lt+\infty$ , the negative sign in the real part guarantees: $${-\infty\lt\alpha^2}-{\beta^2\lt+\infty}$$ and because $\alpha$ can be negative: $$-\infty\lt2\alpha\beta\lt+\infty$$ Since both ${\alpha^2}-{\beta^2}$ and $2\alpha\beta$ are real numbers, positive or negative, ${\alpha^2}-{\beta^2}+i\cdot2\alpha\beta$ is a parametric equation for the whole complex plane. I am not sure of the rigour of this approach, I just know that parametric equations work in most of the cases. Any feedback is appreciated!

Answer 2

The best way to see that is to use polar coordinates, $z=re^{iu}$, where $\in[0,2\pi]$ take $z'=\sqrt re^{iu/2}$, remark that ${z'}^2=z$ and since $u/2\in [0,\pi], z'=\sqrt r(\cos(u/2)+i\sin(u/2))$ and $\sin(u/2)\geq 0$.

Answer 3

It is well-known that each complex number $w = u +iv$ has a root. This can be explicitly computed. If $w \ne 0$, then there are exactly two distinct roots $z_1, z_2$ which differ by a factor $-1$. Thus at least one of the roots has $\text{Im}(z_i) \ge 0$. This shows that the set all all $z^2$ with $\text{Im}(z_i) \ge 0$ gives all $w \in \mathbb C$.