Find the images of $f(z) = z^2$ whose domain is ${z: Re(z) > 0}$ I need to find the image of the function $f(z) = z^2$ whose domain is ${z: Re(z) > 0}$. I first let $z = x + iy$. Then $$w = f(z) = z^2 = (x+iy)^2 = (x^2-y^2) + 2xyi$$ Hence $u(x,y) = x^2-y^2$, and $v(x,y) = 2xy$ Then I first consider the boundary, which is when $Re(z) = 0$. When $Re(z) = 0$, $x = 0$. Then $z = yi$. Then I plug this back into my original function, and get $$f(z) = z^2 = (yi)^2 = -y^2$$, but I'm confused about what to do next, since I need to consider when $Re(z) > 0$. Thanks!
 A: If $\lambda$ is a real number smaller than or equal to $0$, then $\lambda$ does not belong to the image. If $\lambda=x^2-y^2+2xyi$, then, since $\lambda\in\mathbb{R}$, $2xy=0$. But $x\ne0$, and therefore $y=0$. So, $\lambda=x^2>0$.
And this is the only restriction: if $z\in\mathbb{C}$ and $z$ is not a real number smaller than or equal to $0$, then you can write $z$ as $\rho\bigl(\cos(\theta)+\sin(\theta)i\bigr)$, for some $\theta\in(-\pi,\pi)$. But then
$$
z=\left(\sqrt\rho\left(\cos\left(\frac\theta2\right)+\sin\left(\frac\theta2\right)i\right)\right)^2=f\left(\sqrt\rho\left(\cos\left(\frac\theta2\right)+\sin\left(\frac\theta2\right)i\right)\right)
$$
and $\sqrt\rho\cos\left(\frac\theta2\right)>0$, since $\frac\theta2\in\left(-\frac\pi2,\frac\pi2\right)$.
A: If $z=r(\cos(\theta)+i\sin(\theta)$) with $Re(z)>0$, then $r>0$ and $\theta\in (-\pi/2,\pi/2)$. Therefore $f(z)=r^2(\cos(2\theta)+i\sin(2\theta)$) with $2\theta\in (-\pi,\pi)$. As $r$ can vary from $0$ to $\infty$ not including $0$, and $\theta\in (-\pi,\pi)$ is basically all the possible angles without the edges, you get that $Im(f)=\{re^{i\theta}\in\mathbb{C}:r>0,-\pi<\theta<\pi\}=\mathbb{C}\setminus \mathbb{R}_{\leq 0}$.
An intuitive way to think about it is the following - $z^2$ doubles the angle and the radius of the complex number. As the domain doesn't include the $y$ axis and everything left to it, doubling the angle maps the first quadrant (without the $y$ axis part) onto the second, and the fourth (without the $y$ axis part) onto the third, fixing $\mathbb{R}_{\geq 0}$ (only "stretching" it). Since the $y$ axis is mapped exactly to $\mathbb{R}_{\leq 0}$, this is the part the image "misses".
