complex-analysisfunctionstransformation
What is the image of the line $\Re(z)$ = $\Im(z)$ under the transformation $f(z)=z^2$?
My attempt so far: $z=x+iy$; thus if $f(z)=z^2$, then we have $(x+iy)^2= x^2+2ixy-y^2$.
Is this correct: $f(z)=u(x,y)+iv(x,y)$ where $u(x,y) = x^2-y^2$ and $v(x,y)=2xy$
But if $\Re(z)=\Im(z)$ then is $u(x,y) = v(x,y)$ i.e $x^2-y^2=2xy$
I'm not sure if I am doing this correct and if so I'm not sure about how I should continue. Any help would be great; thanks.
You have nearly gone all the steps. As you have obtained, $$\begin{array}{l}u = {x^2} - 1\\v = 2x\end{array}$$Now, $x$ is a dummy variable which we could eliminate in these two relations to obtain $u = \frac{{{v^2}}}{4} - 1$. Note that $u$ plays the same role as $x$ in the mapped plane and $v$ has the role of $y$, so that it might be more clear to write $$v = \pm 2\sqrt {u + 1} $$ 
Every non-singular Möbius transformation maps circles and lines to circles and lines. Equivalently, circles on the Riemann sphere are maps to other such circles. That does not mean that circles are mapped to circles or lines are mapped to lines.
For $c\ne 1$, the line where $-\infty < v < \infty$ is mapped under $$ c+iv \mapsto \frac{c+1+iv}{c-1+iv} $$ to a circle in the complex plane. It cannot be mapped to a line because the point at $\infty$ is not in the image. The image is symmetric about the real axis, $c+i0$ is mapped to $\frac{c+1}{c-1}$, and $c\pm i\infty$ is mapped to $1$. So the center of the circle should be $\frac{1}{2}(1+\frac{c+1}{c-1})=\frac{c}{c-1}$. To verify, \begin{align} \left|\frac{c+1+iv}{c-1+iv}-\frac{c}{c-1}\right|&=\left|\frac{c^2-1+iv(c-1)-c(c-1)-icv}{(c-1)(c-1+iv)}\right| \\ &=\left|\frac{c-1-iv}{(c-1)(c-1+iv)}\right| \\ &=\left|\frac{1}{c-1}\right|. \end{align} So the radius of the circle is given on the right, and you can see the above does define a circle in the complex plane.
Best Answer
You're meant to square numbers of that form, not find which squared numbers are of that form. (The squared complex numbers are just the complex numbers.) The line you start with comprises complex numbers of the form $z:=x+ix=x(1+i)$ with $x\in\Bbb R$. Squaring these gives the image, the set of values of $w:=z^2=2x^2i$ for such $x$. This is the $\Re w=0,\,\Im w\ge0$ half-line on the imaginary axis.