What is the image of the line $Re(z) = Im(z)$ under the transformation $f(z) = z^2$?

Question

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.

Answer 1

If $Re(z)=Im(z),$ then $x=y$, so $u(x,y) = 0$ and $v(x,y)=2x^2$,

so the image is the non-negative imaginary axis.

Answer 2

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.

Answer 3

From the given $Re(z) = Im(z)$, we have $i(z+\bar z) = z - \bar z$, which leads to

$$z = \frac{1+i}{1-i}\bar z=i\bar z$$

Then, the image can be expressed as,

$$f(z) = z^2 = iz\bar z = i|z|^2$$

which represents the positive $y$-axis.