Hi there. I have to find the image under the transformation $\displaystyle w=z^2$ for the region given in the pic.

So, I tried to analyze what would happend. I know that:

$\displaystyle w=z^2=|z|^2 e^{i2\theta}=|z|^2(\cos 2\theta+i\sin 2\theta)$

And I've tried to express the region under consideration, I did it this way, I'm not pretty sure if this is right.

$\displaystyle 1 \geq Re(z) \geq 0$

$\displaystyle 1 \geq Im(z) \geq 1-Re(z)$

$\displaystyle \sqrt{2} \geq |z| \geq 1/2$

$\displaystyle \pi/2 \geq arg(z) \geq 0$

Then I should get

$\displaystyle 2 \geq |w| \geq 1/4$

$\displaystyle \pi \geq arg(w) \geq 0$

But doesn't give enough information to get the image under the transformation. So then I've tried to get the image for the lines.

$\displaystyle y=1\rightarrow{v=1},\alpha \in{[\pi/2,\pi]}$

$\displaystyle u=1\rightarrow{u=1},\alpha \in{[0,\pi/2]}$

Which gives something like semi infinite lines in w, right? but that would be wrong having in mind the limitations for the modulus in w.

And I don't know how to trasnform the other line, y=1-x.