1 Attachment(s)

Image for the region under the transformation w=z^2

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

http://www.mathhelpforum.com/math-he...1&d=1321552043

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.

1 Attachment(s)

Re: Image for the region under the transformation w=z^2

I would do this a little bit differently.

First let $\displaystyle u=\text{Re}(w), v=\text{Im}(x)$

Then $\displaystyle z^2=(x+iy)^2=x^2-y^2+2xyi$

This gives that $\displaystyle u=x^2-y^2, v=2xy$

Now you can just parameterize each of the line segments.

The vertical line segment from (1,0) to (1,1) is given by

$\displaystyle x(t)=1, y(t)=t, t \in [0,1]$

Under the transformation this gives

$\displaystyle u=1-t^2, v=2t$ so the start and end points are (1,0) and (0,2) now if you elminate the paramter you get the curve

$\displaystyle u=1-\left(\frac{v}{2} \right) \iff u=1-\frac{v^2}{4}$ So this is a parabolic arc.

Now for the horizontal line segemtn from (1,1) to (0,1) we have

$\displaystyle x=1-t, y=1 $ under the transformation this gives

$\displaystyle u=(1-t)^2-1^2=t^2-2t, v= 2(1-t)1=-2t+2$

If you eliminate the parameter you get

$\displaystyle u=\frac{1}{4}v^2-1$

we get another parabolic region with start point (0,1) and end point (-1,0)

Finally the line can be paramterized by

$\displaystyle x=t, y=1-t$ under the transformation this gives

$\displaystyle u=t^2-(1-t)^2=-1+2t, v=2t(1-t)=-2t^2+2t$ If we eliminate the parameter we get

$\displaystyle v=-\frac{1}{2}(u^2-1)$ another parabola with start point (-1,0) and end point (1,0)

Here is a plot of the transformation. The first curve is in green, the 2nd in red, and the last in blue.

Re: Image for the region under the transformation w=z^2

Nice. Thank you verymuch :)