Mapping a triangle under $f(z)=z^2$?

Question

If i am given a filled-in triangle with vertices $1$, $-1$, and $i$, how can I find the image of it under $f(z)=z^2$?

I know $w=f(x+iy)=x^2-y^2+2xyi$ so $u=x^2-y^2$ and $v=2xy$, and I know I have to find the image of the three sides but I am not sure how to go about doing that. Any help is greatly appreciated!

Answer 1

For $z=x+iy\ $ write $z=|z|e^{it}, $ where $|z|=\sqrt{x^{2}+y^{2}}\ $ and $\tan t=y/x.\ $Then, $z\mapsto |z|^2e^{2it},\ $ so $|z|\mapsto |z|^2$ and $t\mapsto 2t,\ $ which means that each point in the complex plane doubles its angle, and squares its modulus.

Therefore, the transformed triangle will be the region with vertices $(1,0),(0,i/2),(-1,0),(0,-i/2).$

To find the boundary, consider each line segment joining the vertices of the triangle. For example, the line segment joining $(1,0)$ to $(0,i)$ has equation $y=1-x.\ $ Therefore for $0\le t\le \pi/4,\ $ we have $|z^2|=2x^2-2x+1,\ $ and $0\le 2t\le \pi/2,\ $ which means that the segment from $(1,0)$ to $(1/2,i/2)$ is transformed into a parabolic arc from $(1,0)$ to $(0,i/2).$

Now, consider $\pi/4\le t\le \pi/2$ and repeat the argument, and similarly for the other edges of the triangle.

Answer 2

To get the mapping of the lines, you first to describe the lines by using parameters.

Theline between $-1$ and $1$ ist desribed by $\lambda \in [-1, 1]$ and can easily be mapped onto the positive real numbers upto $1$ After that, you need to transform $\lambda \cdot 1 + (1-\lambda)\cdot i$ for $\lambda \in [0,1]$. This will tell you, where to map the right diagonal line.

You can do similar things with the left diagonal line, to see, where the triangle is mapped to. (Keep in mind to determine, if you actually have the inner or outer part of your figre)