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!
The triangle consists of three segments: $\{-1 \leftrightarrow i, i \leftrightarrow -1$,$-1 \leftrightarrow 1\}$, call them $\{a,b,c\}$. Segment $a$ maps to a parabola in the upper half plane with $1\mapsto 1$, $(1+i)/2\mapsto i/2$ (the vertex of the parabola), $i\mapsto -1$. Segment $b$ maps to a parabola in the lower half plane. The vertex is $(-1+i)/2\mapsto -i/2$. Segment $c$ maps to the positive real axis between $0$ and $1$. The interior of the triangle in the $z$-plane maps to the interior of the transformed region in the $w$-plane (minus the segment from 0 to 1, the image of $c$).
Here is the triangle and its interior in the $z$-plane:
Here is the image of the triangle and its interior in the $w$-plane:
It is relatively straightforward to calculate the equations of the parabolas. One of the other answers has a nice start.
For example: $w=f(z) = f(x+iy) = x^2-y^2 +2 x y i$. Substitute $y=1-x$ (segment a) and we obtain $\xi=\Re\{w\}$ and $\eta=\Im\{w\}$. We can then find the relation between $\xi$ and $\eta$ (the equation of the parabola in the $\xi,\eta$-plane. Similarly for the image of segment $b$. Each half of segment $c$ (from -1 to 0 and from 0 to 1) trivially maps into the segment between 0 and 1 in the $w$-plane.
One interesting point to notice: A mapping such as this (by a holomorphic function) is conformal. Angles are preserved. Notice that the angle of each of the vertices of the triangle maps into an equal angle in the $w$-plane.
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.
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)