Image under billinear transformation

Question

What is a image of $x+y>4$ under billinear transformation $B(z)=\frac{z-4-8i}{z-4}$?

I got that $B(z)=1-\frac{8\sqrt{2}e^{i\frac{\pi}{4}}}{z+4}$, but I cannot conclude image correctly (it should be $|w|<1$).

Any help is welcome. Thanks in advance.

Answer 1

We can do this step by step. Put $w=B(z)=\frac{z-4-8i}{z-4}$.

1. $z_1= z-4$, $w=\frac{z_1-8i}{z_1}=1-\frac{8i}{z_1}$;
2. $z_2=\frac{1}{z_1}$, $w=1-8iz_2$;
3. $z_3=iz_2$, $w=1-8z_3$;
4. $z_4=-8z_3$, $w=1+z_4$;
5. $w=1+z_4$.

Our set is $S=\{x+iy\mid x+y>4\}$ in the $z$-plane.

1. After applying the first mapping, which is just translation, we get $S_1=\{x+iy\mid x+y>0\}$ in $z_1$-plane.
2. Let us apply the second mapping. $\frac{1}{x+iy}=\frac{x}{x^2+y^2}-\frac{y}{x^2+y^2}i$. If we put $a=\frac{x}{x^2+y^2}$ and $b=-\frac{y}{x^2+y^2}$ we get $a^2+b^2=\frac{1}{x^2+y^2}$, so $x=\frac{a}{a^2+b^2}$ and $y=-\frac{b}{a^2+b^2}$, so $x+y>0$ transforms into $\frac{a}{a^2+b^2}-\frac{b}{a^2+b^2}>0$, i.e. into $a-b>0$. Thus $S_2=\{x+yi\mid x-y>0\}$.
3. The third mapping is just $90^\circ$ rotation, so $S_3=\{x+yi\mid x+y>0\}$ in $z_3$-plane.
4. The fourth mapping is homotethy with coefficient $-8$, so $S_4=\{x+yi\mid x+y<0\}$ in $z_4$-plane.
5. Finally, the fifth mapping is again a translation, so $B(S)=\{x+yi\mid x+y<1\}$ in $w$-plane.

Answer 2

Let $x+y=4$ and then $y=4-x$. Thus \begin{eqnarray} w&=&\frac{z-4-8i}{z-4}=\frac{x-4-(4+x)i}{x-4+(4-x)i}\\ &=&\frac{x-4-(4+x)i}{(x-4)(1-i)} =2 \end{eqnarray} Namely $B(z)$ transforms the line to $x+y=4$ the line $w=2$. Note that $z=4+i$ is in the area of $x+y>4$ and $$ \frac{z-4-8i}{z-4}=-7-4i $$ which is in the LHS of $w=2$. Therefore, the image of $x+y>4$ under $B(z)$ is the LHS of $w=2$.

Answer 3

Let us compute the image of the boundary of the domain, $z(t)=t+i(4-t)$. We have

$$B(z(t))=1-\frac{8i}{z-4}=1-\frac{8i}{(t-4)(1-i)}=1+4\frac{1-i}{t-4},$$ which describes a straight line of slope $-1$, through $(1,0)$.

The domain itself is the half-plane above the line $x+y=4$. When you translate it by $-4$, the bonding line is through the origin ($x+y=0$). Then by inversion, you get the half-plane below that line. Next, multiplying by $-8i$, you get the half-plane above $x=y$, which you finally translate horizontally by $1$.