Finding $f(B(0, 1))$ when $f(z) = \frac1 2(z + \frac 1 z)$

Solution 1:

One possible approach is to use polar coordinates. As shown in Maps circles onto ellipses, every circle of radius $0 < r < 1$ is mapped to an ellipse with width and height $$ a = \frac 12 \left( r + \frac 1 r\right) \, , \, b = \frac 12 \left( \frac 1 r - r\right) \, , $$ centered at the origin. With $r$ varying from $1$ to $0$, the width takes all values in $(1, \infty)$, and the height takes all values in $(0, \infty)$.

This shows that the image of the unit disk is $\hat{\Bbb C} \setminus [-1, 1]$, with $z=0$ being mapped to the point at infinity.

Another approach is to write $$ f(z) = \frac 12 \left( z + \frac 1 z\right) = \frac{\left(\frac{z+1}{z-1}\right)^2 +1}{\left(\frac{z+1}{z-1}\right)^2 -1} $$ as the composition of fractional linear mappings (aka Möbius transformations) and a square.

The first transformation $z \to \frac{z+1}{z-1}$ maps the unit disk onto the left halfplane. This is mapped to the “slit plane” $\Bbb C \setminus (-\infty, 0]$ by squaring, and the last transformation maps this to $\hat{\Bbb C} \setminus [-1, 1]$.