Let $a\in \Bbb C$ with $|a|>1$. Find the image of the unit disk under the analytic function $f(z)=1/(z+a)$

Question

Problem 1.4.3: Let $a \in\Bbb C$ with $|a|>1$. Find the image of the unit disk under the analytic function $f(z)=\frac{1}{z+a}$.

My work so far: By the definition of the unit disk \begin{align} D = \{z \in \Bbb C\,∶\,|z|<1\}&\qquad&(i) \end{align}

Suppose $w=f(z)=\frac{1}{z+a}$ \begin{align} w(z+a)&=1&\\ wz+wa&=1&\\ wz&=1-wa&\\ z&=\frac{1-wa}{w}&&(ii)\qquad \text{Put this in }(i)\\ \left\lvert \frac{1-wa}{w}\right\rvert&<1&\\ |1-wa|&<|w|&\\ \end{align} Where $w$: $w$-plane in complex function.

After that, I need help. Thanks

Answer 1

Sketch: Now you can write $w=x+yi,\ a=\alpha+\beta i$ with $x,y,\alpha,\beta\in\Bbb R$ and get $$\lvert 1-(x+yi)(\alpha+\beta i)\rvert^2 <\lvert x+yi\rvert^2\\\lvert 1-\alpha x+\beta y-(\alpha y+\beta x)i\rvert^2<\lvert x+yi\rvert^2$$

Which, after the calcs, turns out to be an inequality in the form $P_{\alpha,\beta}(x,y)<0$, for a suitable real polynomial $P_{\alpha,\beta}(x,y)$ of degree $2$. Then, finding the locus of this inequality in terms of $\alpha$ and $\beta$ is an exercise of elementary analytical geometry of $\Bbb R^2$.

If I were you, I'd observe that $$P_{\alpha,\beta}(x,y)=(\alpha^2+\beta^2-1)x^2+(\alpha^2+\beta^2-1)y^2+r_{\alpha,\beta}(x,y)$$ where $\deg r_{\alpha,\beta}\le1$.

Answer 2

HINT:

$f(z)=f_2(f_1(z)),$ where $f_1(z)=z+a$ and $f_2(z)=\dfrac1z.$

Observations:
1. $f_1$ translates the unit disk to a disk centered at $a$ with radius $1.$
2. Image of the circle $C$ under $f_2$ is either circle or a line according to $0\not\in C$ or $0\in C.$