Find image of complex set:

Question

Find image of set: $$ \{ z \in C : 0 \le Im (z), 0 \le Re(z) \}$$ and $$f(z)=\frac{i-z}{i+z}$$

I caclulate $ w=\frac{i-z}{i+z} $ and then $z=\frac{i(1-w)}{w+1}$ and don't know what to do next... I will be grateful for any help

Answer 1

In your rearranged form for $z$, write $w=u+iv$. Multiply top and bottom of the fraction by the conjugate of the denominator so that you can identify the real and imaginary parts of $z$ in terms of $u$ and $v$. Now you can apply the inequality conditions given in the question, giving you inequalities for $u$ and $v$. You will end up with a region in the complex plane.

Answer 2

$f(ri) = {1 +r \over 1-r}$, hence $\{f(ri)\}_{r \ge 0} = (-1,1]$.

$f(r) = {(1-r^2)+2r i \over 1+r^2}$, from which we see that $|f(r)| = 1$, and $\{f(ri)\}_{r \ge 0} = \{z | |z|=1, \operatorname{im}z >0 \} \cup \{1\}$.

$f(\infty) = -1$.

$f(1+i) = {-1+2i \over 5}$, so we can see that the image of the set (including $f(\infty)$) is $\{ z \in B(0,1) | \operatorname{im}z \ge 0 \}$.