I have to do the following:
Show that the following set:
$D=\lbrace\frac{z-i}{z+i}|z \in C, \Im(z)>0\rbrace$
has a $l \in R^+_0$ and $\phi_1,\phi_2\in[0,2\pi]$ with $\phi_1\le \phi_2$ that $D$ can be described as:
$D=\lbrace r(\cos(\phi)+i\sin(\phi))|r \in [0,l), \phi\in[\phi_1,\phi_2)\rbrace$
I started with the following:
$z=(a+bi), a,b\in R$
$D=\lbrace k=\frac{(a+i(b-1))}{(a+i(b+1))}|k \in C, b>0, a,b\in R\rbrace$
To finde the cap of the magnitudes of the given set of complex numbers, I went with the following:
$0\le|k|=\sqrt{\frac{(a^2+(b-1)^2)}{(a^2+(b+1)^2)}}<1,$ for all $b>0, a\in R$
so I concluded that $l$ might be $1$, because all magnitudes have to lie in
$[0,1) $ also for $b=1 $ and $a=0$ we get $|k| = 0$, so $0$ is as stated an element of all magnitudes of given complex functions. So there has to be some $l\in R^+_0:0<l\le1$ which the magnitudes of the complex functions converge to.
To the upper statement I came with imagining two complex numbers which are getting multiplied..
$z_1= a+i(b-1)\Longrightarrow |z_1|=\sqrt{(a^2+(b-1)^2)}$
$z_2=a+i(b+1)\Longrightarrow |z_2|=\sqrt{(a^2+(b+1)^2)}$
Now I multiplied the $|z_1|$ with $\frac{1}{|z_2|}$
Now the second part of the exercise was to find $\phi_1,\phi_2\in[0,2\pi]$ with $\phi_1\le \phi_2$, so that all $\phi$ will be either $\phi_1$ or greater, and smaller than some $\phi_2$ for sure.
I can rewrite $\frac{z-i}{z+i}$ into $\frac{(a+i(b-1))}{(a+i(b+1))}$
$= \frac{(a+i(b-1))(a-i(b+1))}{a^2+b^2+2b+1} = \frac{a^2-ai(b+1)+ai(b-1)+b^2-1}{a^2+b^2+2b+1} = \frac{a^2+b^2-1}{a^2+b^2+2b+1}-i(\frac{2a}{a^2+b^2+2b+1})$
Since $a$ can be zero and $b$ might be greater than one, so that our complex number lies on the positive real axis, $\phi_1$ should equal $0$, because its the smallest possible angle in $[0,2\pi]$ and it should represent the our smallest angle in $[\phi_1,\phi_2)$
Here I stuck 🙁 When I let $a$ approach zero and $b$ approach infinity (so that the real part stays positive and the imaginary part starts converging to zero from the negative side I. get the feeling there can be the angle $2pi$ … what went wrong, could I get some tipps..
Best Answer
HINT:
I claim that that $D=\{|w|<1\}$, i.e. the open unit disk, which I figured out from graphing the region. You already proved that $D\subset \{|w|<1\}$, so what remains is to prove that the function $f(z)=\frac{z-i}{z+i}$ (on the given domain) is surjective on $\{|w|<1\}$.
To do that, you can find the inverse on $f$, and then check that for any $|w|<1$, we have that $f^{-1}(w)$ exists and $\Im f^{-1}(w)>0$.
Or you can prove that $\frac{|z-i|}{|z+i|}<1 \iff \Im (z)>0$, just by messing around with the inequality.