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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..

CoffeeArabica
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  • This morning I thought, what might happend if we would be allowed to also plugg elements of the negative half-plane into our expression. (so when b doesnt have to be greater 0) and it doenst really change the angles we can reach. $b^2$ will go positive doesnt matter if $b$ is negative, and only the "little" $2b$ is really effected. But at first glance it doesnt seem to effect the angles we are able to reach.. also I still couldnt think of an angle our expression isnt able to reach, even with $b>0$.. – CoffeeArabica Oct 29 '19 at 05:56
  • Also in all my madness I started to wonder: does the an element of the positive real axis has the angle $0$ or $2\pi$? or both? If both, there couldnt be ever an $\phi_2$.. – CoffeeArabica Oct 29 '19 at 05:59
  • It's a very good idea to plot this sort of problems. If you do, you will quickly see that the region should be the open unit disk ${|z|<0}$. Then you know what to try to prove. – Milten Oct 29 '19 at 06:28
  • By the way (and sorry for sounding like a teacher), but in the equations just before "Since $a$ can be zero...", all the $\iff$'s should just be equal signs. – Milten Oct 29 '19 at 06:42

1 Answers1

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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.

Milten
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  • Feel free to ask, if it is unclear. – Milten Oct 29 '19 at 06:37
  • Well, I figured the same this far tbh. But I have to show that there is either $\phi_2$ or there cannot be an open interval for the angles $\Longrightarrow [\phi_1,\phi_2]$ – CoffeeArabica Oct 31 '19 at 10:51
  • And when pluggin in some interesting values into the function, I felt like hitting values on the negative and positive unit disk.. – CoffeeArabica Oct 31 '19 at 10:52
  • What could be a method to find a certain $\phi_x$ which is always greater than any $\phi$ of the function might approach for the values of the upper halfplane elements of $C$? If there would be any $\phi_x$ which fits the properties I stated, I could conclude, that there is some $\phi_2$ which will never be hit (like with the magnitudes $l$) – CoffeeArabica Oct 31 '19 at 10:58
  • The part with $l$ was not so hard I guess.. I just had to show that there is any $l$, and the expression Ive got by showing the multiplicative from the magnitudes is obviously always smaller than 1.. Sorry for all the unordered comments, and the late answer :) I appreciate your answers ;) – CoffeeArabica Oct 31 '19 at 11:06
  • If we know the set is the unit disc, then that determines both $l$ and $\phi_1, \phi_2$. You need to choose angles that ensures that you can hit all of the unit disc, because that is the image we found. – Milten Oct 31 '19 at 11:12
  • The answer is simpler than I think you think. – Milten Oct 31 '19 at 11:13