2

Compute $\operatorname{Re}\left(\frac{1}{z+1}\right)$ when when $|z| = 1$.

The only way I could think to go about this is to simply go by definitions. If $z\in \Bbb C$, then $z\bar z$ = $|z|^2$. Now $$z=\frac{|z|^2}{\bar z}$$and$$z=\frac{1^2}{\bar z}.$$ So $z$ must be the inverse of the conjugate of $z$ which can be written as $$z=\frac{z}{\bar z z}$$ I don't know how to proceed from here. Is this even a good approach?

José Carlos Santos
  • 427,504
In the blind
  • 522

6 Answers6

5

Note that$$\frac1{z+1}=\frac{\overline{z+1}}{\left(z+1\right)\left(\overline{z+1}\right)}=\frac{\overline z+1}{\lvert z+1\rvert^2}$$and therefore\begin{align}\operatorname{Re}\left(\frac1{z+1}\right)&=\frac{\operatorname{Re}(z+1)}{\lvert z+1\rvert^2}\\&=\frac{\operatorname{Re}(z)+1}{\lvert z\rvert^2+2\operatorname{Re}(z)+1}\\&=\frac{\operatorname{Re}(z)+1}{2\operatorname{Re}(z)+2}\text{ (since $\lvert z\rvert=1$).}\\&=\frac12.\end{align}

José Carlos Santos
  • 427,504
  • Would I be correct if I interpreted this as ' the real part of LHS must equal the real part of RHS of the equation - and it is because of this, that we can omit looking at the imaginary part of z in RHS? – In the blind Nov 12 '19 at 15:39
  • I suppose that your question is about how I deduced that$$\frac1{z+1}=\frac{\overline z+1}{\lvert z+1\rvert^2}\implies\operatorname{Re}\left(\frac1{z+1}\right)=\frac{\operatorname{Re}(z)+1}{\lvert z+1\rvert^2}.$$My though process was this: since $\frac{\overline z+1}{\lvert z+1\rvert^2}=\frac{\operatorname{Re}(z)+1}{\lvert z+1\rvert^2}+\frac{\operatorname{Im}(z)}{\lvert z+1\rvert^2}i$, the real part of $\frac{\overline z+1}{\lvert z+1\rvert^2}$ is $\frac{\operatorname{Re}(z)+1}{\lvert z+1\rvert^2}$. – José Carlos Santos Nov 12 '19 at 15:45
5

$$|z|=1\implies z=e^{i\phi}\implies \frac1{1+z}=\frac1{1+e^{i\phi}}= \frac{1+e^{-i\phi}}{2+2\cos\phi}=\frac{1+\cos\phi-i\sin\phi}{2(1+\cos\phi)}\\ \implies \operatorname{Re}\frac1{1+z}=\frac12.$$

user
  • 26,272
3

Geometrically, the map $z\mapsto \frac1{z+1}$ moves the complex plane $1$ to the right, then inverts the plane with respect to the unit circle, and then takes the complex conjugate. The circle of allowable $z$-values becomes the vertical line with real part $\frac12$ under this transformation.

Arthur
  • 199,419
  • 2
    interesting but maybe a bit too "complex" ;-) – lucia de finetti Nov 12 '19 at 14:39
  • 1
    @luciadefinetti Nothing complicated about it. Just geometric. And perhaps a bit more obscure than the algebra of the other answers. – Arthur Nov 12 '19 at 14:47
  • 1
    what i meant - silly "play on words" aside - is that, even though what you state is obviously correct and answers the question, it is quite possibly not answering it in a way that will resort to being useful to the OP (who was clearly after an algebraic solution) and that, in my opinion, should be the main aim of an answer – lucia de finetti Nov 12 '19 at 14:57
  • @luciadefinetti I don't think it's that clear that they are specifically after an algebraic solution. They tried to solve it algebraically. That's a different thing. It's basically a knee-jerk reaction (not just for the OP, but for me and many others, judging by the other answers) to solve a problem like this with algebra, so the fact that that's what the OP tried means nothing. Also, answers here aren't just for the OPs asking questions. They are here for posterity's sake as well. – Arthur Nov 12 '19 at 15:02
  • 1
    agree to disagree. i think it's more than fair to assume that someone who's unable to solve this task algebraically won't be helped by an answer like yours (that surely proves your grasp on geometry though!) but hey, if you're solving $\Re{\left(\frac{1}{z+1}\right)}$ for posterity's sake, that's a whole different story :-P – lucia de finetti Nov 12 '19 at 15:13
  • 1
    @Arthur In the name of posterity, I totally agree - and it's a beautiful answer. – peter a g Nov 12 '19 at 15:19
  • 1
    @peterag thank $\pi$ someone worried about OP as well, i guess! – lucia de finetti Nov 12 '19 at 15:21
  • 1
    @luciadefinetti There were enough standard answers here that I didn't really feel the need to worry about the OP this time. – Arthur Nov 12 '19 at 15:51
2

We have that $\Re(w)=\frac12\left(w+\bar w\right)$ then

$$\Re\left(\frac{1}{z+1}\right)=\frac12\left[\left(\frac{1}{z+1}\right)+\left(\frac{1}{\bar z+1}\right)\right]=$$

$$=\frac12\frac{z+\bar z+2}{(z+1)(\bar z+1)}=\frac12\frac{2\Re (z)+2}{|z+1|^2}=\frac{\Re (z)+1}{|z+1|^2}=\frac{\cos \theta+1}{2+2\cos \theta}=\frac12$$

user
  • 154,566
1

WLOG $z=\cos2t+i\sin2t$

$\dfrac1{z+1}=\dfrac1{2\cos t(\cos t+i\sin t)}=\dfrac{\cos t-i\sin t}{2\cos t}$

So, the real part will be $$\dfrac12$$ unless $\cos t=0$

lab bhattacharjee
  • 274,582
0

Let $z = a+bi$ $$\begin{align}Re\left(\frac{1}{z+1}\right) &= Re\left(\frac{1}{a+1+bi}\right) = Re\left(\frac{1}{a+1+bi}\cdot\frac{a+1-bi}{a+1-bi}\right) = \\ &= Re\left(\frac{a+1-bi}{(a+1)^2-(bi)^2}\right) = Re\left(\frac{a+1-bi}{(a+1)^2+b^2}\right) = \\ &= Re\left(\frac{a+1}{(a+1)^2+b^2} - \frac{b}{(a+1)^2+b^2}i\right) = \frac{a+1}{(a+1)^2+b^2} = \\ &= \frac{Re(z)+1}{(Re(z)+1)^2+Im(z)^2}.\end{align}$$

Daniel P
  • 2,710