1
$\begingroup$

Show $Re(\frac{1}{z+1}) $ 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 $$z=\frac{|z|^2}{\bar z}$$ $$z=\frac{1^2}{\bar z}$$ 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?

$\endgroup$
4
$\begingroup$

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}

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – juhani Nov 12 at 15:39
  • $\begingroup$ 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 $\frac{\operatorname{Re}(z)+1}{\lvert z+1\rvert^2}$. $\endgroup$ – José Carlos Santos Nov 12 at 15:45
5
$\begingroup$

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

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ interesting but maybe a bit too "complex" ;-) $\endgroup$ – lucia de finetti Nov 12 at 14:39
  • 1
    $\begingroup$ @luciadefinetti Nothing complicated about it. Just geometric. And perhaps a bit more obscure than the algebra of the other answers. $\endgroup$ – Arthur Nov 12 at 14:47
  • 1
    $\begingroup$ 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 $\endgroup$ – lucia de finetti Nov 12 at 14:57
  • 1
    $\begingroup$ 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 $\endgroup$ – lucia de finetti Nov 12 at 15:13
  • 1
    $\begingroup$ @Arthur In the name of posterity, I totally agree - and it's a beautiful answer. $\endgroup$ – peter a g Nov 12 at 15:19
2
$\begingroup$

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

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

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

$\endgroup$
1
$\begingroup$

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 $\cos t=0$

$\endgroup$
  • $\begingroup$ (and $\cos t = 0 \Leftrightarrow z=-1$) $\endgroup$ – aschepler Nov 12 at 22:47
0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ A decent start, but you didn't use the "when $|z|=1$", which will simplify the answer quite a bit. $\endgroup$ – aschepler Nov 12 at 22:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.