Real part of a complex number

Question

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?

Answer 1

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}

Answer 2

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

Answer 3

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

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

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

Answer 4

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

Answer 5

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 $\cos t=0$

Answer 6

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.