Question

If $|z|=1$ and $w = \frac{z}{z+1}$ where $z \ne -1$, then Re(w) is

• A. 0
• B. $-\frac{1}{2+|z+1|^2}$
• C. $|\frac{z}{z+1}| \frac{1}{|z+1|^2}$
• D. $\frac{\sqrt{2}}{|z+1|^2}$

Answer 1

$\frac{1}{2}$

Answer 2

Let $z=e^{i\theta}$ (since $|z|=1$). Then

$$z+1 = e^{i\theta}+1 = 2\cos\frac{\theta}{2}\,e^{i\theta/2}.$$

Thus,

$$w=\frac{z}{z+1}=\frac{e^{i\theta}}{2\cos\frac{\theta}{2}\,e^{i\theta/2}}=\frac{e^{i\theta/2}}{2\cos\frac{\theta}{2}}.$$

Taking the real part,

$$\Re(w)=\frac{\cos\frac{\theta}{2}}{2\cos\frac{\theta}{2}}=\frac{1}{2}.$$

None of the provided options match this result.