Question

If $|z|=1$, then ${\left(\frac{1+z}{1+\overline{z}}\right)}^n+{\left(\frac{1+\overline{z}}{1+z}\right)}^n$ is equal to

• A. $2\cos n\left(\arg \right(z\left)\right)$
• B. $2\sin \left(\arg \right(z\left)\right)$
• C. $2\cos n\left(\arg \left(\frac{z}{2}\right)\right)$
• D. $2\sin n\left(\arg \left(\frac{z}{2}\right)\right)$

Answer 1

The correct option is A

$2\cos n\left(\arg \right(z\left)\right)$

If $|z|=1,{\left(\frac{1+z}{1+\overline{z}}\right)}^n+{\left(\frac{1+\overline{z}}{1+z}\right)}^n={\left(\frac{z(1+z)}{z+z\overline{z}}\right)}^n+{\left(\frac{z+z\overline{z}}{z(1+z)}\right)}^n={\left(\frac{z(1+z)}{z+|z{|}^2}\right)}^n+{\left(\frac{z+|z{|}^2}{z(1+2)}\right)}^n={\left(\frac{z(1+z)}{z+1}\right)}^n+{\left(\frac{(z+1)}{z(1+z)}\right)}^n=z^n+\frac{1}{z^n}z=cis\left(\arg z\right)\Rightarrow z^n=cis\left(n\arg z\right)\frac{1}{z}=cis(-\arg z)\Rightarrow \frac{1}{z^n}=cis(-n\arg z)z^n+\frac{1}{z^n}=2\cos n\left(\arg z\right)$