Question

If $z_k=e^{i\theta k}$ for $k=1,2,3,4$ where $i^2=-1$ and if $\left|\stackrel{4}{\sum _{k=1}}\frac{1}{z_k}\right|=1,$ then $\left|\stackrel{4}{\sum _{k=1}}{z}_k\right|$ is equal to :

• A. $4$
• B. $1$
• C. $2$
• D. $3$
• E. $\frac{1}{4}$

Answer 1

Answer: (B)

$1$

Given, $z_k=e^{i\theta k}$ for $k=1,2,3,4$
Also, $\left|{\sum }_{k=1}^4\frac{1}{z_k}\right|=1$
$|\frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3}+\frac{1}{z_4}|=1$
$|\frac{1}{e^{i{\theta }_1}}+\frac{1}{e^{i{\theta }_2}}+\frac{1}{e^{i{\theta }_3}}+\frac{1}{e^{i{\theta }_4}}|=1$
$|e^{-i{\theta }_1}+e^{-i{\theta }_2}+e^{-i{\theta }_3}+e^{-i{\theta }_4}|=1$
Thus $(\cos {\theta }_1-i\sin {\theta }_1)+(\cos {\theta }_2-i\sin {\theta }_2)+(\cos {\theta }_3-i\sin {\theta }_3)+(\cos {\theta }_4-i\sin {\theta }_4)=1$
$=|(\cos {\theta }_1+\cos {\theta }_2+\cos {\theta }_3+\cos {\theta }_4)+i\left(\sin {\theta }_1\right)+\left(\sin {\theta }_2\right)+\left(\sin {\theta }_3\right)+\left(\sin {\theta }_4\right)|$
$=|1+0|=1$