Question

If $z_1,z_2,...,z_n$ lie on $|z|=r$ and $Re\left(\sum _{j=1}^n\sum _{k=1}^n\frac{z_j}{z_k}\right)=0$, then

• A. $\sum _{j=1}^n{z}_j=0$
• B. $\left|\sum _{j=1}^n{z}_j\right|=0$
• C. $\sum _{j=1}^n\frac{1}{z_j}=0$
• D. None of these

Answer 1

Answer: (A), (B), (C)

The correct options are
A $\sum _{j=1}^n{z}_j=0$
B $\sum _{j=1}^n\frac{1}{z_j}=0$
C $\left|\sum _{j=1}^n{z}_j\right|=0$

Let $z_i=r(cos{\theta }_i+i.sin{\theta }_i)$
$Re\left(\sum _{j=1}^n\sum _{k=1}^n\frac{z_j}{z_k}\right)=0$
Thus, $(z_1+z_2+...+z_n)\times (\frac{1}{z_1}+\frac{1}{z_2}...\frac{1}{z_n})=0$
Thus, $\left[\right(cos{\theta }_1+...cos{\theta }_n)+i.(sin{\theta }_1+...s{in\theta }_n\left)\right]\times \left[\right(cos{\theta }_1+...cos{\theta }_n)-i.(sin{\theta }_1+...s{in\theta }_n\left)\right]=0$
So, ${(cos{\theta }_1+...cos{\theta }_n)}^2+{(sin{\theta }_1+...s{in\theta }_n)}^2=0$
So, $(cos{\theta }_1+...cos{\theta }_n)=(sin{\theta }_1+...s{in\theta }_n)=0$
Hence (a), (b), (c) are correct.