Question

If a is a complex nth root of unity and if Z₁ and Z₂ are two complex numbers, then $\sum_{r = 0}^{n - 1}{|Z_{1} + \alpha^{r}Z_{2}|^{2}}$=

• A. n² |Z₁ + Z₂|²
• B. $\left( \frac{Z_{1}}{n} + \frac{Z_{2}}{n} \right)^{2}$
• C. n (|Z₁|² + |Z₂|²)
• D. n² (|Z₁|² + |Z₂|²)

Answer 1

Answer: (C)

n (|Z₁|² + |Z₂|²)

Answer 2

Sol. We have 1 + a + a² + …….. + a^n–1 = 0. It is clear that $\sum_{r - 0}^{n - 1}\alpha^{r}$ = 0

Now$\sum_{r = 0}^{n - 1}{|Z_{1} + \alpha^{r}Z_{2}|^{2}}$=$\sum_{r = 0}^{n - 1}{(Z_{1} + \alpha^{r}Z_{2})({\bar{Z}}_{1} + {\bar{\alpha}}^{r}{\bar{Z}}_{2})}$

= $\sum_{r = 0}^{n - 1}{|Z_{1}|^{2} + Z_{1}{\bar{Z}}_{2}}$

$\sum_{r = 0}^{n - 1}\alpha^{r}$+ $\sum_{r = 0}^{n - 1}{|Z_{2}|^{2}|\alpha|^{2r}}$

= n(|Z₁|² + |Z₂|²), using (1) and |a| = 1.