First slide

Algebra of complex numbers

Question

Let $z_{1}, z_{2}, z_{3}$ be three complex numbers such that $|z_{1}| = |z_{2}| = |z_{3}| = 1$ and $z = (z_{1} + z_{2} + z_{3}) (\frac{1}{z_{1}} + \frac{1}{z_{2}} + \frac{1}{z_{3}})$ , than $| z |$ cannot exceed

Moderate

A

1
B

3
C

6
D

9

Solution

Note that $\frac{1}{z_{1}} = {\bar{z}}_{1}$ etc. thus
$\begin{array}{l} z = (z_{1} + z_{2} + z_{3}) ({\bar{z}}_{1} + {\bar{z}}_{2} + {\bar{z}}_{3}) \\ = {|z_{1} + z_{2} + z_{3}|}^{2} \leq {(|z_{1}| + |z_{2}| + |z_{3}|)}^{2} = 9 \end{array}$
The maximum value is obtained when $z_{1} = z_{2} = z_{3} = 1$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App