First slide

De-moivre's theorem

Question

Suppose $a < 0$ and $z_{1}, z_{2}, z_{3}, z_{4}$ $4$ be the fourth roots of a then $z_{1}^{2} + z_{2}^{2} + z_{3}^{2} + z_{4}^{2}$ is equal

Moderate

A

$- a^{2}$
B

$| a | - a$
C

$a + | a |$
D

${a^{2}}^{}$

Solution

Let $a = - b^{4}$ where $b > 0 .$
Than $z^{4} = a = b^{4} (- 1)$
$\begin{array}{l} \Rightarrow z = b (\pm \cos \frac{π}{4} \pm i \sin \frac{π}{4}) \\ ∴ z_{1}^{2} + z_{2}^{2} + z_{3}^{2} + z_{4}^{2} \\ = 2 b^{2} [\cos (\frac{π}{2}) - i \sin (\frac{π}{2}) + \cos (\frac{π}{2}) + i \sin (\frac{π}{2})] \\ = 0 = a + | a | [∵ a < 0] \end{array}$

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