Question

(cos5 theta+i sin5 theta)/(cos 3 theta-i sin 3 theta)^(2) slove by De Moivres thm...​

Answer 1

Step-by-step explanation:

We have,

$z = \dfrac{( \cos5 \theta+i \sin5 \theta)}{{ \{ \cos( 3 \theta)-i \sin( 3 \theta) \}}^{2}}$

$\implies \: z = \dfrac{ {e}^{i(5 \theta)} }{{ \{ \cos( - 3 \theta) + i \sin( - 3 \theta) \}}^{2}}$

$\implies \: z = \dfrac{ {e}^{i(5 \theta)} }{ \{ {e}^{ i( - 3 \theta)} \}^{2}}$

$\implies \: z = \dfrac{ {e}^{i(5 \theta)} }{ {e}^{ i( - 6 \theta)} }$

$\implies \: z = {e}^{i \{(5 \theta) - (- 6 \theta) \} }$

$\implies \: z = {e}^{i (5 \theta + 6 \theta) }$

$\implies \: z = {e}^{i (11\theta ) }$

$\implies \: z = \cos(11 \theta) + i \sin( 11\theta)$