# DE-MOVIRE’S THEOREM (REVISION)

$1.\ Prove\ that:\ [\frac{cos⁡\ θ + i sin⁡\ θ} {sin⁡\ θ – i cos\ θ}]^4\ = 1\ \hspace{18cm}$
$\color {black}{Solution:}\ Question\ No.\ 8\ Solution\ \hspace{2cm}$

https://yanamtakshashila.com/2021/10/13/unit-ii-complex-numbers-2/

$2:\ Simplify\ using\ DeMoivre’s\ theorem:\ \frac{(cos⁡\ 2θ – i sin⁡\ 2θ)^7\ (cos⁡\ 3θ + i sin⁡\ 3θ)^{-5}} {(cos⁡\ 4θ + i sin⁡\ 4θ)^2\ (cos⁡\ 5θ – i sin⁡\ 5θ)^{-6}}\ \hspace{10cm}$
$\color {black}{Solution:}\ \frac{(cos⁡\ 2θ – i sin⁡\ 2θ)^7\ (cos⁡\ 3θ + i sin⁡\ 3θ)^{-5}} {(cos⁡\ 4θ + i sin⁡\ 4θ)^2\ (cos⁡\ 5θ – i sin⁡\ 5θ)^{-6}}\ \hspace{18cm}$
$= \frac{(cos⁡\ θ + i sin⁡\ θ)^{7 \times -2}\ (cos⁡\ θ + i sin⁡\ θ)^{3 \times -5}} {(cos⁡\ θ + i sin⁡\ θ)^{4 \times 2}\ (cos⁡\ θ + i sin⁡\ θ)^{-5 \times -6 }}\ \hspace{8cm}$
$= \frac{(cos⁡\ θ + i sin⁡\ θ)^{-14}\ (cos⁡\ θ + i sin⁡\ θ)^{-15}} {(cos⁡\ θ + i sin⁡\ θ)^{8}\ (cos⁡\ θ + i sin⁡\ θ)^{30}}\ \hspace{8cm}$
$= (cos\ θ + i sin⁡\ θ )^{-\ 14\ -\ 15\ -\ 8\ -\ 30}\ \hspace{10cm}$
$= (cos\ θ + i sin⁡\ θ )^{-67}\ \hspace{10cm}$
$= cos\ 67\ θ\ -\ i sin⁡\ 67\ θ\ \hspace{10cm}$