\[1.\ Prove\ that:\ [\frac{cos\ θ + i sin\ θ} {sin\ θ – i cos\ θ}]^4\ = 1\ \hspace{18cm}\]
\[\color {black}{Solution:}\ Question\ No.\ 8\ Solution\ \hspace{2cm}\]
https://yanamtakshashila.com/2021/10/13/unit-ii-complex-numbers-2/
\[2:\ Simplify\ using\ DeMoivre’s\ theorem:\ \frac{(cos\ 2θ – i sin\ 2θ)^7\ (cos\ 3θ + i sin\ 3θ)^{-5}} {(cos\ 4θ + i sin\ 4θ)^2\ (cos\ 5θ – i sin\ 5θ)^{-6}}\ \hspace{10cm}\]
\[\color {black}{Solution:}\ \frac{(cos\ 2θ – i sin\ 2θ)^7\ (cos\ 3θ + i sin\ 3θ)^{-5}} {(cos\ 4θ + i sin\ 4θ)^2\ (cos\ 5θ – i sin\ 5θ)^{-6}}\ \hspace{18cm}\]
\[= \frac{(cos\ θ + i sin\ θ)^{7 \times -2}\ (cos\ θ + i sin\ θ)^{3 \times -5}} {(cos\ θ + i sin\ θ)^{4 \times 2}\ (cos\ θ + i sin\ θ)^{-5 \times -6 }}\ \hspace{8cm}\]
\[= \frac{(cos\ θ + i sin\ θ)^{-14}\ (cos\ θ + i sin\ θ)^{-15}} {(cos\ θ + i sin\ θ)^{8}\ (cos\ θ + i sin\ θ)^{30}}\ \hspace{8cm}\]
\[= (cos\ θ + i sin\ θ )^{-\ 14\ -\ 15\ -\ 8\ -\ 30}\ \hspace{10cm}\]
\[= (cos\ θ + i sin\ θ )^{-67}\ \hspace{10cm}\]
\[= cos\ 67\ θ\ -\ i sin\ 67\ θ\ \hspace{10cm}\]
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