# DE-MOIVRE’S THEOREM (Excercise Problems)

$\LARGE{\color {purple} {PART- A}}$
$\color {purple} {1.}\ \color {red} {Find\ the\ value\ of}\ (cos\ 30^0 + i sin⁡\ 30^0 )^3 \hspace{18cm}$
$\color {blue}{Solution:}\ (cos\ 30^0 + i sin⁡\ 30^0 )^3\ =\ cos\ 90^0 + i sin⁡\ 90^0 \hspace{18cm}$
$= 0 + i (1)\ \hspace{10cm}$
$= i\ \hspace{10cm}$
$\LARGE{\color {purple} {PART- B}}$
$\color {purple} {2.}\ If\ x = cos⁡\ \theta + i sin⁡\ \theta,\ \color {red} {find\ x^m\ +\ \frac{1}{x^m}}\ \hspace{18cm}$
$\color {blue}{Solution:}\ x = cos⁡\ \theta + i sin⁡\ \theta\ \hspace{18cm}$
$x^m = (cos⁡\ \theta + i sin⁡\ \theta)^m =\ cos⁡\ m \theta\ +\ i sin⁡\ m \theta\ \hspace{10cm}$
$\frac{1}{x^m}\ = \frac{1}{cos⁡\ m \theta\ +\ i sin⁡\ m \theta} =\ cos⁡\ m \theta\ -\ i sin⁡\ m \theta\ \hspace{10cm}$
$x^m +\ \frac{1}{x^m}\ =\ cos⁡\ m \theta\ +\ i sin⁡\ m \theta\ +\ cos⁡\ m \theta\ -\ i sin⁡\ m \theta\ \hspace{10cm}$
$= 2cos⁡\ m \theta\ \hspace{10cm}$
$\color {purple} {3:}\ \color {red} {Simplify:}\ \frac{cos⁡\ 5 θ + i sin⁡\ 5 θ} {cos⁡\ 3 θ – i sin⁡\ 3 θ}\ \hspace{18cm}$
$\color {blue}{Solution:}\ \frac{cos⁡\ 5 θ + i sin⁡\ 5 θ} {cos⁡\ 3 θ + i sin⁡\ 3 θ}= \frac{(cos⁡\ θ + i sin⁡\ θ)^5} {(cos⁡\ θ – i sin⁡\ θ)^{-3}} \hspace{18cm}$
$= (cos\ θ + i sin⁡\ θ )^{5 +3 }\ \hspace{10cm}$
$= (cos\ θ + i sin⁡\ θ )^8\ \hspace{10cm}$
$= cos\ 8θ + i sin⁡\ 8θ\ \hspace{10cm}$
$\color {purple} {4.}\ \color {red} {Simplify}\ \frac{cos⁡\ 3 θ – i sin⁡\ 3 θ} {cos⁡\ 2 θ – i sin⁡\ 2 θ}\ \hspace{18cm}$
$\color {blue}{Solution:}\ \frac{cos⁡\ 3 θ – i sin⁡\ 3 θ} {cos⁡\ 2 θ – i sin⁡\ 2 θ}= \frac{(cos⁡\ θ + i sin⁡\ θ)^{-3}} {(cos⁡\ θ + i sin⁡\ θ)^{-2}} \hspace{18cm}$
$= (cos\ θ + i sin⁡\ θ )^{-3\ +\ 2 }\ \hspace{10cm}$
$= (cos\ θ + i sin⁡\ θ )^{-1}\ \hspace{10cm}$
$= cos\ θ – i sin⁡\ θ\ \hspace{10cm}$
$\LARGE{\color {purple} {PART- C}}$
$\color {purple} {5.}\ \color {red} {Simplify\ using\ DeMoivre’s\ theorem:}\ \frac{(cos⁡\ 3θ + i sin⁡\ 3θ)^2\ (cos⁡\ 4θ + i sin⁡\ 4θ)^5} {(cos⁡\ 2θ + i sin⁡\ 2θ)^4\ (cos⁡\ 3θ + i sin⁡\ 3θ)^2}\ \hspace{10cm}$
$\color {blue}{Solution:}\ \frac{(cos⁡\ 3θ + i sin⁡\ 3θ)^2\ (cos⁡\ 4θ + i sin⁡\ 4θ)^5} {(cos⁡\ 2θ + i sin⁡\ 2θ)^4\ (cos⁡\ 3θ + i sin⁡\ 3θ)^2}\ \hspace{18cm}$
$= \frac{(cos⁡\ θ + i sin⁡\ θ)^{3 \times 2}\ (cos⁡\ θ + i sin⁡\ θ)^{4 \times 5}} {(cos⁡\ θ + i sin⁡\ θ)^{2 \times 4}\ (cos⁡\ θ + i sin⁡\ θ)^{3 \times 2 }}\ \hspace{8cm}$
$= \frac{(cos⁡\ θ + i sin⁡\ θ)^6\ (cos⁡\ θ + i sin⁡\ θ)^{20}} {(cos⁡\ θ + i sin⁡\ θ)^8\ (cos⁡\ θ + i sin⁡\ θ)^6}\ \hspace{8cm}$
$= (cos\ θ + i sin⁡\ θ )^{6\ +\ 20\ -\ 8\ -\ 6 }\ \hspace{10cm}$
$= (cos\ θ + i sin⁡\ θ )^{12}\ \hspace{10cm}$
$= cos\ 12θ\ +\ i sin⁡\ 12θ\ \hspace{10cm}$
$\color {purple} {6.}\ \color {red} {Simplify\ using\ DeMoivre’s\ theorem:}\ \frac{(cos⁡\ 2θ\ -\ i sin⁡\ 2θ)^4\ (cos⁡\ 4θ\ + i sin⁡\ 4θ)^{-5}} {(cos⁡\ 3θ\ +\ i sin⁡\ 3θ)^2\ (cos⁡\ 5θ\ – i sin⁡\ 5θ)^{-3}}\ \hspace{10cm}$
$\color {blue}{Solution:}\ \frac{(cos⁡\ 2θ\ -\ i sin⁡\ 2θ)^4\ (cos⁡\ 4θ\ + i sin⁡\ 4θ)^{-5}} {(cos⁡\ 3θ\ +\ i sin⁡\ 3θ)^2\ (cos⁡\ 5θ\ – i sin⁡\ 5θ)^{-3}}\ \hspace{18cm}$
$= \frac{(cos⁡\ θ + i sin⁡\ θ)^{-2 \times 4}\ (cos⁡\ θ + i sin⁡\ θ)^{4 \times -5}} {(cos⁡\ θ + i sin⁡\ θ)^{3 \times 2}\ (cos⁡\ θ + i sin⁡\ θ)^{-3 \times -5 }}\ \hspace{8cm}$
$= \frac{(cos⁡\ θ + i sin⁡\ θ)^{-8\ (cos⁡\ θ + i sin⁡\ θ)^{-20}} {(cos⁡\ θ + i sin⁡\ θ)^6\ (cos⁡\ θ + i sin⁡\ θ)^15}\ \hspace{8cm}$
$= (cos\ θ + i sin⁡\ θ )^{-8\ -\ 20\ -\ 6\ -\ 15 }\ \hspace{10cm}$
$= (cos\ θ + i sin⁡\ θ )^{-49}\ \hspace{10cm}$
$= cos\ 49θ\ -\ i sin⁡\ 49θ\ \hspace{910cm}$

$\color {purple} {7.}\ \color {red} {Simplify\ using\ DeMoivre’s\ theorem:}\ \frac{(cos⁡\ 2θ – i sin⁡\ 2θ)^7\ (cos⁡\ 3θ + i sin⁡\ 3θ)^{-5}} {(cos⁡\ 4θ + i sin⁡\ 4θ)^2\ (cos⁡\ 5θ – i sin⁡\ 5θ)^{-6}}\ \hspace{10cm}$
$\color {blue}{Solution:}\ Question\ No.\ 2\ Solution\ \hspace{18cm}$

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