# DE-MOIVRE’S THEOREM (Text)

$\color {royalblue} {De-Moivre’s\ Theorem( Statement\ only)}:\ \hspace{20cm}$
$( i )\ If\ n\ is\ an\ integer\ positive\ or\ negative\ then\ (cos\ θ + i sin⁡\ θ )^n = cos⁡\ n θ + i sin⁡\ n θ$
$( ii )\ If\ n\ is\ a\ fraction,\ then\ cos⁡\ n θ + i sin⁡\ n θ\ is\ one\ of\ the\ values\ of\ (cos\ θ + i sin⁡\ θ )^n$
$\color {royalblue} {Results}:\ \hspace{20cm}$
$1 )\ (cos\ θ + i sin⁡\ θ )^{-n} = cos⁡\ n θ- i sin⁡\ n θ$
$2)\ \frac{1}{cos⁡\ θ + i sin⁡\ θ} = (cos\ θ + i sin⁡\ θ )^{-1 } = cos⁡\ θ- i sin⁡\ θ$
$3)\ \frac{1}{cos⁡\ θ – i sin⁡\ θ} = (cos\ θ – i sin⁡\ θ )^{-1 } = cos⁡\ θ + i sin⁡\ θ$
$\color {royalblue} {Note}:\ \hspace{20cm}$
$1)\ (cos\ θ_1 + i sin⁡\ θ_1) (cos⁡\ θ_2 + i sin\ θ_2) = cos(⁡θ_1+θ_2) + i sin⁡ (θ_1 +θ_2)$
$2)\ \frac{cos⁡\ θ_1 + i sin⁡\ θ_1}{cos⁡\ θ_2 + i sin⁡\ θ_2} = cos(θ_1-θ_2) + i sin⁡ (θ_1 -θ_2)$
$\color {purple} {Example:\ 1}\ \color {red} {Simplify}\ (cos\ 15^0 + i sin⁡\ 15^0 )^6\ \hspace{18cm}$
$\color {blue}{Solution:}\ (cos\ 15^0 + i sin⁡\ 15^0 )^6\ =\ cos\ 90^0 + i sin⁡\ 90^0 \hspace{18cm}$
$= 0 + i (1)\ \hspace{10cm}$
$= i\ \hspace{10cm}$
$\boxed{(cos\ 15^0 + i sin⁡\ 15^0 )^6\ =\ i}$
$\color {purple} {Example\ 2}:\ \color {red} {Simplify:}\ (cos\ θ + i sin⁡\ θ )^2\ (cos\ 3θ + i sin⁡\ 3θ )\ \hspace{18cm}$
$\color {blue}{Solution:}\ (cos\ θ + i sin⁡\ θ )^2\ (cos\ 3θ + i sin⁡\ 3θ )\ =\ (cos\ θ + i sin⁡\ θ )^2\ (cos\ θ + i sin⁡\ θ )^3\ \hspace{18cm}$
$= (cos\ θ + i sin⁡\ θ )^{2 +3 }\ \hspace{10cm}$
$= (cos\ θ + i sin⁡\ θ )^5\ \hspace{10cm}$
$= cos\ 5θ + i sin⁡\ 5θ\ \hspace{10cm}$
$\boxed{(cos\ θ + i sin⁡\ θ )^2\ (cos\ 3θ + i sin⁡\ 3θ )\ =\ cos\ 5θ + i sin⁡\ 5θ}$
$\color {purple} {Example\ 3:}\ If\ a = cos⁡\ x + i sin⁡\ x,\ b = cos⁡\ y + i sin⁡\ y,\ \color {red} {find\ ab\ and\ \frac{1}{ab}}\ \hspace{18cm}$
$\color {blue}{Solution:}\ a = cos⁡\ x + i sin⁡\ x,\ b = cos⁡\ y + i sin⁡\ y\ \hspace{16cm}$
$ab = (cos⁡\ x + i sin⁡\ x)\ (cos⁡\ y + i sin⁡\ y)\ \hspace{12cm}$
$\boxed{ab\ = cos⁡\ (x + y) + i sin⁡\ (x + y)}\ \hspace{10cm}$
$\frac{1}{ab}\ = \frac{1}{cos⁡\ (x + y) + i sin⁡\ (x + y)}\ \hspace{10cm}$
$\boxed{\frac{1}{ab}\ = cos⁡\ (x + y) – i sin⁡\ (x + y)}\ \hspace{10cm}$
$\color {purple} {Example\ 4:}\ If\ a = cos\ α + i sin\ α,\ b = cos\ β + i sin\ β ,\ \color {red} {find\ ab\ – \frac{1}{ab}}\ \hspace{18cm}$
$\color {blue}{Solution:}\ a = cos\ α + i sin\ α,\ b = cos\ β + i sin\ β\ \hspace{16cm}$
$ab = (cos\ α + i sin\ α)\ (cos\ β + i sin\ β)\ \hspace{12cm}$
$ab\ = cos⁡\ (α + β) + i sin⁡\ (α + β)\ \hspace{10cm}$
$\frac{1}{ab}\ = \frac{1}{cos⁡\ (α + β) + i sin⁡\ (α + β)}\ \hspace{10cm}$
$\frac{1}{ab}\ = cos⁡\ (α + β) – i sin⁡\ (α + β)\ \hspace{10cm}$
$ab\ – \frac{1}{ab} = cos⁡\ (α + β) + i sin⁡\ (α + β)\ – [ cos⁡\ (α + β) – i sin⁡\ (α + β)] \hspace{6cm}$
$= cos⁡\ (α + β) + i sin⁡\ (α + β)\ – cos⁡\ (α + β) + i sin⁡\ (α + β) \hspace{7cm}$
$= 2i sin⁡\ (α + β)\ \hspace{10cm}$
$\boxed{ab\ – \frac{1}{ab}\ =\ 2i sin⁡\ (α + β)}\ \hspace{10cm}$
$\color {purple} {Example\ 5:}\ If\ x = cos⁡\ α + i sin⁡\ α,\ \color {red} {find\ the\ value\ of\ x^2\ +\ \frac{1}{x^2}}\ \hspace{18cm}$
$\color {blue}{Solution:}\ x = cos⁡\ α + i sin⁡\ α \hspace{18cm}$
$x^2 = (cos⁡\ α + i sin⁡\ α)^2 =\ cos⁡\ 2α + i sin⁡\ 2α \hspace{10cm}$
$\frac{1}{x^2}\ = \frac{1}{cos⁡\ 2α + i sin⁡\ 2α}\ = cos⁡\ 2α – i sin⁡\ 2α \hspace{10cm}$
$x^2 +\ \frac{1}{x^2}\ =\ cos⁡\ 2α + i sin⁡\ 2α\ +cos⁡\ 2α – i sin⁡\ 2α\hspace{10cm}$
$= 2cos⁡\ 2α\ \hspace{10cm}$
$\boxed{x^2\ +\ \frac{1}{x^2}\ =\ 2cos⁡\ 2α}$
$\color {purple}{Example\ 6:}\ \color {red} {Simplify\ :}\ \frac{cos⁡\ 5 θ + i sin⁡\ 5 θ} {cos⁡\ 3 θ + i sin⁡\ 3 θ}\ \hspace{18cm}$
$\color {blue}{Solution:}\ \frac{cos⁡\ 5 θ + i sin⁡\ 5 θ} {cos⁡\ 3 θ + i sin⁡\ 3 θ}= \frac{(cos⁡\ θ + i sin⁡\ θ)^5} {(cos⁡\ θ + i sin⁡\ θ)^3} \hspace{18cm}$
$= (cos\ θ + i sin⁡\ θ )^{5 -3 }\ \hspace{10cm}$
$= (cos\ θ + i sin⁡\ θ )^2\ \hspace{10cm}$
$= cos\ 2θ + i sin⁡\ 2θ\ \hspace{10cm}$
$\boxed{\frac{cos⁡\ 5 θ + i sin⁡\ 5 θ} {cos⁡\ 3 θ + i sin⁡\ 3 θ}\ =\ cos\ 2θ + i sin⁡\ 2θ}$
$\color {purple} {Example\ 7:}\ \color {red} {Simplify\ using\ DeMoivre’s\ theorem:}\ \frac{(cos⁡\ 3θ – i sin⁡\ 3θ)^5\ (cos⁡\ 4θ + i sin⁡\ 4θ)^4} {(cos⁡\ 2θ + i sin⁡\ 2θ)^7\ (cos⁡\ 3θ – i sin⁡\ 3θ)^6}\ \hspace{10cm}$
$\color {blue}{Solution:}\ \frac{(cos⁡\ 3θ – i sin⁡\ 3θ)^5\ (cos⁡\ 4θ + i sin⁡\ 4θ)^4} {(cos⁡\ 2θ + i sin⁡\ 2θ)^7\ (cos⁡\ 3θ – i sin⁡\ 3θ)^6}\ \hspace{18cm}$
$= \frac{(cos⁡\ θ + i sin⁡\ θ)^{5 \times -3}\ (cos⁡\ θ + i sin⁡\ θ)^{4 \times 4}} {(cos⁡\ θ + i sin⁡\ θ)^{2 \times 7}\ (cos⁡\ θ + i sin⁡\ θ)^{6 \times -3 }}\ \hspace{8cm}$
$= \frac{(cos⁡\ θ + i sin⁡\ θ)^{-15}\ (cos⁡\ θ + i sin⁡\ θ)^{16}} {(cos⁡\ θ + i sin⁡\ θ)^{14}\ (cos⁡\ θ + i sin⁡\ θ)^{-18}}\ \hspace{8cm}$
$= (cos\ θ + i sin⁡\ θ )^{-15 + 16 + 14 – 18 }\ \hspace{10cm}$
$= (cos\ θ + i sin⁡\ θ )^5\ \hspace{10cm}$
$= cos\ 5θ + i sin⁡\ 5θ\ \hspace{10cm}$
$\boxed{\frac{(cos⁡\ 3θ – i sin⁡\ 3θ)^5\ (cos⁡\ 4θ + i sin⁡\ 4θ)^4} {(cos⁡\ 2θ + i sin⁡\ 2θ)^7\ (cos⁡\ 3θ – i sin⁡\ 3θ)^6}\ =\ cos\ 5θ + i sin⁡\ 5θ}$
$\color {purple} {Example\ 8:}\ \color {red} {Simplify\ using\ DeMoivre’s\ theorem:}\ \frac{(cos⁡\ 3θ + i sin⁡\ 3θ)^{-5}\ (cos⁡\ 2θ + i sin⁡\ 2θ)^4} {(cos⁡\ 4θ – i sin⁡\ 4θ)^{-2}\ (cos⁡\ 5θ – i sin⁡\ 5θ)^3}\ \hspace{10cm}$
$\color {blue}{Solution:}\ \frac{(cos⁡\ 3θ + i sin⁡\ 3θ)^{-5}\ (cos⁡\ 2θ + i sin⁡\ 2θ)^4} {(cos⁡\ 4θ – i sin⁡\ 4θ)^{-2}\ (cos⁡\ 5θ – i sin⁡\ 5θ)^3}\ \hspace{18cm}$
$= \frac{(cos⁡\ θ + i sin⁡\ θ)^{-5 \times 3}\ (cos⁡\ θ + i sin⁡\ θ)^{4 \times 2}} {(cos⁡\ θ + i sin⁡\ θ)^{-2 \times -4}\ (cos⁡\ θ + i sin⁡\ θ)^{3 \times -5 }}\ \hspace{8cm}$
$= \frac{(cos⁡\ θ + i sin⁡\ θ)^{-15}\ (cos⁡\ θ + i sin⁡\ θ)^8} {(cos⁡\ θ + i sin⁡\ θ)^8\ (cos⁡\ θ + i sin⁡\ θ)^{-15}}\ \hspace{8cm}$
$= (cos\ θ + i sin⁡\ θ )^{-15 + 8 – 8 + 15 }\ \hspace{10cm}$
$= (cos\ θ + i sin⁡\ θ )^0\ \hspace{10cm}$
$= cos\ 0θ + i sin⁡\ 0θ\ = 1\ \hspace{10cm}$
$\boxed{\frac{(cos⁡\ 3θ + i sin⁡\ 3θ)^{-5}\ (cos⁡\ 2θ + i sin⁡\ 2θ)^4} {(cos⁡\ 4θ – i sin⁡\ 4θ)^{-2}\ (cos⁡\ 5θ – i sin⁡\ 5θ)^3}\ =\ 1}$
$\color {purple} {Example\ 9:}\ \color {red} {Find\ the\ value\ of}\ \frac{(cos⁡\ x\ -\ i sin⁡\ x)^3\ (cos⁡\ 3x\ +\ i sin⁡\ 3x)^5} {(cos⁡\ 2x\ – i sin⁡\ 2x)^5\ (cos⁡\ 5x\ + i sin⁡\ 5x)^7}\ \color {red} {when\ x\ =\ \frac{2\pi}{13}}\hspace{10cm}$
$\color {blue}{Solution:}\ \frac{(cos⁡\ x\ -\ i sin⁡\ x)^3\ (cos⁡\ 3x\ +\ i sin⁡\ 3x)^5} {(cos⁡\ 2x\ – i sin⁡\ 2x)^5\ (cos⁡\ 5x\ + i sin⁡\ 5x)^7}\ \hspace{18cm}$
$= \frac{(cos⁡\ x + i sin⁡\ x)^{3 \times -1}\ (cos⁡\ x + i sin⁡\ x)^{5 \times 3}} {(cos⁡\ x + i sin⁡\ x)^{5 \times -2}\ (cos⁡\ θ + i sin⁡\ θ)^{5 \times 7}}\ \hspace{8cm}$
$= \frac{(cos⁡\ x + i sin⁡\ x)^{-3}\ (cos⁡\ x + i sin⁡\ x)^{15}} {(cos⁡\ x + i sin⁡\ x)^{-10}\ (cos⁡\ x + i sin⁡\ x)^{35}}\ \hspace{8cm}$
$= (cos\ x + i sin⁡\ x )^{-3 + 15 + 10 – 35 }\ \hspace{10cm}$
$= (cos\ x + i sin⁡\ x)^{13}\ \hspace{10cm}$
$= cos\ 13x\ +\ i sin⁡\ 13x\ \hspace{10cm}$
$= cos\ 13(\frac{2\pi}{13})\ +\ i sin⁡\ 13(\frac{2\pi}{13})\ \hspace{2cm}\ when\ x\ =\ \frac{2\pi}{13}\ \hspace{3cm}$
$= cos\ 2\ \pi\ +\ i sin⁡\ 2\ \pi\ \hspace{10cm}$
$=\ 1\ \hspace{2cm}\ \because\ cos\ n\ \pi\ = {-\ 1}^n\ and\ sin\ n\ \pi\ =\ 0\ \hspace{3cm}$
$\boxed{\frac{(cos⁡\ x\ -\ i sin⁡\ x)^3\ (cos⁡\ 3x\ +\ i sin⁡\ 3x)^5} {(cos⁡\ 2x\ – i sin⁡\ 2x)^5\ (cos⁡\ 5x\ + i sin⁡\ 5x)^7}\ =\ 1}$
$\color {purple} {Example\ 10:}\ \color {red} {Prove\ that:}\ [\frac{cos⁡\ θ + i sin⁡\ θ} {sin⁡\ θ – i cos\ θ}]^4\ = 1\ \hspace{18cm}$
$\color {blue}{Solution:}\ LHS = [\frac{cos⁡\ θ + i sin⁡\ θ} {sin⁡\ θ – i cos\ θ}]^4\ \hspace{18cm}$
$=\ [\frac{cos⁡\ θ + i sin⁡\ θ} {-i(cos⁡\ θ + i sin\ θ)}]^4\ \hspace{5cm}\ \because sin⁡\ θ – i cos\ θ\ =\ -i(cos⁡\ θ + i sin\ θ)$
$= [\frac{(cos⁡\ θ + i sin⁡\ θ)^4} {(-i)^4(cos⁡\ θ + i sin\ θ)^4}]\ \hspace{10cm}$
$= \frac{(cos⁡\ θ + i sin⁡\ θ)^4} {(1)(cos⁡\ θ + i sin\ θ)^4}\ \hspace{5cm}\ \because (-i)^4\ =\ 1$
$= \frac{(cos⁡\ θ + i sin⁡\ θ)^4} {(cos⁡\ θ + i sin\ θ)^4}\ \hspace{10cm}$
$= (cos\ θ + i sin⁡\ θ )^{4 -4 }\ \hspace{10cm}$
$= (cos\ θ + i sin⁡\ θ )^{0}\ \hspace{10cm}$
$= cos\ 0 + i sin⁡\ 0\ \hspace{10cm}$
$= 1\ + i(0)\ \hspace{5cm}\ \because cos\ 0\ =\ 1,\ sin\ 0\ =\ 0$
$= 1\ =\ RHS\ \hspace{10cm}$

### Exercise Problems

$\LARGE{\color {purple} {PART- A}}$
$\color {purple} {1.}\ \color {red} {Find\ the\ value\ of}\ (cos\ 30^0 + i sin⁡\ 30^0 )^3 \hspace{18cm}$
$\LARGE{\color {purple} {PART- B}}$
$\color {purple} {2.}\ If\ x = cos⁡\ \theta + i sin⁡\ \theta,\ \color {red} {find\ x^m\ +\ \frac{1}{x^m}}\ \hspace{18cm}$
$\color {purple} {3:}\ \color {red} {Simplify:}\ \frac{cos⁡\ 5 θ + i sin⁡\ 5 θ} {cos⁡\ 3 θ – i sin⁡\ 3 θ}\ \hspace{18cm}$
$\color {purple} {4.}\ \color {red} {Simplify}\ \frac{cos⁡\ 3 θ – i sin⁡\ 3 θ} {cos⁡\ 2 θ – i sin⁡\ 2 θ}\ \hspace{18cm}$
$\LARGE{\color {purple} {PART- C}}$
$\color {purple} {5.}\ \color {red} {Simplify\ using\ DeMoivre’s\ theorem:}\ \frac{(cos⁡\ 3θ + i sin⁡\ 3θ)^2\ (cos⁡\ 4θ + i sin⁡\ 4θ)^5} {(cos⁡\ 2θ + i sin⁡\ 2θ)^4\ (cos⁡\ 3θ + i sin⁡\ 3θ)^2}\ \hspace{10cm}$
$\color {purple} {6.}\ \color {red} {Simplify\ using\ DeMoivre’s\ theorem:}\ \frac{(cos⁡\ 2θ – i sin⁡\ 2θ)^7\ (cos⁡\ 3θ + i sin⁡\ 3θ)^{-5}} {(cos⁡\ 4θ + i sin⁡\ 4θ)^2\ (cos⁡\ 5θ – i sin⁡\ 5θ)^{-6}}\ \hspace{10cm}$