DE-MOIVRE’S THEOREM (Text)
\[\color {royalblue} {De-Moivre’s\ Theorem( Statement\ only)}:\ \hspace{20cm}\]
\[( i )\ If\ n\ is\ an\ integer\ positive\ or\ negative\ then\ (cos\ θ + i sin\ θ )^n = cos\ n θ + i sin\ n θ\]
\[( ii )\ If\ n\ is\ a\ fraction,\ then\ cos\ n θ + i sin\ n θ\ is\ one\ of\ the\ values\ of\ (cos\ θ + i sin\ θ )^n\]
\[\color {royalblue} {Results}:\ \hspace{20cm}\]
\[1 )\ (cos\ θ + i sin\ θ )^{-n} = cos\ n θ- i sin\ n θ\]
\[2)\ \frac{1}{cos\ θ + i sin\ θ} = (cos\ θ + i sin\ θ )^{-1 } = cos\ θ- i sin\ θ\]
\[3)\ \frac{1}{cos\ θ – i sin\ θ} = (cos\ θ – i sin\ θ )^{-1 } = cos\ θ + i sin\ θ\]
\[\color {royalblue} {Note}:\ \hspace{20cm}\]
\[1)\ (cos\ θ_1 + i sin\ θ_1) (cos\ θ_2 + i sin\ θ_2) = cos(θ_1+θ_2) + i sin (θ_1 +θ_2)\]
\[2)\ \frac{cos\ θ_1 + i sin\ θ_1}{cos\ θ_2 + i sin\ θ_2} = cos(θ_1-θ_2) + i sin (θ_1 -θ_2)\]
\[Ex1:\ Find\ the\ value\ of\ (cos\ 15^0 + i sin\ 15^0 )^6 \hspace{18cm}\]
\[\color {black}{Solution:}\ (cos\ 15^0 + i sin\ 15^0 )^6\ =\ cos\ 90^0 + i sin\ 90^0 \hspace{18cm}\]
\[= 0 + i (1)\ \hspace{10cm}\]
\[= i\ \hspace{10cm}\]
\[Ex2:\ Simplify:\ (cos\ θ + i sin\ θ )^2\ (cos\ 3θ + i sin\ 3θ )\ \hspace{18cm}\]
\[\color {black}{Solution:}\ (cos\ θ + i sin\ θ )^2\ (cos\ 3θ + i sin\ 3θ )\ =\ (cos\ θ + i sin\ θ )^2\ (cos\ θ + i sin\ θ )^3\ \hspace{18cm}\]
\[= (cos\ θ + i sin\ θ )^{2 +3 }\ \hspace{10cm}\]
\[= (cos\ θ + i sin\ θ )^5\ \hspace{10cm}\]
\[= cos\ 5θ + i sin\ 5θ\ \hspace{10cm}\]
\[Ex3:\ If\ a = cos\ x + i sin\ x,\ b = cos\ y + i sin\ y,\ find\ ab\ and\ \frac{1}{ab}\ \hspace{18cm}\]
\[\color {black}{Solution:}\ a = cos\ x + i sin\ x,\ b = cos\ y + i sin\ y\ \hspace{16cm}\]
\[ab = (cos\ x + i sin\ x)\ (cos\ y + i sin\ y)\ \hspace{12cm}\]
\[ab\ = cos\ (x + y) + i sin\ (x + y)\ \hspace{10cm}\]
\[\frac{1}{ab}\ = \frac{1}{cos\ (x + y) + i sin\ (x + y)}\ \hspace{10cm}\]
\[\frac{1}{ab}\ = cos\ (x + y) – i sin\ (x + y)\ \hspace{10cm}\]
\[Ex4:\ If\ a = cos\ α + i sin\ α,\ b = cos\ β + i sin\ β ,\ find\ ab\ – \frac{1}{ab}\ \hspace{18cm}\]
\[\color {black}{Solution:}\ a = cos\ α + i sin\ α,\ b = cos\ β + i sin\ β\ \hspace{16cm}\]
\[ab = (cos\ α + i sin\ α)\ (cos\ β + i sin\ β)\ \hspace{12cm}\]
\[ab\ = cos\ (α + β) + i sin\ (α + β)\ \hspace{10cm}\]
\[\frac{1}{ab}\ = \frac{1}{cos\ (α + β) + i sin\ (α + β)}\ \hspace{10cm}\]
\[\frac{1}{ab}\ = cos\ (α + β) – i sin\ (α + β)\ \hspace{10cm}\]
\[ab\ – \frac{1}{ab} = cos\ (α + β) + i sin\ (α + β)\ – [ cos\ (α + β) – i sin\ (α + β)] \hspace{6cm}\]
\[ = cos\ (α + β) + i sin\ (α + β)\ – cos\ (α + β) + i sin\ (α + β) \hspace{7cm}\]
\[= 2i sin\ (α + β)\ \hspace{10cm}\]
\[Ex5:\ If\ x = cos\ α + i sin\ α,\ find\ the\ value\ of\ x^2\ +\ \frac{1}{x^2}\ \hspace{18cm}\]
\[\color {black}{Solution:}\ x = cos\ α + i sin\ α \hspace{18cm}\]
\[x^2 = (cos\ α + i sin\ α)^2 =\ cos\ 2α + i sin\ 2α \hspace{10cm}\]
\[\frac{1}{x^2}\ = \frac{1}{cos\ 2α + i sin\ 2α}\ = cos\ 2α – i sin\ 2α \hspace{10cm}\]
\[x^2 +\ \frac{1}{x^2}\ =\ cos\ 2α + i sin\ 2α\ +cos\ 2α – i sin\ 2α\hspace{10cm}\]
\[= 2cos\ 2α\ \hspace{10cm}\]
\[Ex6:\ Simplify\ \frac{cos\ 5 θ + i sin\ 5 θ} {cos\ 3 θ + i sin\ 3 θ}\ \hspace{18cm}\]
\[\color {black}{Solution:}\ \frac{cos\ 5 θ + i sin\ 5 θ} {cos\ 3 θ + i sin\ 3 θ}= \frac{(cos\ θ + i sin\ θ)^5} {(cos\ θ + i sin\ θ)^3} \hspace{18cm}\]
\[= (cos\ θ + i sin\ θ )^{5 -3 }\ \hspace{10cm}\]
\[= (cos\ θ + i sin\ θ )^2\ \hspace{10cm}\]
\[= cos\ 2θ + i sin\ 2θ\ \hspace{10cm}\]
\[Ex7:\ Simplify\ using\ DeMoivre’s\ theorem:\ \frac{(cos\ 3θ – i sin\ 3θ)^5\ (cos\ 4θ + i sin\ 4θ)^4} {(cos\ 2θ + i sin\ 2θ)^7\ (cos\ 3θ – i sin\ 3θ)^6}\ \hspace{10cm}\]
\[\color {black}{Solution:}\ \frac{(cos\ 3θ – i sin\ 3θ)^5\ (cos\ 4θ + i sin\ 4θ)^4} {(cos\ 2θ + i sin\ 2θ)^7\ (cos\ 3θ – i sin\ 3θ)^6}\ \hspace{18cm}\]
\[= \frac{(cos\ θ + i sin\ θ)^{5 \times -3}\ (cos\ θ + i sin\ θ)^{4 \times 4}} {(cos\ θ + i sin\ θ)^{2 \times 7}\ (cos\ θ + i sin\ θ)^{6 \times -3 }}\ \hspace{8cm}\]
\[= \frac{(cos\ θ + i sin\ θ)^{-15}\ (cos\ θ + i sin\ θ)^{16}} {(cos\ θ + i sin\ θ)^{14}\ (cos\ θ + i sin\ θ)^{-18}}\ \hspace{8cm}\]
\[= (cos\ θ + i sin\ θ )^{-15 + 16 + 14 – 18 }\ \hspace{10cm}\]
\[= (cos\ θ + i sin\ θ )^5\ \hspace{10cm}\]
\[= cos\ 5θ + i sin\ 5θ\ \hspace{10cm}\]
\[Ex8:\ 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
\[\color {black}{Solution:}\ \frac{(cos\ 3θ + i sin\ 3θ)^{-5}\ (cos\ 2θ + i sin\ 2θ)^4} {(cos\ 4θ – i sin\ 4θ)^{-2}\ (cos\ 5θ – i sin\ 5θ)^3}\ \hspace{18cm}\]
\[= \frac{(cos\ θ + i sin\ θ)^{-5 \times 3}\ (cos\ θ + i sin\ θ)^{4 \times 2}} {(cos\ θ + i sin\ θ)^{-2 \times -4}\ (cos\ θ + i sin\ θ)^{3 \times -5 }}\ \hspace{8cm}\]
\[= \frac{(cos\ θ + i sin\ θ)^{-15}\ (cos\ θ + i sin\ θ)^8} {(cos\ θ + i sin\ θ)^8\ (cos\ θ + i sin\ θ)^{-15}}\ \hspace{8cm}\]
\[= (cos\ θ + i sin\ θ )^{-15 + 8 – 8 + 15 }\ \hspace{10cm}\]
\[= (cos\ θ + i sin\ θ )^0\ \hspace{10cm}\]
\[= cos\ 0θ – i sin\ 0θ\ = 1\ \hspace{10cm}\]
\[Ex9:\ Prove\ that:\ [\frac{cos\ θ + i sin\ θ} {sin\ θ – i cos\ θ}]^4\ = 1\ \hspace{18cm}\]
\[\color {black}{Solution:}\ LHS = [\frac{cos\ θ + i sin\ θ} {sin\ θ – i cos\ θ}]^4\ \hspace{18cm}\]
\[=\ [\frac{cos\ θ + i sin\ θ} {-i(cos\ θ + i sin\ θ)}]^4\ \hspace{5cm}\ \because sin\ θ – i cos\ θ\ =\ -i(cos\ θ + i sin\ θ)\]
\[ = [\frac{(cos\ θ + i sin\ θ)^4} {(-i)^4(cos\ θ + i sin\ θ)^4}]\ \hspace{10cm}\]
\[= \frac{(cos\ θ + i sin\ θ)^4} {(1)(cos\ θ + i sin\ θ)^4}\ \hspace{5cm}\ \because (-i)^4\ =\ 1\]
\[ = \frac{(cos\ θ + i sin\ θ)^4} {(cos\ θ + i sin\ θ)^4}\ \hspace{10cm}\]
\[= (cos\ θ + i sin\ θ )^{4 -4 }\ \hspace{10cm}\]
\[= (cos\ θ + i sin\ θ )^{0}\ \hspace{10cm}\]
\[= cos\ 0 + i sin\ 0\ \hspace{10cm}\]
\[= 1\ + i(0)\ \hspace{5cm}\ \because cos\ 0\ =\ 1,\ sin\ 0\ =\ 0\]
\[= 1\ =\ RHS\ \hspace{10cm}\]
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