Unit – II – COMPLEX NUMBERS

rajuviswa
\[\LARGE{\color {red}{CHAPTER\ 2.2:\ DE-MOIVRE’S\ THEOREM\ (Text)}}\]
\[\color {royalblue} {De-Moivre’s\ Theorem( Statement\ only)}:\ \hspace{20cm}\]
\[( i )\ If\ n\ is\ an\ integer\ positive\ or\ negative\ then\ (cos\ θ + i sin⁡\ θ )^n = cos⁡\ n θ + i sin⁡\ n θ\]
\[( ii )\ If\ n\ is\ a\ fraction,\ then\ cos⁡\ n θ + i sin⁡\ n θ\ is\ one\ of\ the\ values\ of\ (cos\ θ + i sin⁡\ θ )^n\]
\[\color {royalblue} {Results}:\ \hspace{20cm}\]
\[1 )\ (cos\ θ + i sin⁡\ θ )^{-n} = cos⁡\ n θ- i sin⁡\ n θ\]
\[2)\ \frac{1}{cos⁡\ θ + i sin⁡\ θ} = (cos\ θ + i sin⁡\ θ )^{-1 } = cos⁡\ θ- i sin⁡\ θ\]
\[3)\ \frac{1}{cos⁡\ θ – i sin⁡\ θ} = (cos\ θ – i sin⁡\ θ )^{-1 } = cos⁡\ θ + i sin⁡\ θ\]
\[\color {royalblue} {Note}:\ \hspace{20cm}\]
\[1)\ (cos\ θ_1 + i sin⁡\ θ_1) (cos⁡\ θ_2 + i sin\ θ_2) = cos(⁡θ_1+θ_2) + i sin⁡ (θ_1 +θ_2)\]
\[2)\ \frac{cos⁡\ θ_1 + i sin⁡\ θ_1}{cos⁡\ θ_2 + i sin⁡\ θ_2} = cos(θ_1-θ_2) + i sin⁡ (θ_1 -θ_2)\]
\[Ex1:\ Find\ the\ value\ of\ (cos\ 15^0 + i sin⁡\ 15^0 )^6 \hspace{18cm}\]
\[\color {black}{Solution:}\ (cos\ 15^0 + i sin⁡\ 15^0 )^6\ =\ cos\ 90^0 + i sin⁡\ 90^0 \hspace{18cm}\]
\[= 0 + i (1)\ \hspace{10cm}\]
\[= i\ \hspace{10cm}\]
\[Ex2:\ Simplify:\ (cos\ θ + i sin⁡\ θ )^2\ (cos\ 3θ + i sin⁡\ 3θ )\ \hspace{18cm}\]
\[\color {black}{Solution:}\ (cos\ θ + i sin⁡\ θ )^2\ (cos\ 3θ + i sin⁡\ 3θ )\ =\ (cos\ θ + i sin⁡\ θ )^2\ (cos\ θ + i sin⁡\ θ )^3\ \hspace{18cm}\]
\[= (cos\ θ + i sin⁡\ θ )^{2 +3 }\ \hspace{10cm}\]
\[= (cos\ θ + i sin⁡\ θ )^5\ \hspace{10cm}\]
\[= cos\ 5θ + i sin⁡\ 5θ\ \hspace{10cm}\]
\[Ex3:\ If\ a = cos⁡\ x + i sin⁡\ x,\ b = cos⁡\ y + i sin⁡\ y,\ find\ ab\ and\ \frac{1}{ab}\ \hspace{18cm}\]
\[\color {black}{Solution:}\ a = cos⁡\ x + i sin⁡\ x,\ b = cos⁡\ y + i sin⁡\ y\ \hspace{16cm}\]
\[ab = (cos⁡\ x + i sin⁡\ x)\ (cos⁡\ y + i sin⁡\ y)\ \hspace{12cm}\]
\[ab\ = cos⁡\ (x + y) + i sin⁡\ (x + y)\ \hspace{10cm}\]
\[\frac{1}{ab}\ = \frac{1}{cos⁡\ (x + y) + i sin⁡\ (x + y)}\ \hspace{10cm}\]
\[\frac{1}{ab}\ = cos⁡\ (x + y) – i sin⁡\ (x + y)\ \hspace{10cm}\]
\[Ex4:\ If\ a = cos\ α + i sin\ α,\ b = cos\ β + i sin\ β ,\ find\ ab\ – \frac{1}{ab}\ \hspace{18cm}\]
\[\color {black}{Solution:}\ a = cos\ α + i sin\ α,\ b = cos\ β + i sin\ β\ \hspace{16cm}\]
\[ab = (cos\ α + i sin\ α)\ (cos\ β + i sin\ β)\ \hspace{12cm}\]
\[ab\ = cos⁡\ (α + β) + i sin⁡\ (α + β)\ \hspace{10cm}\]
\[\frac{1}{ab}\ = \frac{1}{cos⁡\ (α + β) + i sin⁡\ (α + β)}\ \hspace{10cm}\]
\[\frac{1}{ab}\ = cos⁡\ (α + β) – i sin⁡\ (α + β)\ \hspace{10cm}\]
\[ab\ – \frac{1}{ab} = cos⁡\ (α + β) + i sin⁡\ (α + β)\ – [ cos⁡\ (α + β) – i sin⁡\ (α + β)] \hspace{6cm}\]
\[ = cos⁡\ (α + β) + i sin⁡\ (α + β)\ – cos⁡\ (α + β) + i sin⁡\ (α + β) \hspace{7cm}\]
\[= 2i sin⁡\ (α + β)\ \hspace{10cm}\]
\[Ex5:\ If\ x = cos⁡\ α + i sin⁡\ α,\ find\ the\ value\ of\ x^2\ +\ \frac{1}{x^2}\ \hspace{18cm}\]
\[\color {black}{Solution:}\ x = cos⁡\ α + i sin⁡\ α \hspace{18cm}\]
\[x^2 = (cos⁡\ α + i sin⁡\ α)^2 =\ cos⁡\ 2α + i sin⁡\ 2α \hspace{10cm}\]
\[\frac{1}{x^2}\ = \frac{1}{cos⁡\ 2α + i sin⁡\ 2α}\ = cos⁡\ 2α – i sin⁡\ 2α \hspace{10cm}\]
\[x^2 +\ \frac{1}{x^2}\ = (cos⁡\ α + i sin⁡\ α)^2 =\ cos⁡\ 2α + i sin⁡\ 2α\ +cos⁡\ 2α – i sin⁡\ 2α\hspace{10cm}\]
\[= 2cos⁡\ 2α\ \hspace{10cm}\]
\[Ex6:\ Simplify\ \frac{cos⁡\ 5 θ + i sin⁡\ 5 θ} {cos⁡\ 3 θ + i sin⁡\ 3 θ}\ \hspace{18cm}\]
\[\color {black}{Solution:}\ \frac{cos⁡\ 5 θ + i sin⁡\ 5 θ} {cos⁡\ 3 θ + i sin⁡\ 3 θ}= \frac{(cos⁡\ θ + i sin⁡\ θ)^5} {(cos⁡\ θ + i sin⁡\ θ)^3} \hspace{18cm}\]
\[= (cos\ θ + i sin⁡\ θ )^{5 -3 }\ \hspace{10cm}\]
\[= (cos\ θ + i sin⁡\ θ )^2\ \hspace{10cm}\]
\[= cos\ 2θ + i sin⁡\ 2θ\ \hspace{10cm}\]
\[Ex7:\ Simplify\ using\ DeMoivre’s\ theorem:\ \frac{(cos⁡\ 3θ – i sin⁡\ 3θ)^5\ (cos⁡\ 4θ + i sin⁡\ 4θ)^4} {(cos⁡\ 2θ + i sin⁡\ 2θ)^7\ (cos⁡\ 3θ – i sin⁡\ 3θ)^6}\ \hspace{10cm}\]
\[\color {black}{Solution:}\ \frac{(cos⁡\ 3θ – i sin⁡\ 3θ)^5\ (cos⁡\ 4θ + i sin⁡\ 4θ)^4} {(cos⁡\ 2θ + i sin⁡\ 2θ)^7\ (cos⁡\ 3θ – i sin⁡\ 3θ)^6}\ \hspace{18cm}\]
\[= \frac{(cos⁡\ θ + i sin⁡\ θ)^{5 \times -3}\ (cos⁡\ θ + i sin⁡\ θ)^{4 \times 4}} {(cos⁡\ θ + i sin⁡\ θ)^{2 \times 7}\ (cos⁡\ θ + i sin⁡\ θ)^{6 \times -3 }}\ \hspace{8cm}\]
\[= \frac{(cos⁡\ θ + i sin⁡\ θ)^{-15}\ (cos⁡\ θ + i sin⁡\ θ)^{16}} {(cos⁡\ θ + i sin⁡\ θ)^{14}\ (cos⁡\ θ + i sin⁡\ θ)^{-18}}\ \hspace{8cm}\]
\[= (cos\ θ + i sin⁡\ θ )^{-15 + 16 + 14 – 18 }\ \hspace{10cm}\]
\[= (cos\ θ + i sin⁡\ θ )^5\ \hspace{10cm}\]
\[= cos\ 5θ + i sin⁡\ 5θ\ \hspace{10cm}\]
\[Ex8:\ Prove\ that:\ [\frac{cos⁡\ θ + i sin⁡\ θ} {sin⁡\ θ – i cos\ θ}]^4\ = 1\ \hspace{18cm}\]
\[\color {black}{Solution:}\ LHS = [\frac{cos⁡\ θ + i sin⁡\ θ} {sin⁡\ θ – i cos\ θ}]^4\ \hspace{18cm}\]
\[=\ [\frac{cos⁡\ θ + i sin⁡\ θ} {-i(cos⁡\ θ + i sin\ θ)}]^4\ \hspace{5cm}\ \because sin⁡\ θ – i cos\ θ\ =\ -i(cos⁡\ θ + i sin\ θ)\]
\[ = [\frac{(cos⁡\ θ + i sin⁡\ θ)^4} {(-i)^4(cos⁡\ θ + i sin\ θ)^4}]\ \hspace{10cm}\]
\[= \frac{(cos⁡\ θ + i sin⁡\ θ)^4} {(1)(cos⁡\ θ + i sin\ θ)^4}\ \hspace{5cm}\ \because (-i)^4\ =\ 1\]
\[ = \frac{(cos⁡\ θ + i sin⁡\ θ)^4} {(cos⁡\ θ + i sin\ θ)^4}\ \hspace{10cm}\]
\[= (cos\ θ + i sin⁡\ θ )^{4 -4 }\ \hspace{10cm}\]
\[= (cos\ θ + i sin⁡\ θ )^{0}\ \hspace{10cm}\]
\[= cos\ 0 + i sin⁡\ 0\ \hspace{10cm}\]
\[= 1\ + i(0)\ \hspace{5cm}\ \because cos\ 0\ =\ 1,\ sin\ 0\ =\ 0\]
\[= 1\ =\ RHS\ \hspace{10cm}\]