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Syllabus:
\[\text{Definition of a complex number- Real and imaginary parts – Modulus and}\ \hspace{10cm}\\ \text{argument – Polar form of a complex number – Conjugate of a complex number -}\ \hspace{10cm}\\ \text{Representation of complex numbers on Argand plane – Addition, subtraction, multiplication}\ \hspace{10cm}\\ \text{and division of complex numbers – De-Moivre’s theorem (without proof) – Simple problems.}\ \hspace{5cm}\]
\[\color {royalblue} {Introduction}:\ \hspace{20cm}\]
\[In\ earlier\ classes,\ we\ have\ studied\ linear\ equations\ in\ one\ and\ two\ variables\ and\ quadratic\]
\[equations\ in\ one\ variable.\ We\ have\ seen\ that\ the\ equation\ x^2 + 1 = 0\ has\ no\ real solution\ as\ x^2 + 1 = 0\]
\[gives\ x^2 = – 1\ and square\ of\ every\ real\ number\ is\ non-negative.\ So,\ we\ need\ to\ extend\ the\ real\ number\]
\[system\ to\ a\ larger\ system\ so\ that\ we\ can\ find\ the\ solution\ of\ the\ equation\ x^2 = – 1.\]
\[Let\ us\ denote\ \sqrt{-1}\ by\ the\ symbol\ ‘i ‘.\ Then,\ we\ have\ i^2 = -1.\ This\ means\ that\]
\[‘i ‘\ is\ a\ solution\ of\ the\ equation\ x^2 + 1 = 0.\]
\[\color {purple}{Example\ 1:}\ \color {red}{Find\ the\ value\ of}\ i^3\ +\ i^5\ \hspace{15cm}\]
\[\color {blue}{Solution:}\ i^3\ +\ i^5\ \hspace{19cm}\]
\[ =\ i^2\ .\ i\ +\ i^4\ .\ i\ \hspace{15cm}\]
\[ =\ (-1)\ .\ i\ +\ (1)\ .\ i\ \hspace{15cm}\]
\[ =\ -i\ +\ i\ \hspace{15cm}\]
\[ =\ 0\ \hspace{15cm}\]
\[\color {royalblue} {Definition\ of\ complex\ number}:\ \hspace{20cm}\]
\[A\ number\ which\ is\ of\ the\ form\ a + ib\ where\ a, b ∈ R\ and\ i^2 = -1\ is\ called\ a\ complex\]
\[number\ and\ it\ is\ denoted\ by\ Z.\]
\[If\ Z = a + ib ,\ then\ a\ is\ called\ the\ real\ part\ of\ Z\ and\ b\ is\ called\ the\ imaginary\ part\ of\ Z.\]
\[Re ( Z ) = a\ and\ Im ( Z ) = b\]
\[Ex:\ If\ Z = 3 + 4i\ then\ Re ( Z ) = 3\ and\ Im ( Z ) = 4\]
\[\color {royalblue} {Conjugate\ of\ a\ complex\ number}:\ \hspace{20cm}\]
\[If\ Z = a + ib\ then\ the\ conjugate\ of\ Z\ is\ denoted\ by\ \bar z\ and\ is\ defined\ as\ \bar z\ = a – ib\]
\[\color {purple} {Example\ 2:}\ \color {red} {Find\ the\ Real\ and\ Imaginary\ parts\ of}\ \hspace{18cm}\]
\[\hspace{1cm}\ a)\ \frac{2-3i}{5}\ \hspace{0.5cm}\ b)\ i\ \hspace{0.5cm}\ c)\ 2\ \hspace{0.5cm}\ d)\ (1-i)(2-i)\]
\[\color {blue}{Solution:}\ \hspace{19cm}\]
\[a).\ \hspace{0.5cm}\ Let\ z\ =\ \frac{2-3i}{5}\ =\ \frac{2}{5}\ -\ \frac{3}{5}\ i\ \implies\ RP\ =\ \frac{2}{5}\ and\ IP\ =\ -\ \frac{3}{5}\]
\[b).\ \hspace{0.5cm}\ Let\ z\ =\ i\ =\ 0\ + 1\ i\ \implies\ RP\ =\ 0\ and\ IP\ =\ 1\ \hspace{3cm}\]
\[c).\ \hspace{0.5cm}\ Let\ z\ =\ 2\ =\ 2\ + 0\ i\ \implies\ RP\ =\ 2\ and\ IP\ =\ 0\ \hspace{3cm}\]
\[d).\ \hspace{0.5cm}\ Let\ z\ =\ (1-i)(1-2i)\ =\ 1\ -\ 2i\ -i\ +\ 2\ i^2\ =\ 1\ -\ 3i\ -\ 2\ \hspace{1cm}\\ \hspace{1cm}\ =\ -\ 1\ -\ 3i\ \implies\ RP\ =\ -\ 1:\ and\ IP\ =\ -\ 3\]
\[\color {purple} {Example\ 3:}\ \color {red} {Find\ the\ conjugate\ of}\ (a)\ 2\ -\ 3\ i\ \hspace{0.5cm}\ (b)\ \frac{3-5i}{2}\ \hspace{10cm}\]
\[\color {blue}{Solution:}\ \hspace{19cm}\]
\[a)\ \hspace{1cm}\ Let\ z\ =\ 2\ -\ 3i\ \hspace{15cm}\]
\[\hspace{1.3cm}\ \therefore\ \bar z\ =\ 2\ +\ 3i\ \hspace{14cm}\]
\[b)\ \hspace{1cm}\ Let\ z\ =\ \frac{3\ -\ 5i}{2}\ =\ \frac{3}{2}\ -\ \frac{5}{2}\ i\ \hspace{12cm}\]
\[\hspace{1.3cm}\ \therefore\ \bar z\ =\ \frac{3}{2}\ +\ \frac{5}{2}\ i \hspace{14cm}\]
\[\color {brown} {Addition\ of\ two\ Complex\ numbers}:\ \hspace{20cm}\]
\[Let\ Z_1= a + ib,\ Z_2 = c + id\ be\ any\ two\ complex\ numbers.\ Then\]
\[Z_1 + Z_2 = a + ib\ +\ c + id\]
\[=\ a + c + i(b + d )\]
\[\color {brown} {Difference\ of\ two\ Complex\ numbers}:\ \hspace{20cm}\]
\[Let\ Z_1= a + ib,\ Z_2 = c + id\ be\ any\ two\ complex\ numbers.\ Then\]
\[Z_1 – Z_2 = a + ib\ -\ (c + id)\]
\[ = a + ib – c – id\]
\[=\ a – c + i(b – d )\]
\[\color {purple} {Example\ 4:}\ If\ Z_1 = (-1 , 2),\ Z_2 = (-3 , 4),\ \color {red} {find\ 2\ Z_1 -\ 3\ Z_2}\ \hspace{18cm}\]
\[\color {blue}{Solution:}\ Z_1 = (-1\ ,\ 2)\ =\ -1\ +\ 2i\ \hspace{19cm}\]
\[ Z_2 = (-3\ ,\ 4)\ =\ -3\ +\ 4i\ \hspace{19cm}\]
\[2Z_1 – 3Z_2 = 2(-1 + 2i ) – 3 (-3 + 4i)\ \hspace{10cm}\]
\[2Z_1 – 3 Z_2\ =\ -2\ +\ 4\ i +\ 9\ – 12i\ \hspace{10cm}\]
\[2Z_1\ -\ 3Z_2\ =\ -2\ +\ 9\ +\ 4i\ -\ 12i\ \hspace{10cm}\]
\[2Z_1 – 3Z_2\ =\ 7\ -\ 8i\ =\ (7,\ -\ 8)\ \hspace{10cm}\]
\[\color {brown} {Multiplication\ of\ two\ Complex\ numbers}:\ \hspace{20cm}\]
\[Let\ Z_1= a + ib,\ Z_2 = c + id\ be\ any\ two\ complex\ numbers.\ Then\]
\[Z_1 Z_2 = (a + ib) (c + id)\]
\[ = ac + iad + ibc + i^2bd\]
\[ = ac + i(ad + bc) – bd\]
\[ = (ac – bd) + i(ad + bc)\]
\[\color {brown} {(ii)\ Division\ of\ two\ Complex\ numbers}:\ \hspace{20cm}\]
\[Let\ Z_1= a + ib,\ Z_2 = c + id\ be\ any\ two\ complex\ numbers.\ Then\]
\[\frac{z_1}{z_2}\ =\frac{a+ib}{c+id}\ ×\ \frac{c-id}{c-id}\]
\[= \frac{ac- iad + ibc – i^2bd}{c^2 – i^2d^2}\]
\[= \frac{ac + i (bc – ad) + bd}{c^2 – i^2d^2}\]
\[= \frac{ac + bd + i(bc – ad)}{c^2 + d^2}\]
\[= \frac{ac + bd} {c^2 + d^2}\ +\ i\ \frac{bc – ad}{c^2 + d^2}\]
\[\color {purple} {Example\ 5:}\ \color {red} {Find\ the\ Real\ and\ Imaginary\ parts\ of}\ \frac{1}{2+ 3i} \hspace{18cm}\]
\[\color {blue}{Solution:}\ z = \frac{1}{2+ 3i}\ ×\ \frac{2 – 3i}{2 – 3i}\ \hspace{18cm}\]
\[ = \frac{2 – 3i}{(2)^2 + (3)^2}\ \hspace{3cm}\ \because [(a +ib)(a – ib)\ =\ (a)^2 + (b)^2]\]
\[ = \frac{2 – 3i}{4+ 9}\ \hspace{15cm}\]
\[ = \frac{2 – 3i}{13}\ \hspace{15cm}\]
\[ Z = \frac{2}{13}\ -\ i\ \frac{3}{13}\ \hspace{15cm}\]
\[Re(Z)\ =\ \frac{2}{13};\ Im(Z)\ =\ -\frac{3}{13}\ \hspace{15cm}\]
\[\color {purple} {Example\ 6:}\ \color {red} {Find\ the\ Real\ and\ Imaginary\ parts\ of}\ \frac{1+ i}{1- i}\ \hspace{18cm}\]
\[\color {blue}{Solution:}\ z = \frac{1+ i}{1- i}\ ×\ \frac{1+ i}{1+i}\ \hspace{18cm}\]
\[ = \frac{1 + i + i + (i)^2}{(1)^2 + (1)^2}\ \hspace{3cm}\ \because [(a +ib)(a – ib)\ =\ (a)^2 + (b)^2]\]
\[ = \frac{1 + 2i – 1}{2}\ \hspace{15cm}\]
\[ = \frac{2i}{2}\ \hspace{15cm}\]
\[ = i\ =\ 0\ +\ 1\ i\ \hspace{15cm}\]
\[Re(Z)\ =\ 0;\ Im(Z)\ =\ 1\ \hspace{15cm}\]
\[\color {purple} {Example\ 7:}\ \color {red} {Find\ the\ conjugate\ of}\ \frac{7}{5+ 2i} \hspace{18cm}\]
\[\color {blue}{Solution:}\ z = \frac{7}{5+ 2i}\ ×\ \frac{5 – 2i}{5 – 2i}\ \hspace{18cm}\]
\[ = \frac{35 – 14i}{(5)^2 + (2)^2}\ \hspace{3cm}\ \because [(a +ib)(a – ib)\ =\ (a)^2 + (b)^2]\]
\[ = \frac{35 – 14i}{25+ 4}\ \hspace{15cm}\]
\[ = \frac{35 – 14i}{29}\ \hspace{15cm}\]
\[ Z = \frac{35}{29}\ -\ i\ \frac{14}{29}\ \hspace{15cm}\]
\[ conjugate\ of\ Z\ is\ \bar z\ = \frac{35}{29}\ +\ i\ \frac{14}{29}\ \hspace{10cm}\]
\[\color {purple} {Example\ 8:}\ \color {red} {Express}\ \frac{(1+ i)(1 + 2i)}{1+ 3i}\ \color{red}{in\ the\ form\ of\ a\ +\ ib}\ \hspace{18cm}\]
\[\color {blue}{Solution:} \hspace{20cm}\]
\[Let\ z = \frac{(1+ i)(1 + 2i)}{1+ 3i}\ \hspace{18cm}\]
\[ = \frac{1+ 2i + i + 2i^2}{1+ 3i}\ \hspace{15cm}\]
\[ = \frac{1+ 3i – 2}{1+ 3i}\ \hspace{15cm}\]
\[ = \frac{-1+ 3i}{1+ 3i}\ \hspace{15cm}\]
\[= \frac{-1+3i}{1+ 3i}\ ×\ \frac{1 – 3i}{1 – 3i}\ \hspace{15cm}\]
\[ = \frac{-1 + 3i + 3i – 9i^2}{(1)^2 + (3)^2}\ \hspace{3cm}\ \because [(a +ib)(a – ib)\ =\ (a)^2 + (b)^2]\]
\[ = \frac{-1 + 6i + 9 }{1+ 9}\ \hspace{4cm}\ \because [i^2\ =\ -1]\]
\[ = \frac{8 + 6i}{10}\ \hspace{10cm}\]
\[ Z = \frac{4}{5}\ +\ i\ \frac{3}{5}\ \hspace{15cm}\]
\[Re(Z)\ =\ \frac{4}{5};\ Im(Z)\ =\ \frac{3}{5}\ \hspace{15cm}\]
\[\color {royalblue} {Polar\ form\ of\ a\ Complex\ number}:\ \hspace{20cm}\]
\[Z = r cosθ + i r sin θ = r ( cosθ + i sin θ )\]
\[\color {royalblue} {Modulus\ and\ Amplitude\ (or)\ Argument\ of\ a\ Complex\ number}:\ \hspace{20cm}\]
\[If\ Z = a + ib\ then\ Modulus is |z| = \sqrt{a^2 + b^2}\ and\ Amplitude\ is\ θ = tan^{-1} (\frac{b}{a})\]
\[\color {purple} {Example\ 9:}\ \color {red} {Find\ the\ modulus\ and\ amplitude\ of}\ 1 + \sqrt{3}i\ \hspace{18cm}\]
\[\color {blue}{Solution:} \hspace{20cm}\]
\[Let\ z = 1 + \sqrt{3}i\ = a\ + ib\ \hspace{15cm}\]
\[a = 1,\ b\ = \sqrt{3}\ \hspace{15cm}\]
\[\color {brown} {T0\ find\ modulus}:\ \hspace{18cm}\]
\[|z| = \sqrt{a^2 + b^2}\ \hspace{12cm}\]
\[ = \sqrt{(1)^2 + (\sqrt{3})^2}\ \hspace{12cm}\]
\[ = \sqrt{1 + 3}\ =\ \sqrt{4}\ \hspace{12cm}\]
\[|z| = 2\ \hspace{15cm}\]
\[\color {brown} {To\ find\ amplitude}:\ \hspace{18cm}\]
\[θ = tan^{-1} (\frac{b}{a})\ =\ tan^{-1} \frac{\sqrt{3}}{1}\ =\ tan^{-1} \sqrt{3}\]
\[θ = 60^0\ \hspace{15cm}\]
\[\color {purple} {Example\ 10:}\ \color {red} {Find\ the\ modulus\ and\ amplitude\ of}\ 1 + i\ \hspace{18cm}\]
\[\color {blue}{Solution:} \hspace{20cm}\]
\[Let\ z = 1 + i\ = a\ + ib\ \hspace{15cm}\]
\[a = 1,\ b\ = 1\ \hspace{15cm}\]
\[\color {brown} {T0\ find\ modulus}:\ \hspace{18cm}\]
\[|z| = \sqrt{a^2 + b^2}\ \hspace{12cm}\]
\[ = \sqrt{(1)^2 + (1)^2}\ \hspace{12cm}\]
\[ = \sqrt{1 + 1}\ =\ \sqrt{2}\ \hspace{12cm}\]
\[|z| =\sqrt{2}\ \hspace{15cm}\]
\[\color {brown} {To\ find\ amplitude}:\ \hspace{18cm}\]
\[θ = tan^{-1} (\frac{b}{a})\ =\ tan^{-1} \frac{1}{1}\ =\ tan^{-1} {1}\]
\[θ = 45^0\ \hspace{15cm}\]
DE-MOIVRE’S THEOREM
\[\color {royalblue} {De-Moivre’s\ Theorem( Statement\ only)}:\ \hspace{20cm}\]
\[( i )\ If\ n\ is\ an\ integer\ positive\ or\ negative\ then\ (cos\ θ + i sin\ θ )^n = cos\ n θ + i sin\ n θ\]
\[( ii )\ If\ n\ is\ a\ fraction,\ then\ cos\ n θ + i sin\ n θ\ is\ one\ of\ the\ values\ of\ (cos\ θ + i sin\ θ )^n\]
\[\color {royalblue} {Results}:\ \hspace{20cm}\]
\[1 )\ (cos\ θ + i sin\ θ )^{-n} = cos\ n θ- i sin\ n θ\]
\[2)\ \frac{1}{cos\ θ + i sin\ θ} = (cos\ θ + i sin\ θ )^{-1 } = cos\ θ- i sin\ θ\]
\[3)\ \frac{1}{cos\ θ – i sin\ θ} = (cos\ θ – i sin\ θ )^{-1 } = cos\ θ + i sin\ θ\]
\[\color {royalblue} {Note}:\ \hspace{20cm}\]
\[1)\ (cos\ θ_1 + i sin\ θ_1) (cos\ θ_2 + i sin\ θ_2) = cos(θ_1+θ_2) + i sin (θ_1 +θ_2)\]
\[2)\ \frac{cos\ θ_1 + i sin\ θ_1}{cos\ θ_2 + i sin\ θ_2} = cos(θ_1-θ_2) + i sin (θ_1 -θ_2)\]
\[\color {purple} {Example\ 11:}\ If\ a = cos\ α + i sin\ α,\ b = cos\ β + i sin\ β ,\ \color {red} {find\ \frac{a}{b}}\ \hspace{18cm}\]
\[\color {blue}{Solution:}\ a = cos\ α + i sin\ α,\ b = cos\ β + i sin\ β\ \hspace{16cm}\]
\[\frac{a}{b}\ = \frac{cos\ α + i sin\ α}{cos\ β + i sin\ β}\ \hspace{10cm}\]
\[\frac{a}{b}\ = cos\ (α – β) + i sin\ (α – β)\ \hspace{10cm}\]
\[\color {purple} {Example\ 12:}\ If\ x = cos\ \theta + i sin\ \theta,\ \color {red} {find\ the\ value\ of\ x^m\ +\ \frac{1}{x^m}}\ \hspace{18cm}\]
\[\color {blue}{Solution:}\ x = cos\ \theta + i sin\ \theta, \hspace{18cm}\]
\[x^m = (cos\ \theta + i sin\ \theta,)^m =\ cos\ m\ \theta + i sin\ m\ \theta\ \hspace{10cm}\]
\[\frac{1}{x^m}\ = \frac{1}{cos\ m\ \theta + i sin\ m\ \theta}\ = cos\ m\ \theta – i sin\ m\ \theta \hspace{10cm}\]
\[x^m +\ \frac{1}{x^m}\ =\ cos\ m \theta + i sin\ m\ \theta\ +\ cos\ m \theta – i sin\ m\ \theta\ \hspace{10cm}\]
\[= 2 cos\ m \theta\ \hspace{10cm}\]
\[\boxed{x^m\ +\ \frac{1}{x^m}\ =\ 2\ cos\ m \theta}\]
\[\color {purple}{Example\ 13:}\ \color {red} {Simplify\ :}\ \frac{cos\ 5 θ – i sin\ 5 θ} {cos\ 3 θ + i sin\ 3 θ}\ \hspace{18cm}\]
\[\color {blue}{Solution:}\ \frac{cos\ 5 θ – i sin\ 5 θ} {cos\ 3 θ + i sin\ 3 θ}= \frac{(cos\ θ + i sin\ θ)^-5} {(cos\ θ + i sin\ θ)^3} \hspace{18cm}\]
\[= (cos\ θ + i sin\ θ )^{-5 -3 }\ \hspace{10cm}\]
\[= (cos\ θ + i sin\ θ )^-8\ \hspace{10cm}\]
\[= cos\ 8θ – i sin\ 8θ\ \hspace{10cm}\]
\[\boxed{\frac{cos\ 5 θ – i sin\ 5 θ} {cos\ 3 θ + i sin\ 3 θ}\ =\ cos\ 8θ – i sin\ 8θ}\]
\[\color {purple} {Example\ 14:}\ \color {red} {Simplify:}\ \frac{cos\ 5 θ + i sin\ 5 θ} {cos\ 3 θ – i sin\ 3 θ}\ \hspace{18cm}\]
\[\color {blue}{Solution:}\ \frac{cos\ 5 θ + i sin\ 5 θ} {cos\ 3 θ + i sin\ 3 θ}= \frac{(cos\ θ + i sin\ θ)^5} {(cos\ θ – i sin\ θ)^{-3}} \hspace{18cm}\]
\[= (cos\ θ + i sin\ θ )^{5 +3 }\ \hspace{10cm}\]
\[= (cos\ θ + i sin\ θ )^8\ \hspace{10cm}\]
\[= cos\ 8θ + i sin\ 8θ\ \hspace{10cm}\]
\[\color {purple} {Example\ 15:}\ \color {red} {Simplify\ using\ DeMoivre’s\ theorem:}\ \frac{(cos\ 3θ – i sin\ 3θ)^5\ (cos\ 4θ + i sin\ 4θ)^4} {(cos\ 2θ + i sin\ 2θ)^7\ (cos\ 3θ – i sin\ 3θ)^6}\ \hspace{10cm}\]
\[\color {blue}{Solution:}\ \frac{(cos\ 3θ – i sin\ 3θ)^5\ (cos\ 4θ + i sin\ 4θ)^4} {(cos\ 2θ + i sin\ 2θ)^7\ (cos\ 3θ – i sin\ 3θ)^6}\ \hspace{18cm}\]
\[= \frac{(cos\ θ + i sin\ θ)^{5 \times -3}\ (cos\ θ + i sin\ θ)^{4 \times 4}} {(cos\ θ + i sin\ θ)^{2 \times 7}\ (cos\ θ + i sin\ θ)^{6 \times -3 }}\ \hspace{8cm}\]
\[= \frac{(cos\ θ + i sin\ θ)^{-15}\ (cos\ θ + i sin\ θ)^{16}} {(cos\ θ + i sin\ θ)^{14}\ (cos\ θ + i sin\ θ)^{-18}}\ \hspace{8cm}\]
\[= (cos\ θ + i sin\ θ )^{-15 + 16 + 14 – 18 }\ \hspace{10cm}\]
\[= (cos\ θ + i sin\ θ )^5\ \hspace{10cm}\]
\[= cos\ 5θ + i sin\ 5θ\ \hspace{10cm}\]
\[\boxed{\frac{(cos\ 3θ – i sin\ 3θ)^5\ (cos\ 4θ + i sin\ 4θ)^4} {(cos\ 2θ + i sin\ 2θ)^7\ (cos\ 3θ – i sin\ 3θ)^6}\ =\ cos\ 5θ + i sin\ 5θ}\]
\[\color {purple} {Example\ 16:}\ \color {red} {Simplify\ using\ DeMoivre’s\ theorem:}\ \frac{(cos\ 3θ + i sin\ 3θ)^{-5}\ (cos\ 2θ + i sin\ 2θ)^4} {(cos\ 4θ – i sin\ 4θ)^{-2}\ (cos\ 5θ – i sin\ 5θ)^3}\ \hspace{10cm}\]
\[\color {blue}{Solution:}\ \frac{(cos\ 3θ + i sin\ 3θ)^{-5}\ (cos\ 2θ + i sin\ 2θ)^4} {(cos\ 4θ – i sin\ 4θ)^{-2}\ (cos\ 5θ – i sin\ 5θ)^3}\ \hspace{18cm}\]
\[= \frac{(cos\ θ + i sin\ θ)^{-5 \times 3}\ (cos\ θ + i sin\ θ)^{4 \times 2}} {(cos\ θ + i sin\ θ)^{-2 \times -4}\ (cos\ θ + i sin\ θ)^{3 \times -5 }}\ \hspace{8cm}\]
\[= \frac{(cos\ θ + i sin\ θ)^{-15}\ (cos\ θ + i sin\ θ)^8} {(cos\ θ + i sin\ θ)^8\ (cos\ θ + i sin\ θ)^{-15}}\ \hspace{8cm}\]
\[= (cos\ θ + i sin\ θ )^{-15 + 8 – 8 + 15 }\ \hspace{10cm}\]
\[= (cos\ θ + i sin\ θ )^0\ \hspace{10cm}\]
\[= cos\ 0θ + i sin\ 0θ\ = 1\ \hspace{10cm}\]
\[\boxed{\frac{(cos\ 3θ + i sin\ 3θ)^{-5}\ (cos\ 2θ + i sin\ 2θ)^4} {(cos\ 4θ – i sin\ 4θ)^{-2}\ (cos\ 5θ – i sin\ 5θ)^3}\ =\ 1}\]
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