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2.8.1 De Moivre's Theorem

Chapter 4: Chapter 2 · MATHEMATICS-VOLUME 1

. . De Moivre's Theorem De Moivre’s Theorem Given any complex number cos and any integer n , (cos sin ) Corollary ( ) (cos sin ) ( ) (cos sin ) ( ) (cos sin ) ( ) sin Now let us apply De Moivre's theorem to simplify complex numbers and to find solution of equations. De Moivre – Example .

If z  , show that z Let z  . By De Moivre's theorem , z n = cos z n = z Therefore, z + = cos + = 2cos n θ . Similarly, − = cos − = i sin θ . Example .

Simplify sin   We have, sin = i   . Raising to the power on both sides gives,   = i               i . Therefore, sin   = . Example .

Simplify       Let z = cos As | | z = | | z zz = , we get  z Complex Numbers Therefore, = z z z . Therefore,       = z = cos Example . Simplify (i) ( + i (ii) ( (i) ( + i Let + i = r . Then, we get r = ;       tan , θ = (  + i lies in the first Quadrant) Therefore + i =   Raising to power on both sides, + i            By De Moivre's theorem, + i = cos   =          + i = ( ) i (ii) ( Let i = r .

Then, we get r = α = tan θ = (  i lies in II Quadrant) Therefore, i =   Raising power on both sides, =        =                    =        =                     =   

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