1/04/2023

How can De Moivre's theorem be described? What is the scope of this theorem? Examples for roots and powers.

Finding powers of complex numbers is greatly simplified using De Moivre’s Theorem. 

According to the De Moivre’s Theorem 

If Z = r(cos θ + isin θ) is a complex number, then Zn = rn[cos(nθ) + isin(nθ)] 

where n is a positive integer. Zn = rn cis(nθ) 

How does it come from? To understand this theorem, we must know the products of complex numbers in polar form and the quotients of complex numbers in polar form first.

Recall that : 

    sin(α + β) = sin(α)cos(β) + cos(α)sin(β)

    sin(α – β) = sin(α)cos(β) – cos(α)sin(β)

    cos(α + β) = cos(α)cos(β) – sin(α)sin(β)

    cos(α – β) = cos(α)cos(β) + sin(α)sin(β)


 


(1+ \sqrt[]{3}i)^3

r= \sqrt[]{ 1^{2}+( \sqrt[]{3})^2 } =2

tan \theta= \sqrt[]{3}  ,  3(\frac{ \pi }{3})= \pi

(1+ \sqrt[]{3}i)^3=2^3(cos\pi +sin \pi)=-8


(\sqrt[]{2}+ \sqrt[]{2}i)^3

r= \sqrt[]{ (\sqrt[]{2})^{2}+( \sqrt[]{2})^2 } =2

tan \theta= 1  ,  \theta= \pi/4

(\sqrt[]{2}+ \sqrt[]{2}i)^3=2^3(cos3\pi/4 +sin3\pi/4)=\sqrt[]{2}(-4+4i)


(1+ \sqrt[]{3}i)^{ \frac{1}{2}}

r= \sqrt[]{(1)^2+ (\sqrt[]{3})^2 } =2

tan \theta= \sqrt[]{3} ,   \theta= \frac{ \pi }{3}

(1+ \sqrt[]{3}i)^{ \frac{1}{2}}   =(2^{ \frac{1}{2} })(cos(\frac{1}{2})(\frac{ \pi }{3})+isin(\frac{1}{2})(\frac{ \pi }{3}))  =  \sqrt[]{2} ( \frac{\sqrt[]{3}}{2} + \frac{1}{2} i)


(\sqrt[]{2}+ \sqrt[]{2}i)^{ \frac{1}{2}}

r= \sqrt[]{( \sqrt[]{2})^2+(\sqrt[]{2})^2} =2

tan \theta=1  ,  \theta= \frac{ \pi }{4}

(\sqrt[]{2}+ \sqrt[]{2}i)^{ \frac{1}{2}}   = \sqrt[]{2}(cos \frac{ \pi }{8} +isin \frac{ \pi }{8})= \sqrt[]{2}( \sqrt[]{\frac{1+ \frac{\sqrt[]{2}}{2} }{2} } +\sqrt[]{\frac{2-\sqrt[]{2} }{2} })

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