Complex Number

2. De Moivre’s Theorem

De Moivre’s Theorem

De Moivre’s theorem is a fundamental result in complex number theory that expresses the power of a complex number in polar form:

$$(\cos \theta + i \sin \theta)^n = \cos(n\theta) + i\sin(n\theta)$$(cosθ+isinθ)n=cos(nθ)+isin(nθ)

where $n$n is any integer (positive, negative, or zero).

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Proof by Mathematical Induction

Base Case ( $n = 1$n=1 )

For $n = 1$n=1:

$$(\cos \theta + i \sin \theta)^1 = \cos \theta + i \sin \theta$$(cosθ+isinθ)1=cosθ+isinθ

which is trivially true.

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Inductive Step

Assume that the theorem holds for $n = k$n=k, i.e.,

$$(\cos \theta + i \sin \theta)^k = \cos(k\theta) + i \sin(k\theta)$$(cosθ+isinθ)k=cos(kθ)+isin(kθ)

Now, for $n = k+1$n=k+1:

$$(\cos \theta + i \sin \theta)^{k+1} = (\cos \theta + i \sin \theta) \cdot (\cos k\theta + i \sin k\theta)$$(cosθ+isinθ)k+1=(cosθ+isinθ)⋅(coskθ+isinkθ)

Expanding using the distributive property:

$$\cos \theta \cos k\theta + i \cos \theta \sin k\theta + i \sin \theta \cos k\theta + i^2 \sin \theta \sin k\theta$$cosθcoskθ+icosθsinkθ+isinθcoskθ+i2sinθsinkθ

Since $i^2 = -1$i2=−1, we simplify:

$$\cos \theta \cos k\theta - \sin \theta \sin k\theta + i (\cos \theta \sin k\theta + \sin \theta \cos k\theta)$$cosθcoskθ−sinθsinkθ+i(cosθsinkθ+sinθcoskθ)

Using the trigonometric identities:

$$\cos(A + B) = \cos A \cos B - \sin A \sin B$$cos(A+B)=cosAcosB−sinAsinB $$\sin(A + B) = \sin A \cos B + \cos A \sin B$$sin(A+B)=sinAcosB+cosAsinB

we get:

$$\cos((k+1)\theta) + i \sin((k+1)\theta)$$cos((k+1)θ)+isin((k+1)θ)

Thus, by induction, De Moivre’s theorem holds for all positive integers $n$n.

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Extension to Negative Powers

For $n = -p$n=−p, where $p$p is positive:

$$(\cos \theta + i \sin \theta)^{-p} = \frac{1}{(\cos \theta + i \sin \theta)^p}$$(cosθ+isinθ)−p=(cosθ+isinθ)p1​

Using the identity:

$$\cos(-p\theta) = \cos(p\theta), \quad \sin(-p\theta) = -\sin(p\theta)$$cos(−pθ)=cos(pθ),sin(−pθ)=−sin(pθ)

we obtain:

$$(\cos \theta + i \sin \theta)^{-p} = \cos(-p\theta) + i\sin(-p\theta) = \cos(p\theta) - i\sin(p\theta)$$(cosθ+isinθ)−p=cos(−pθ)+isin(−pθ)=cos(pθ)−isin(pθ)

Thus, De Moivre’s theorem is valid for all integer values of $n$n.

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Applications of De Moivre’s Theorem

1. Finding Powers of Complex Numbers

   • Example: Compute $(1 + i)^4$(1+i)4.

   • Convert to polar form:

     $$1 + i = \sqrt{2} (\cos 45^\circ + i \sin 45^\circ)$$1+i=2​(cos45∘+isin45∘)

   • Apply De Moivre’s theorem:

     $$(1+i)^4 = (\sqrt{2})^4 (\cos 180^\circ + i \sin 180^\circ) = 4(-1 + 0i) = -4$$(1+i)4=(2​)4(cos180∘+isin180∘)=4(−1+0i)=−4

2. Finding Roots of Complex Numbers

   • To find the $n$nth roots of a complex number $z = r (\cos \theta + i \sin \theta)$z=r(cosθ+isinθ), use:

     $$z^{1/n} = r^{1/n} \left[ \cos \left( \frac{\theta + 2k\pi}{n} \right) + i \sin \left( \frac{\theta + 2k\pi}{n} \right) \right], \quad k = 0,1,2,\dots, n-1$$z1/n=r1/n[cos(nθ+2kπ​)+isin(nθ+2kπ​)],k=0,1,2,…,n−1