How to Apply De Moivre’s Theorem: Methods & Examples
De Moivre's Theorem is a fundamental tool in Mathematics that is used to connect two different branches of Mathematics, i.e., trigonometry and complex numbers. De Moivre's Theorem is significant in complex analysis and forms a basis of number theory. In this article, we will discuss the detailed proof of the theorem and also about the law of rational indices.
Table of Contents
Introduction
History of Abraham de Moivre
Statement of De Moivre's Theorem
Proof of De Moivre's Theorem
Applications of De Moivre's Theorem
Limitations of De Moivre's Theorem
History of Abraham De Moivre
Abraham De Moivre
Image Credit: Wikimedia
Name: Abraham de Moivre
Born: 26 May 1667
Died: 27 November 1754
Field: Mathematics
Nationality: French
Statement of De Moivre's Theorem
De Moivre's Theorem states the relationship between complex numbers and trigonometry. De Moivre’s Theorem for integral and rational powers has relation as follows:
$(\cos x+i \sin x)^{n}=\cos (n x)+i \sin (n x)$
Proof of De Moivre's Theorem
Here, we will use the method of Mathematical Induction to prove De Moivre's Theorem.
To Prove: $(\cos x+i \sin x)^{n}=\cos (n x)+i \sin (n x) \ldots(i)$
Firstly, we will see whether the result holds for $n=1$
So, For $n=1$,
we have
$(\cos x+i \sin x)^{1}=\cos (1 x)+i \sin (1 x)=\cos (x)+i \sin (x)$
which is true.
The result holds true for $n=1$.
Now let us assume that result is true for $n=k$.
Then, we have
$(\cos x+i \sin x)^{k}=\cos (k x)+i \sin (k x) \ldots$ (ii)
So, Now we will prove that result is also true for $n=k+1$.
$(\cos x+i \sin x)^{k+1}=(\cos x+i \sin x)^{k}(\cos x+i \sin x)$
$=(\cos (k x)+i \sin (k x))(\cos x+i \sin x) \quad[U \operatorname{sing}(i)]$
$=\cos (k x) \cos x-\sin (k x) \sin x+i(\sin (k x) \cos x+\cos (k x) \sin x)$
$=\cos \{(k+1) x\}+i \sin \{(k+1) x\}$
$=(\cos x+i \sin x)^{k+1}=\cos \{(k+1) x\}+i \sin \{(k+1) x\}$
Hence the result is proved.
Since the theorem is true for $n=1$ and $n=k+1$, it is true $\forall n \geq 1$.
De Moivre's Theorem When Power is in Fraction Form
If $z=r(\cos \theta+i \sin \theta)$ and $n$ is a positive integer
then De Moivre's Theorem states that: $z^{\dfrac{1}{n}}=r^{\dfrac{1}{n}}\left[\cos \left(\dfrac{2 k \pi+\theta}{n}\right)+i \sin \left(\dfrac{2 k \pi+\theta}{n}\right)\right]$
where $k=0,1,2,3, \ldots,(n-1)$.
Applications of De Moivre's Theorem
De Moivre's Theorem is used to find the nth root of complex numbers.
De Moivre's Theorem is used in computing and programming.
De Moivre's Theorem connects complex numbers with trigonometry.
Limitations of De Moivre's Theorem
De Moivre's theorem gives multiple valued results in the case of non-integers power.
De Moivre's Theorem does not work for all non-integers powers.
Solved Examples
1: Find the value of $(1-\sqrt{3} i)^{5}$ using the De Moivre formula.
Ans. Let $z=1-\sqrt{3} i=a+i b$
Its modulus is, $r=\sqrt{\left(a^{2}+b^{2}\right)}=\sqrt{(1+3)}=2$.
$\alpha=\tan ^{-1}\left|\dfrac{b}{a}\right|=\tan ^{-1} \sqrt{3}=\dfrac{\pi}{3}$
Since $a>0$ and $b<0, \theta$ is in the $4^{t h}$ quadrant.
So,
$\theta=2 \pi-\dfrac{\pi}{3}=\dfrac{5 \pi}{3}$
Thus,
$z=r(\cos \theta+i \sin \theta)=2\left(\cos \dfrac{5 \pi}{3}+i \sin \dfrac{5 \pi}{3}\right)$
Now,
$z^{5}=\left(2\left(\cos \dfrac{5 \pi}{3}+i \sin \dfrac{5 \pi}{3}\right)\right)^{5}$
By De Moivre formula,
$z^{5}=2^{5}\left(\cos \dfrac{25 \pi}{3}+i \sin \dfrac{25 \pi}{3}\right)$
$=32\left(\dfrac{1}{2}+\dfrac{\sqrt{3}}{2 i}\right)$
$=16+16 \sqrt{3} i$
2. If $z=(\cos \theta+i \sin \theta)$, show that $z^{n}+\dfrac{1}{z^{n}}=2 \cos n \theta$ and $z^{n}-\left[\dfrac{1}{z^{n}}\right]=2 i \sin n \theta$
Ans. Let $z=(\cos \theta+i \sin \theta)$.
By De Moivre's Theorem,
$z^{n}=(\cos \theta+i \sin \theta)^{n}=\cos n \theta+i \sin n \theta$
$z^{n}=(\cos \theta+i \sin \theta)^{n}=\cos n \theta+i \sin \theta$
$\dfrac{1}{z^{n}}=z^{-n}=\operatorname{cosn} \theta-i \sin \theta$
Therefore,
$z^{n}+\dfrac{1}{z^{n}}=(\operatorname{cosn} \theta+i \sin \theta)+(\operatorname{cosn} \theta-i \sin \theta \theta) $
$z^{n}+\dfrac{1}{z^{n}}=2 \operatorname{cosn} \theta$
Similarly,
$z^{n}-\dfrac{1}{z^{n}}=(\operatorname{cosn} \theta+i \sin n \theta)+(\cos n \theta-i \sin n \theta) $
$z^{n}-\dfrac{1}{z^{n}}=2 i \sin n \theta$
3. Solve $\left(\sin \dfrac{\pi}{6}+i \cos \dfrac{\pi}{6}\right)^{18}$.
Ans. We have,
$\sin \dfrac{\pi}{6}+i \cos \dfrac{\pi}{6}=i\left(\cos \dfrac{\pi}{6}-i \sin \dfrac{\pi}{6}\right)$.
Raising to the power of 18 on both sides it gives,
$\left(\sin \dfrac{\pi}{6}+i \cos \dfrac{\pi}{6}\right)^{18}=(i)^{18}\left(\cos \dfrac{\pi}{6}-i \sin \dfrac{\pi}{6}\right)^{18} $
$=(-1)\left(\cos \dfrac{18 \pi}{6}-i \sin \dfrac{18 \pi}{6}\right) $
$=-(\cos 3 \pi-i \sin 3 \pi)=1+0 i$
Therefore, $\left(\sin \dfrac{\pi}{6}+i \cos \dfrac{\pi}{6}\right)^{18}=1$.
Conclusion
In the article, we have discussed the detailed proof of De Moivre's Theorem for natural and rational roots. De Moivre’s Theorem is an important component of number theory and has a wide range of applications in computing. So, we conclude that De Moivre Theorem, which acts as a connecting link between complex analysis and trigonometry, is an important component of mathematics.
Important Points to Remember
If $n$ is integer: $(\cos x+i \sin x)^{n}=\cos (n x)+i \sin (n x)$.
If the power is in fraction form: $z^{\dfrac{1}{n}}=r^{\dfrac{1}{n}}\left[\cos \left(\dfrac{2 k \pi+\theta}{n}\right)+i \sin \left(\dfrac{2 k \pi+\theta}{n}\right)\right]$.
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FAQs on De Moivre’s Theorem for Integral and Rational Powers
1. What is De Moivre's Theorem and what is its formula?
De Moivre's theorem provides a fundamental formula for calculating powers of complex numbers. It states that for any complex number in polar form (cos θ + i sin θ) and any integer n, the following relationship holds true: (cos θ + i sin θ)ⁿ = cos(nθ) + i sin(nθ). This theorem elegantly connects complex numbers with trigonometry, simplifying the process of raising a complex number to a power.
2. What are the main applications of De Moivre's Theorem in mathematics?
De Moivre's theorem is a powerful tool with several important applications in mathematics, primarily within the topic of complex numbers. Its main uses include:
Calculating Powers: Efficiently finding large integer powers of complex numbers without repeated multiplication.
Finding Roots: Determining the 'n' distinct nth roots of a complex number, including the roots of unity.
Deriving Trigonometric Identities: Establishing formulas for multiple angles, such as expressing cos(nθ) and sin(nθ) in terms of cos(θ) and sin(θ).
3. What components are necessary to express a complex number before applying De Moivre's Theorem?
To apply De Moivre's theorem, a complex number must be expressed in its polar form, which is z = r(cos θ + i sin θ). The essential components are:
The modulus (r): The distance of the complex number from the origin in the Argand plane, calculated as r = |z|.
The argument (θ): The angle the line connecting the complex number to the origin makes with the positive real axis.
4. How does De Moivre's theorem simplify the process of finding the nth roots of a complex number?
Finding the nth roots of a complex number algebraically can be extremely difficult. De Moivre's theorem provides a clear and systematic method. By extending the theorem to fractional exponents (1/n), it generates a formula that gives all 'n' distinct roots. Geometrically, these roots are shown to be equally spaced on the circumference of a circle in the complex plane, which provides a powerful visual and computational shortcut.
5. What is the relationship between Euler's formula and De Moivre's theorem?
De Moivre's theorem can be seen as a direct consequence of Euler's formula (eⁱᶿ = cos θ + i sin θ). According to the laws of exponents, (eⁱᶿ)ⁿ = eⁱ⁽ⁿᶿ⁾. If we substitute the trigonometric forms back into this equation, we get (cos θ + i sin θ)ⁿ on the left side and cos(nθ) + i sin(nθ) on the right side. This demonstrates that De Moivre's theorem is consistent with the exponential properties of complex numbers.
6. Why is the polar form of a complex number essential for using De Moivre's theorem?
Raising a complex number in Cartesian form (a + bi) to a large power involves tedious binomial expansion. The polar form transforms this difficult multiplication problem into a simple geometric operation. When you multiply complex numbers in polar form, you multiply their moduli and add their arguments. De Moivre's theorem is the generalisation of this principle for 'n' multiplications, making it far more efficient and conceptually clear than the Cartesian approach.
7. Does De Moivre's theorem apply to non-integer powers?
The standard statement of De Moivre's theorem is for integer exponents. However, its logic can be extended to rational exponents (like p/q), which is how we find the roots of complex numbers. When the exponent 'n' is a rational number, the expression cos(nθ) + i sin(nθ) is not the only result but becomes one of the possible values. For an exponent like 1/q, there will be 'q' distinct values or roots.
8. How can you use De Moivre's theorem to derive a trigonometric identity for cos(3θ)?
You can derive the identity for cos(3θ) by equating two different expansions. First, apply De Moivre's theorem: (cos θ + i sin θ)³ = cos(3θ) + i sin(3θ). Next, expand (cos θ + i sin θ)³ using the binomial theorem, which results in (cos³θ - 3cosθsin²θ) + i(3cos²θsinθ - sin³θ). By equating the real parts of both results, you get the identity: cos(3θ) = cos³θ - 3cosθsin²θ, which can be further simplified to 4cos³θ - 3cosθ.
