De Moivre's Theorem | Proof | examples | complex numbers| bsc 1st year| engineering mathematics

This video lecture De Moivre's Theorem | Proof | application examples | complex numbers| bsc 1st year| engineering mathematics is an important topic of complex numbers complex analysis method by SM Yousuf calculas and analytic geometry ADP CS. In this video we learn de Moivre's Theorem Proof and application examples of de Moivre's Theorem.
De Moivre’s Theorem Statement
De Moivre’s Theorem is a special theorem of complex numbers which is used to expand the complex number raised to any integer. De Moivre’s Formula states that for all real values of x,
(cos x + i.sinx)n = cos(nx) + i.sin(nx)
where, n is any integer
De Moivre’s Formula
De Moivre’s Formula for complex numbers is, for any real value of x,
(cos x + i.sinx)n = cos(nx) + i.sin(nx)
Also, we know that,
cos x + i.sinx = eix
Now,
(eix)n = einx
Note, “n” in the above formula is an integer, and “i” is an imaginary number iota. Such that,
i = √(-1)
De Moivre’s Formula is shown in the image added below,
De Moivre’s Formula
De Moivre’s Theorem Proof
DeMoivre’s Theorem can be proved with the help of mathematical induction as follows:

P(n) = (cos x + i.sinx)n = cos(nx) + i.sin(nx) ⇢ (1)

Step 1: For n = 1,

(cos x + i sin x)1 = cos(1x) + i sin(1x) = cos(x) + i sin(x),

Which is true. thus, P(n) is true for n = 1.

Step 2: Assume P(k) is true

(cos x + i.sin x)k = cos(kx) + i.sin(kx) ⇢ (2)

Step 3: Now we have to prove P(k1) is true.

(cos x + i.sin x)k+1 = (cos x + i.sin x)k(cos x + i sin x)
= [cos (kx) + i.sin (kx)].[cos x + i.sin x] ⇢ [Using (1)]
= cos (kx).cos x − sin (kx).sin x + i [sin (kx).cos x + cos (kx). = cos {(k+1)x} + i.sin {(k+1)x}
= cos {(k+1)x} + i.sin {(k+1)x}
Thus, P(k+1) is also true and by the principle of mathematical induction P(n) is true.
Uses of De Moivre’s Theorem
De Moivre’s Theorem is used for various purposes. Some of its most important uses are,
Finding the Roots of Complex Numbers.
Finding the relationships between Powers of
Trigonometric Functions
and Trigonometric Angles.
Solving the Power of Complex Numbers.
Now, let’s learn about them with the help of examples.
Finding the Roots of Complex Numbers
The polar form of the complex number is,
z = r(cos x + i sin x)
Then for nth root of the complex number
z1/n = r1/n(cos x + i sin x)1/n
z1/n = r1/n[cos (x + 2kπ)/n + i sin (x + 2kπ)/n]
where k = 0, 1, 2, 3, …
Power of Complex Numbers
We can easily solve the power of Complex numbers using De Moivre’s Theorem. This can be understood using the example as follows,
Example: Evaluate (√3 + i)200
Solution:
Let, z = √3 + i comparing with z = x + iy
x = √3, y = 1
Also, z = r(cos θ + i sin θ)
r = √(x2 + y2) = √[(√3)2 + 12]
r = 2
θ = tan-1(y/x) = tan-1(1/√3) = π/6
z = r(cos θ + i sin θ)
z = 2(cos π/6 +i.sin π/6)
z200 = [2(cos π/6 +i.sin π/6)]200
z200 = [2]200[(cos π/6 +i.sin π/6)]200
Using DeMoivre’s Theorem
z200 = [2]200[(cos 200π/6 +i.sin 200π/6)]
z200 = [2]200[-1/2 – i√3/2]
z200 = [2]200[1/2 + i√3/2]
Solved Examples on De Moivre’s Theorem
Example : Expand (1 + i)100.
Solution:
Let, z = 1 + i comparing with z = x + iy
x = 1, y = 1
Also, z = r(cos θ + i sin θ)
r = √(x2 + y2) = √[12 + 12]
r = √2
θ = tan-1(y/x) = tan-1(1/1) = π/4
z = r(cos θ + i sin θ)
z = √2(cos π/4 +i.sin π/4)
z100 = [2(cos π/4 +i.sin π/4)]100
z100 = [2]50[(cos π/4 +i.sin π/4)]100
Using DeMoivre’s Theorem
z100 = [2]50[(cos 100π/4 +i.sin 100π/4)]
z100 = [2]50[-1 + i.0]
z100 = [2]50
FAQs on De Moivre’s Theorem
Q: What is De Moivre’s Theorem?
Answer:The De Moivre’s theorem is the basic theorem used in complex numbers for solving various problems. The De Moivre’s theorem states that,
(cos x + i sin x)n = cos(nx) + i sin(nx)
Q: What are the uses of De Moivre’s Theorem?
Answer:The various uses of De Moivre’s theorem include the solving of complex roots, finding the power of the complex number and others.
Q: Is De Moivre’s Theorem work for Non-Integer Powers?
Answer:No, De Moivre’s formula does not work for non-integer powers. The result for non-negative integers is the multiple-value different from the original results.
Q: Who invented De Moivre’s Theorem?
Answer:The De Moivre’s theorem was first introduced by the French mathematician Abraham De Moivre's
This is Also an important topic for CSS PMS Ppsc gate and all other competitive exams.

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