Complex Numbers and Demoivres Theorem Review Answer Key

In the field of complex numbers, DeMoivre's Theorem is ane of the most important and useful theorems which connects circuitous numbers and trigonometry. As well helpful for obtaining relationships betwixt trigonometric functions of multiple angles. DeMoivre's Theorem too known as "De Moivre'southward Identity" and "De Moivre'southward Formula". The proper noun of the theorem is after the name of cracking Mathematician De Moivre, who made many contributions to the field of mathematics, mainly in the areas of theory of probability and algebra.

Table of Content:

• Formula
• Proof
• Uses
• Problems

De Moivre'due south Formula

Mathematical Argument: For whatsoever existent number x, nosotros have

(cos x + i sin x)^{due north} = cos(nx) + i sin(nx)

OR

\(\brainstorm{array}{l}(e^{i \theta})^due north = due east^{in \theta}\cease{array} \)

Where n is a positive integer and " i " is the imaginary part, and i = √(-1). Also assume i² = -i.

Remark: Consequence can be shown truthful when n is a negative integer and even

north is a rational number.

Likewise Read =>

• Complex Numbers
• Inverse Trigonometric Functions

Aspirants are advised, before starting this section should revise and get familiar with the argand diagram and polar form of complex numbers. This department introduces -De Moivre's theorem statement, proof of theorem and some of its consequences.

De Moivre's Theorem Proof

Apply Mathematical Induction to bear witness De Moivre'southward Theorem.

We know, (cos x + i sin x)ⁿ = cos(nx) + i sin(nx) …(i)

Step 1: For due north = ane, nosotros have

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

Which is true.

Pace two: Assume that formula is true for north = k.

(cos x + i sin 10)^g = cos(kx) + i sin(kx) ….(2)

Footstep 3: Prove that result is true for northward = one thousand + i.

(cos x + i sin x)^k+i = (cos x + i sin ten)^k (cos x + i sin x)

= (cos (kx) + i sin (kx)) (cos x + i sin ten) [Using (i)]

= cos (kx) cos x − sin(kx) sinx + i (sin(kx) cosx + cos(kx) sinx)

= cos {(g+1)x} + i sin {(m+1)ten}

=> (cos x + i sin 10)^thou+1 = cos {(thousand+1)x} + i sin {(k+one)x}

Hence the result is proved.

Since the theorem is true for n = 1 and n = m + 1, it is true ∀ n ≥ one.

Uses of De Moivre'southward Theorem

To discover the roots of complex numbers

If z is a circuitous number, and z = r(cos x + i sin ten) [In polar course]

And then, the nth roots of z are:

\(\brainstorm{array}{50}r^{\frac {i}{n}}\left (cos(\frac{x+2k \pi}{n}) + i\ sin (\frac{ten+2k \pi}{north}) \right )\end{array} \)

where chiliad = 0, 1, ii,….., (n − 1)

If k = 0, above formula reduce to

\(\begin{assortment}{fifty}r^{\frac {1}{n}}\left (cos(\frac{x}{n}) + i\ sin (\frac{x}{due north}) \right )\cease{assortment} \)

To obtain relationships between powers of trigonometric functions and trigonometric angles.

We utilize polar form of circuitous numbers to represent a circuitous number using trigonometry.

The parameters of the rectangular and polar form are as:

a = r cos x and b = r sin x

Where r =

\(\begin{array}{fifty}\sqrt{a^ii+b^2}\terminate{array} \)

and tan ten = (b/a)

This implies, z = a + ib = r(cos 10 + i sin x)

Raising to a Power

Example: Evaluate ( 1 + i )^m.

Solution:

Permit z = 1 + i

We have to represent z in the form of r(cos θ + i sin θ).

Here,

Argument = θ = arc(tan (one/1) = arc tan(1) = π/four

Absolute value = r =

\(\begin{assortment}{l}\sqrt{(1)^ii + (1)^2}= \sqrt{two}\stop{array} \)

Applying DeMoivre's theorem, we get

z¹⁰⁰⁰ = [√2{cos(π/four) + i sin(π/iv)}]¹⁰⁰⁰

= two¹⁰⁰⁰ {cos(1000π/four) + i sin(1000π/4)}

= ii¹⁰⁰⁰ {i + i (0)}

= 2¹⁰⁰⁰

Bug on De Moivre's Identity

Problem i: Evaluate (2 + 2i)⁶

Solution: Let z = 2 + 2i

Hither, r = ii√2 and θ = 45 degrees

Since z lies in the starting time quadrant, sinθ and cosθ functions are positive.

Applying De Moivre'southward Theorem:

z^half-dozen = (ii + 2i)⁶ = (ii√ii)⁶ [cos 45⁰ + i sin 45⁰]⁶

= (2√2)^six [cos 270⁰ + i sin 270⁰]^{half dozen}

= – 512i

Problem 2: Express 5 5th‐roots of (√3 + i) in trigonometric form.

Solution:

We know, z = a + ib = r(cos x + i sin 10)

Where r =

\(\begin{array}{l}\sqrt{a^ii+b^two}\end{array} \)

and tan ten = (b/a)

So,

Here r = ii and θ = xxx degrees

Therefore, z = ii[cos(30⁰ + 360⁰ k) + i sin cos(xxx⁰ + 360⁰ k)]

Applying nth root theorem:

z^1/5 = {2[cos(30⁰ + 360⁰ k) + i sin cos(xxx⁰ + 360⁰ k)]}^1/5

= 2^i/v [cos((xxx⁰ + 360⁰ k)/v) + i sin cos((thirty⁰ + 360⁰ one thousand)/five)] …(ane)

Where k = 0,1,two,3,four

At thou = 0; (1)=> z₁ = 2^1/v [cos half dozen⁰ + i sin 6⁰]

At g = ane; (one)=> z₁ = 2^1/5 [cos 78⁰ + i sin 78⁰]

At chiliad = ii; (1)=> z₁ = 2^1/5 [cos 150⁰ + i sin 150⁰]

At k = iii; (1)=> z₁ = 2^1/5 [cos 222⁰ + i sin 222⁰]

At m = 4; (1)=> z₁ = 2^1/5 [cos 294⁰ + i sin 294⁰]

Problem 3: Express

\(\brainstorm{array}{l}\left(\frac{cos \theta + i\ sin \theta}{sin \theta + i\ cos \theta}\right)^iv\end{array} \)

in a+ib course.

Solution:

\(\begin{array}{50}\left(\frac{cos \theta + i\ sin \theta}{sin \theta + i\ cos \theta}\right)^iv\end{array} \)

=

\(\begin{assortment}{l}\frac{\left ( cos\ \theta + i\ sin\theta\right )^4}{i^iv\left ( \ cos \theta – i sin \theta\correct )^4}\end{array} \)

= (cos 4θ + i sin 4θ ) / (cos 4θ – i sin 4θ )

By rationalising the fraction, we have

= (cos 4θ + i sin 4θ )² / (cos² 4θ – i sin^two 4θ )

= cos 8θ + i sin 8θ

Problem four: If

\(\begin{array}{fifty}{{ten}_{r}}=\cos \left( \frac{\pi }{{{2}^{r}}} \correct)+i\sin \left( \frac{\pi }{{{ii}^{r}}} \right), and then \ {{ten}_{1}}\ .\ {{10}_{2}}……\infty \ is\terminate{array} \)

Solution:

\(\begin{array}{l}{{ten}_{1}},{{x}_{2}},{{ten}_{three}}…..upto\infty =\left( \cos \frac{\pi }{2}+i\sin \frac{\pi }{2} \right)\,\,\left( \cos \frac{\pi }{{{2}^{two}}}+i\sin \frac{\pi }{{{2}^{2}}} \right) upto …..\infty \\=\cos \left( \frac{\pi }{2}+\frac{\pi }{{{2}^{ii}}}+….. \right)+i\sin \left( \frac{\pi }{two}+\frac{\pi }{{{2}^{two}}}+….. \right) \\=\cos \left( \frac{\frac{\pi }{2}}{1-\frac{1}{2}} \correct)+i\sin \left( \frac{\frac{\pi }{2}}{ane-\frac{1}{2}} \correct)\\=\cos \pi +i\sin \pi \\=-ane\finish{assortment} \)

Problem 5: The value of

\(\begin{array}{l}\frac{4(\cos {{75}^{o}}+i\sin {{75}^{o}})}{0.four(\cos {{30}^{o}}+i\sin {{thirty}^{o}})} \ is\terminate{array} \)

Solution:

\(\begin{array}{l}\frac{4(\cos {{75}^{o}}+i\sin {{75}^{o}})}{0.iv(\cos {{30}^{o}}+i\sin {{xxx}^{o}})} \\=ten(\cos {{75}^{o}}+i\sin {{75}^{o}})(\cos {{30}^{o}}-i\sin {{thirty}^{o}}) \\=ten(\cos {{45}^{o}}+i\sin {{45}^{o}})\\=\frac{10}{\sqrt{ii}}(1+i)\end{array} \)

Problem vi: If

\(\begin{array}{50}(\cos \theta +i\sin \theta )(\cos 2\theta +i\sin 2\theta )…….. (\cos n\theta +i\sin northward\theta )=1,\end{array} \)

then what is the value of

\(\brainstorm{assortment}{l}\theta\cease{array} \)

?

Solution:

\(\begin{array}{fifty}(\cos \theta +i\sin \theta )(\cos 2\theta +i\sin two\theta ) ……(\cos north\theta +i\sin n\theta )=1 \\ \cos (\theta +2\theta +three\theta +…+n\theta )+i\sin (\theta +2\theta +.+n\theta )=1 \\ \cos \left( \frac{north(n+1)}{2}\theta \right)+i\sin \left( \frac{northward(n+ane)}{ii}\theta \correct)=one\\ \cos \left( \frac{northward\,(n+ane)}{two}\theta \right)\,=\,i\text{ and }\sin \left( \frac{due north(n+1)}{2}\theta \right)\,=0 \\ \frac{n(n+one)}{2}\theta =2m\pi \\\Rightarrow \theta =\frac{4m\pi }{n(n+1)},\text where \ k\in I.\end{assortment} \)

De Moivre's Theorem – Video Lesson

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