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Математика 10-11 класс Syed Waqas Ali

Q: State And Prove De - Moivres Theorem?


In De Moivre's Theorem applies fundamental processes of algebra such as powers and taking roots in complex numbers. The theorem was formulated by the famous French mathematician Abraham de Moivre (1730), who related complex numbers to trigonometry.

Abraham De Moivre created this association through the expressions of sine and cosine. This mathematician created a kind of formula that can be used to raise a complex number z to the power of n, which is a positive integer greater than or equal to 1.

What is De Moivre's theorem?

De Moivre's theorem states the following:

If we have a complex number in the polar form z = rƟ where r is the modulus of the complex number z and the angle Ɵ is called the amplitude or argument of any complex number with 0 ≤ Ɵ ≤ 2π, to calculate its nth power there is no need to multiply it on yourself n times; that is, it is not necessary to produce the following product:

Zp \\u003d z * z * z *. . .*r = rƟ*rƟ*rƟ*. . .* рƟ n times.

On the contrary, the theorem states that, writing z in its trigonometric form, to calculate the nth power, we proceed as follows:

If z = r (cos Ɵ + i * sin Ɵ), then zp = rp (cos n * Ɵ + i * sen n * Ɵ).

For example, if n = 2, then z2 = z2[cos 2 (Ɵ) + i sin 2 (Ɵ)]. If n = 3, then z3 = z2 * z. Further:

z3 = z2[cos 2 (Ɵ) + i sin 2 (Ɵ)] * z[cos 2 (Ɵ) + i sin 2 (Ɵ)] = z3[cos 3 (Ɵ) + i sin 3 ( Ɵ)].

Thus, the trigonometric ratios of sine and cosine can be obtained for multiple angles if the trigonometric ratios of the angle are known.

In the same way, it can be used to find more precise and less confusing expressions for the nth root of a complex number z, so that zn = 1.

To prove Moivre's theorem, the principle of mathematical induction is used: if an integer \"a\" has the property \"P\", and if for any integer \"n\" greater than \"a\" has the property \"P\", This means that n + 1 also has the \"P\" property, then all integers greater than or equal to \"a\" have the \"P\" property.


Thus, the proof of the theorem is carried out in the following steps:

inductive base

First it is checked for n = 1.

Since z1 = (r (cos Ɵ + i * sen Ɵ))1 = r1 (cos Ɵ + i * sen Ɵ)1 = r1 [cos (1* Ɵ) + i * sen (1* Ɵ)] it follows that for n = 1 the theorem is fulfilled.

Inductive hypothesis

The formula is assumed to be true for some positive integer, i.e. n = k.

zk = (r (cos Ɵ + i * sen Ɵ))k = rk (cos k Ɵ + i * sin to Ɵ).


This has been proven to be true for n = k + 1.

Since zk + 1= zk * z, then zk + 1 = (r (cos Ɵ + i * sen Ɵ))k + 1 = rk (cos kƟ + i * sen kƟ) * r (cos Ɵ + i * sen Ɵ) .

The expressions are then multiplied:

zk + 1 = gk + 1((cos kƟ)*(cosƟ) + (cos kƟ)*(i*senƟ) + (i * sen kƟ)*(cosƟ) + (i * sen kƟ)*(i* senƟ )).

For a moment, the factor r is ignored to + 1, and the common factor i is taken:

(cos kƟ)*(cosƟ) + i (cos kƟ)*(sinƟ) + i (sin kƟ)*(cosƟ) + i2(sen kƟ)*(senƟ).

Like n2 = -1, we substitute into the expression and get:

(cos kƟ)*(cosƟ) + i (cos kƟ)*(sinƟ) + i (sin kƟ)*(cosƟ) - (sin kƟ)*(senƟ).

Now the real and imaginary parts are ordered:

(cos kƟ)*(cosƟ) - (sin kƟ)*(sinƟ) + i [(sin kƟ)*(cosƟ) + (cos kƟ)*(senƟ)].

To simplify the expression, the trigonometric identities for the sum of angles are used for cosine and sine:

cos(A + B) = cos A * cos B - sin A * sen B.

sin(A + B) = sin A * cos B - cos A * cos B.

In this case, the variables are the angles Ɵ and kƟ. Applying trigonometric identities, we have:

cos kƟ * cosƟ - sen ko * sinƟ = cos (kƟ + Ɵ)

sen ko *cosƟ + cos kƟ *sin = sin(kƟ + Ɵ)

So the expression looks like this:

zk + 1 = gk + 1 (cos (kƟ + Ɵ) + i * sin (kƟ + Ɵ))

zk + 1 = gk + 1(cos [(k +1) Ɵ] + i * sin [(k +1) Ɵ]).

Thus, it can be shown that the result is true for n = k + 1. By the principle of mathematical induction, it is concluded that the result is true for all positive integers; i.e. n ≥ 1.

Negative integer

De Moivre's theorem also applies when n ≤ 0. Consider a negative integer \"n\"; then \"n\" can be written as \"-m\" i.e. n = -m where \"m\" is a positive integer. In this way:

(cos Ɵ + i * sen Ɵ)p = (cos Ɵ + i * sen Ɵ) -m

To get the exponent \"m\" in a positive way, the expression is written in reverse order:

(cos Ɵ + i * sen Ɵ)p = 1 ÷ (cos Ɵ + i * sen Ɵ) m

(cos Ɵ + i * sen Ɵ)n = 1 ÷ (cos mƟ + i * sen mƟ)

Now it is used that if z = a + b * i is a complex number, then 1 ÷ z = a-b * i. In this way:

(cos Ɵ + i * sen Ɵ)n \\u003d cos (mƟ) - i * sen (mƟ).

Using cos(x) = cos(-x) and -sen(x) = sin(-x) we have:

(cos Ɵ + i * sen Ɵ)n = [cos (mƟ) - i * sen (mƟ)]

(cos Ɵ + i * sen Ɵ)n = cos (- mƟ) + i * sen (-mƟ)

(cos Ɵ + i * sen Ɵ)n = cos (nƟ) - i * sen (n).

Thus, we can say that the theorem applies to all integer values ​​of \"n\".

Solved Exercises

Calculation of positive powers

One of the operations on complex numbers in their polar form is multiplication by two of them; in this case, modules are multiplied and arguments are added.

If we have two complex numbers z1 and z2 and you want to calculate (z1 * z2)2 then proceed as follows:

z1z2 = [r1 (cos Ɵ1 + i * sen Ɵ1)] * [r2 (cos Ɵ2 + i * sen Ɵ2)]

The distributive property extends:

z1z2 = r1 p2 (cos Ɵ1* cos Ɵ2 + i * cos Ɵ1* i * sen Ɵ2 + i * sen Ɵ1* cos Ɵ2 + i2* sen Ɵ1* sen Ɵ2).

They are grouped by taking the term \"i\" as a common expression factor:

z1z2 = r1 p2 [cos Ɵ1* cos Ɵ2 + i (cos Ɵ1* sen Ɵ2 + sen Ɵ1* cos Ɵ2) + i2* sen Ɵ1* sen Ɵ2]

Like n2 = -1, it is substituted into the expression:

z1z2 = r1 p2 [cos Ɵ1* cos Ɵ2 + i (cos Ɵ1* sen Ɵ2 + sen Ɵ1* cos Ɵ2) - sin Ɵ1* sen Ɵ2]

The real members are regrouped with the real ones, and the imaginary ones with the imaginary ones:

z1z2 = r1 p2 [(cos Ɵ1* cos Ɵ2 - sen Ɵ1* sen Ɵ2) + i (cos Ɵ1* sen Ɵ2 + sen Ɵ1* cos Ɵ2)]

Finally, the trigonometric properties are applied:

z1z2 = r1 p2 [cos (Ɵ1 + Ɵ2) + i sin (Ɵ1 + Ɵ2)].

In custody:

(r1 * z2)2= (r1 p2 [cos (Ɵ1 + Ɵ2) + i sin (Ɵ1 + Ɵ2)])2

\\u003d g12r22 [cos 2 * (Ɵ1 + Ɵ2) + i sin 2 * (Ɵ1 + Ɵ2)].

Отв. дан Truess
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