# De Moivre’s Theorem

De Moivre’s Theorem is a mathematical theorem related to complex numbers, unlike the prime numbers you see when it comes to the best personal loans or simple pricing. The power of any complex number can be expanded, similar to how one would expand the power of a binomial. In the case of De Moivre’s Theorem, the process is simplified to find the power of any complex number. If you want to apply De Moivre’s Theorem formula, you need to convert the complex numbers to their polar forms. Let us learn more about this formula in this article.

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**What Is De Moivre’s Theorem?**

The first step here is to understand De Moivre’s Theorem. Also known as the De Moivre’s Identity, the De Moivre’s Theorem is any real number (denoted by x) and integer (denoted by n) where:

(cos x + i sin x)n = cos nx + i sin nx

In the formula above, the ‘i’ stands for the imaginary unit (i2 = −1). At times, the expression cos x + i sin x; however, this expression can also be shortened for cis x.

This formula has a lot of importance in the world of mathematics. This is because you can easily connect trigonometry and complex numbers. Suppose you can simply expand the LHS (left-hand side) and the imaginary numbers under the assumption that x is real. Thanks to this formula, you can also derive useful expressions for sin nx and cos nx in terms of sin x and cos x, respectively.

However, this peculiar formula is not valid for any ‘n’ that has non-integer powers. But, you can use some generalizations of the formula for other types of exponents. With this, you can develop some explicit expressions for the roots of unity (nth), which includes complex numbers z, as in zn = 1.

To further understand De Moivre’s Theorem, let us consider this De Moivre’s Theorem example.

z2= [r(cos x + i sin x)]2

= r2 (cos x + i sin x)(cos x + i sin x)

= r2 (cos x cos x + i sin x cos x + i sin x cos x +i2 sin x sin x)

= r2 [(cos x cos x – sin x sin x) + i (sin x cos x + sin x cos x)]

= r2 (cos 2x + i sin 2x)

From the above, you can easily see that you can manually expand (r (cos x + i sin x))3 = r3 (cos 3x + i sin 3x). This is the overall idea behind this formula – any complex number in a polar form raised to a certain power ‘n.’

**De Moivre’s Theorem Proof**

Now that you understand the theorem, let us look into De Moivre’s Theorem proof. Let us consider that S(n): (r (cos x + i sin x))n = rn (cos nx + i sin nx).

The first step towards proving the theorem is to prove that S(n) for n = 1.

To do this, the LHS needs to be equal to (r (cos x + i sin x))1 = r (cos x + i sin x)

On the other hand, the RHS should be equal to r1 (cos (1)x + i sin (1)x) = r (cos x + i sin x)

As for the second step, we need to assume that S(n) is true for any natural numerical value n = k. In this case, the formula becomes

(r (cos x + i sin x))k = rk (cos kx + i sin kx)

Finally, the last step has to prove S(n) for n = k + 1

LHS = (r (cos x + i sin x))k + 1

= (r (cos x + i sin x))k • (r (cos x + i sin x))

= rk (cos kx + i sin kx) • (r (cos x + i sin x)) (By Step 2)

= rk + 1 [(cos kx cos x – sin kx sin x) + i (cos kx sin x + sin kx cos x)]

= rk + 1 [ cos (kx + x) + i sin (kx + x) ]

= rk + 1 [ cos (k + 1)x + i sin (k + 1)x ]

= RHS

So S(n) is true for n = k + 1

Judging by these formulas, S(n) is true for all ‘n’ values. Additionally, we also already know that cos x + i sin x can also be denoted as cis x. Therefore, De Moivre’s Theorem roots can also be written as (r cis x)n = rn cis nx, where n ∈ Z.

**How to Use De Moivre’s Theorem?**

Now that you have a fair understanding of this theorem, you can now understand how to use De Moivre’s theorem. The first use of this theorem is to find the roots of the complex number. For instance, let us consider a complex number denoted by z. In this case, the polar form can be denoted by z = r(cos x + i sin x).

In this case, z’s nth roots are:

[latex]r^{\frac {1}{n}}\left (cos(\frac{x+2k \pi}{n}) + i\ sin (\frac{x+2k \pi}{n}) \right )[/latex] where k = 0, 1, 2,….., (n − 1)The formula gets reduced to [latex]r^{\frac {1}{n}}\left (cos(\frac{x}{n}) + i\ sin (\frac{x}{n}) \right )[/latex], if the value of k is zero.

Another way of using De Moivre’s Theorem is to obtain the relationship between the powers of trigonometric angles and functions. In this case, as well, we need to utilize the polar form of the complex numbers. The parameters of the polar and rectangular form as:

a = r cos x and b = r sin x

Where r = [latex]\sqrt{a^2+b^2}[/latex] and tan x = (b/a)

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

**What Is The Relationship Between De Moivre’s Theorem And Euler’s Formula?**

When you look into both the formulas, you may notice that De Moivre’s Theorem is a precursor of Euler’s formula. In this case, the formula is:

E^{ix} = cos x + i sin x

From the above-mentioned formula, the fundable relationship between complex exponential function and trigonometric functions can be established.

You can also derive De Moivre’s Theorem with the help of Euler’s formula, which is the exponential law for integer power. This is denoted by.

(e^{ix})^{n} = e^{inx}

Since the LHS of Euler’s formula is (cos x + isin x), the RHS will be e^{inx} = cos nx + i sin nx.

**Final Thoughts**

From the above, it is pretty easy to understand how the De Moivre’s Theorem calculator works. Of course, you may feel overwhelmed after seeing all these mathematical formulas. But once you start to understand it, you will definitely learn. This theorem has found a lot of uses, especially in the fields of engineering. All you need to do is remember the basic formula for multiplying complex numbers in their polar forms.

Would you add anything? Where do you think this is useful? Let us know in the comments!