De Moivre’s Theorem is one of the important aspects of the Complex number series and plays a major role in calculating complex numbers with an order of 2 and above. Now in our discussion on the polar form of complex numbers, we discussed how the multiplication of complex numbers could be carried out in their polar forms. Also, we have discussed the multiplication of the standard form of complex numbers. However, our discussion of these respective multiplications only considered them when multiplying other complex numbers just once. It is important to note that this is not the only pattern of multiplication we will need to deal with when we talk about multiplication.

Now it could be possible for a student to be asked to determine the fourth-order of a complex number. i.e. (a + ib)^{4 }or (r[Cosϴ + isinϴ])^{4}. There are two ways to determine this. Either the multiplication will be done in this form (r[Cosϴ + isinϴ]) x (r[Cosϴ + isinϴ]) x (r[Cosϴ + isinϴ]) x (r[Cosϴ + isinϴ]) or we choose an easier way. Going with the first will definitely take more time, and it will only get worse if we have to handle a complex number with an order of 8 or more.

On the other hand, the easier way will see us apply the De Moivre Theorem. The De Moivre theorem is very easy to understand theorem and formula. It was born from the Euler Formula and is very important in establishing the relationship between complex numbers and trigonometric functions. However, its major importance is to find the order of complex numbers with less confounding steps.

This article will discuss the role of De Moivre’s theorem in helping students find how they can multiply the polar and standard forms of complex numbers with high orders and get their answers speedily.

**The De Moivre Theorem and Formula**

De Moivre theorem states that for any real number and imaginary [Cosϴ + isinϴ]; then

[r(Cosϴ + isinϴ)]^{n }= [r^{n}(Cosnϴ + isinnϴ)]

Note that [Cosrϴ + isinrϴ]^{n }= [r(Cosϴ + isinϴ)]^{n}

So whichever appears, know that it will be equal to what we have on the right-hand side

As seen from the above, the order n on the left-hand side was then placed inside each trigonometric function. The aim of this De Moivre formula is designed to make the order form of multiplication easy to compute.

Now it is important to understand that it is quite easy to compute the order of a polar form complex number.

Now n is a natural number, meaning it could be anything from 1 to an infinite positive number.

Proving the De Moivre theorem is often not necessary. However, it is quite easy to determine. Since the order of anything is often from 1 and above, the mathematical induction can also be used to find the De Movre theorem. This is because n will always be at least 1 and other higher orders. Proving by mathematical induction, however, is under another scope and is usually taught even before complex numbers. The bottom line of Mathematical Induction is to show constancy in order of natural numbers. The De Move Theorem satisfied this condition. However, students will need to be vast in mathematical induction topics to perform the proof and computation needed in respect to complex numbers.

**How to Calculate the De Moivre Theorem?**

Step 1: Ensure that the modulus and argument are available

Step 2: Use the formula given strictly and arrive at your answer

Step 3: Express the answer in Radians or degrees as required.

**Examples of De Moivre Theorem Application**

Now let’s consider some examples of the De Moivre theorem application

**Example 1**

Find the answer to the above [5(cos30^{0 }+ isin30^{0})]^{2}

Now we can decide to calculate in the normal multiplication form which is

5(cos30^{0 }+ isin30^{0}) x 5(cos30^{0 }+ isin30^{0})

= 5 x 5 [cos(30^{0} + 30^{0}) + isin(30^{0} + 30^{0})]

=25 (COS60^{0 }+ isin 60^{0})

We could, however, use the De Moivre theorem which is given as

[r(Cosϴ + isinϴ)]^{n }= [r^{n}(Cosnϴ + isinnϴ)]

This means that

[5(cos30^{0 }+ isin30^{0})]^{2} = [5^{2}(cos(2 x 30^{0}) + Isin(2 x 30^{0}))]

[5(cos30^{0 }+ isin30^{0})]^{2} = [25cos(60^{0}) + isin(60^{0})]

The above expression is in the standard form of a complex number. If we are to leave it at the polar form, then we won’t need to go past the part where it is

[5(cos30^{0 }+ isin30^{0})]^{2} = [25cos(60^{0}) + isin(60^{0})]

As can be seen, it is obvious that the De Moivre theorem will always provide the right answer in a much faster way. While it seems like the time of computation is the same when dealing with an order of 2, the benefit of the formula will become more obvious when dealing with higher orders.

**Example 2**

Calculate the complex number [4(cos25^{0 }+ isin25^{0})]^{7}

**Solution**

Using the De Moivre theorem, which is given as

[r(Cosϴ + isinϴ)]^{n }= [r^{n}(Cosnϴ + isinnϴ)], we will have the following

[4(cos25^{0 }+ isin25^{0})]^{7} = [4^{7}(Cos(7 x 25^{0}) + isin(7 x 25^{0}))]

[4(cos25^{0 }+ isin25^{0})]^{7} = [16384(Cos175^{0} + isin175^{0})]

[4(cos25^{0 }+ isin25^{0})]^{7} = [16384(-0.996 + 0.087i)]

[4(cos25^{0 }+ isin25^{0})]^{7} = [16384(-0.996) + 16384(0.087i)]

[4(cos25^{0 }+ isin25^{0})]^{7} = -16318.46 + 1425.4i

**Example 3**

Find the solution to this complex number (4 – 5i)^{6} and ensure that your answer remains in the standard form

**Solution**

Now we may decide to compute the multiplication of the complex number by 6 times or change the standard form to the polar form and use the De Moivre theorem. We will go for the second option

So these are the steps to using this method

- Calculate the modulus and the argument and convert to the polar form
- Use the De Moivre theorem and

Now the modulus is given as

now for the argument, which is ϴ; calculating it depends on where the complex number will be plotted in the Argand diagram and its reference angle, which is given as

Now plotting the complex number in the argand diagram will show that it is in the fourth quadrant

And based on the complex number theory;

ϴ = 360 – ϴ_{ref }when in the fourth quadrant of the Argand diagram

**ϴ _{ref }= _{ }**51.34

ϴ = 360 – 51.34

ϴ = 308.66^{0}

Since we have the modulus and the argument determined, we can now express the standard form of the complex number to its polar for of [r(Cosϴ + isinϴ)]

Therefore

Since we now have the polar form of the complex number, we will have to use the De Moivre formula its 6^{th} term.

De Moivre theorem is given as

[r(Cosϴ + isinϴ)]^{n }= [r^{n}(Cosnϴ + isinnϴ)]

As such;

The above answer is an expression in standard form.

Everything about the De Moivre theorem has been considered with respect to its importance to Complex numbers.

**Conclusion**

The De Moivre Theorem is a very important part of complex numbers. It is not the first stage of the complex number series to learn, so students seeing all the computations done here may be a bit confused. We have previously considered some other major topics that led to this one, and checking them out will help you get in line with all that has been carried out here. Some of the important topics that you should check out include

The Algebra of Complex Number

The Conjugates of Complex Number

The Modulus and Arguments of Complex Number

The Polar Form of Complex Numbers

Checking out these topics in the order they are arranged will make it easy for students to comprehend what De Moivre theorem is all about, its motive and why it can quicken computation that involves the order of complex numbers.

This article considered three examples with the aim of showing students how the De Moivre theorem works. Also, by virtue of computing the De Moivre theorem, we also highlighted exactly how to convert standard complex numbers to polar complex numbers and vice versa. As was revealed in this article, the process is very simple, and students who follow them will be able to reach solutions more quickly and easily.