**de Moivre’s theorem** states that:

for all real values of

n,(cos θ + i sin θ)^{n}= cos nθ + i sin nθ.

This is a very important relationship we need to know about complex numbers. before we start using it, let’s try to prove it first.

** **

__ PROOF__When

**n = 1**,

**(cos θ + i sin θ)**

and so the theorem hold for

^{1}= cos θ + i sin θ**n = 1**.

Now, we assume that the theorem is true for

**n = k**, so

**(cos θ + i sin θ)**

^{k}= cos kθ + i sin kθif the equation is true for

**n = k**, it should be true for

**n = k + 1**, and therefore

**(cos θ + i sin θ)**

= cos kθ cos θ + 2i cos kθ sin θ – sin kθ sin θ

= cos (k + 1)θ + i sin (k + 1)θ

which is true.

^{k+1}= (cos kθ + i sin kθ)(cos θ + i sin θ)= cos kθ cos θ + 2i cos kθ sin θ – sin kθ sin θ

= cos (k + 1)θ + i sin (k + 1)θ

.·. by mathematical induction, de Moivre’s Theorem is true for all integers

**n > 0**.

Let’s try proving for negative numbers too.

Let **n = –p**.

since **p = –n, cos (-n)θ – i sin (-n)θ = cos nθ + i sin nθ**.

.·. once again, this theorem is proven.

So we see that actually de Moivre’s Theorem is true for all values of **n**, where** n** is any integer. We can also show that it is true for fractions, but this is beyond what we can learn. However, one thing to note that if **n** is not an integer, **cos nθ + i sin nθ** is only **one of the possible **values. I will elaborate more on the next post in the section on **roots of unity**.

The most important thing for this section, is that you need to remember how to prove this theorem, and know how to use it. You will be able to simplify a lot of complex number equations by changing the exponents into just multiplication of numbers.

Another thing you should note is the relations of negative angles.**cos (-θ) = cos θsin (-θ) = - sin θ**

You will be dealing with all these a lot. It is good to memorize it, and be careful not to make mistakes.

** APPLICATIONS**I’ll show you an example how de Moivre’s Theorem help you in proving trigonometric identities.

1.

*Express*

**sin 3A**in terms of**sin A**.**sin 3A**

**= Im (cos 3A + i sin 3A)**[here, “

**Im**” means imaginary, while “

**Re**” means real.]

**= Im (cos A + i sin A)**

^{3}**= Im ( cos**

= 3 cos

= 3 sin A – 4 sin

^{3}A + 3 cos^{2}A i sin A - 3 cos A sin^{2}A – i sin^{3}A)= 3 cos

^{2}A sin A – sin^{3}A= 3 sin A – 4 sin

^{3}AOkay, I need to explain this. Here, we are trying to project the term

**sin 3A**in terms of a complex number, which can be dealt with using de Moivre’s Theorem. So

**sin 3A**, is actually the imaginary part of

**cos 3A + i sin 3A**, and we put the “

**Im**” there because

**sin 3A**belongs to the imaginary part (this means that if our question was

**cos 3A**, we have to put “

**Re**” in front of it instead). We evaluate it, and when we remove the “

**Im**” sign, we remove all the real parts (terms without the

**‘i’**), leaving the imaginary part without the

**‘i’**in it. Try using this method to solve

**cos 3A**, you will understand more by then.

**2.**If we set

**z =**

**cos θ + i sin θ**, then

From here, you can further deduce that

With all these, we can do the above example backwards.

*Express*

**sin**in terms of sines of multiple angles.^{3}A

That’s all this time. Try doing lots of exercises on this kind of questions, it might get very complicated when it’s greater than power of 3. ☺

[2-√3i]^6, i saw this question in one of the reference book, how can i simplify it with de Moivre's theorem? i get decimal places for the θ.

ReplyDeleteYou should get √7^6(cos 6 (tan^-1 (-√3/2))+i sin 6 (tan^-1 (-√3/2))). It will then be -143 +180√3i

ReplyDeleteReally thank you..this helps alot..

ReplyDeleteHow to solve 5z^4 - z^3 + 4z^2 - z+ 5 = 0 by using de Moivre theorem?

ReplyDelete