Teorema De Moivre's

A common situation in algebra is when we have to find the expansion of a complex number, i.e. $(r\operatorname{cis}\theta)^n$ (where $\operatorname{cis} \theta = \cos \theta + i\sin\theta$). An interesting result known as de Moivre's Formula actually says that $(r\operatorname{cis}\theta)^n = r^n\operatorname{cis}n\theta.$

We'll proceed by proving a more generalized statement of the multiplication of two complex numbers, and then we'll extend that to get de Moivre's Formula.

Let $z_1 = r_1\operatorname{cis}\theta_1$ and $z_2 = r_2\operatorname{cis}\theta_2.$ Multiplying the two complex numbers gives: $$r_1r_2(\cos \theta_1 + i \sin\theta_1)(\cos \theta_2+i\sin\theta_2)$$ $$=r_1r_2((\cos\theta_1\cos\theta_2-\sin\theta_1\sin\theta_2)+i(\sin\theta_1\cos\theta_2+\cos\theta_1\sin\theta_2))$$ $$= r_1r_2(\cos(\theta_1+\theta_2) + i\sin(\theta_1+\theta_2))$$ $$=r_1r_2\operatorname{cis}(\theta_1+\theta_2)$$ So, we have just found that multiplying two complex numbers in polar form multiplies their magnitudes, and more interestingly, adds their arguments.

This makes de Moivre's Formula almost obvious. If we let $z_1 = z_2 = z,$ then we get $$z_1z_2 = z^2 = r^2\operatorname{cis}(\theta + \theta) = r^2 \operatorname{cis}2\theta.$$ Similarly, $$z^3 = z^2z = r^2r\operatorname{cis}(2\theta + \theta) = r^3\operatorname{cis}3\theta$$ We can continue in this process to find that $z^n = (r\operatorname{cis}\theta)^n = r^n\operatorname{cis}n\theta.$ A formal proof would be by induction, but this should give you a general idea of how it works.

Ultimately, we have proved two very strong statements: $$(r_1\operatorname{cis}\theta_1)(r_2\operatorname{cis}\theta_2) = r_1r_2\operatorname{cis}(\theta_1+\theta_2)$$ $$(r\operatorname{cis}\theta)^n = r^n\operatorname{cis}n\theta$$

The former is very useful in say, coordinate geometry, because it means that multiplying by $\operatorname{cis}\theta$ rotates a point counterclockwise by an angle of $\theta$. The latter is strongly relevant in the next section, when we are faced with the challenge of finding roots to the equation $z^n - 1 = 0.$

De Moivre’s Theorem is commonly used when finding roots of unity.

De Moivre's Theorem, named after the French statistician Abraham de Moivre, is simple enough to state, but the power of the formula lies in its ability to convert complex numbers to trigonometry and vice versa.

Here it is: $$(\cos(x) + i \sin(x))^n = \cos(nx) + i \sin(nx)$$ You'll very occasionally see $\cos(x) + i \sin(x)$ written as $\text{cis}(x)$, but this doesn't appear to be very popular ("cis" stands for "cos i sin"). We can convert this formula to exponentials with Euler's formula, and then it looks really easy to prove: $$e^{(xi)n} = e^{(nx)i}$$

If you think that looks a little too easy, there's another proof by induction. Unfortunately, induction only works here on positive integers (there's such a thing as transfinite induction, but that is way, way beyond our scope).

First, note $(\cos(x) + i \sin(x))^1 = \cos(1x) + i \sin(1x)$

Assume $(\cos(x) + i \sin(x))^n = \cos(nx) + i \sin(nx)$. Then, $$\begin{align*} (\cos(x) + i \sin(x))^{n+1} &= (\cos(x) + i \sin(x))^n(\cos(x) + i \sin(x))\\ &= (\cos(nx) + i \sin(nx))(\cos(x) + i \sin(x))\\ &= \cos(nx)\cos(x) - \sin(nx)\sin(x) \\&+ i(\cos(nx)\sin(x)+\cos(x)\sin(nx))\\ &= \cos((n+1)x) + i \sin((n+1)x) \end{align*}$$ And we're done (for the positive integers...)!

Having proved this formula with one line, and then proved a much less general portion with much more lines, it's natural to ask what the purpose of this formula is. Well, for one thing, we can let $x=\frac{2k\pi}{n}$, for some integer k. Then,

$$\left(\cos \left(\frac{2k\pi}{n} \right) + i \sin \left(\frac{2k\pi}{n} \right) \right)^n = \cos(2k\pi) + i \sin(2k\pi) = 1$$

If we can calculate $\cos \left(\frac{2k\pi}{n} \right)$ and $\sin \left(\frac{2k\pi}{n} \right)$, we can find the $n$th roots of unity.

We just let $n = 6$ and realize the sixth roots of unity are of the form $\cos \left(\frac{k\pi}{3} \right) + i\sin \left(\frac{k\pi}{3} \right)$. If we let $k=4$, we get the number $-\frac{1}{2}-\frac{\sqrt{3}}{2}i$.

We can also use de Moivre's formula to exponentiate other number, simply by factoring out the magnitude and exponentiating the complex part by itself.

$(-1-\sqrt{3}i)^6 = 2^6 \left(-\frac{1}{2}-\frac{\sqrt{3}}{2}i \right)^6 = 64(1)= \fbox{64}$

de Moivre's formula has a few other applications that can be used to evaluate various expressions in $\cos(nx)$ or $\sin(nx)$. But it's important to remember that the formula also allows us to convert between complex numbers and trigonometry, which makes it ultimately very important in the grand scheme of mathematics.

De Moivre's Formula

Using this formula, we can connect trigonometry with complex numbers. Instead of multiplying a complex number by itself over and over, the theorem provides a simple rule for raising it to any positive integer power.

It is especially useful when working with large powers and when finding the roots of complex numbers.

Suppose a complex number is written in polar form as:

$$z=r(\cos\theta + i\sin\theta)$$

Then, De Moivre's Theorem states that:

$$z^n = r^n (\cos(n\theta) + i\sin(n\theta))$$

Here,

• $r$ is the modulus
• $\theta$ is the argument
• $n$ is any integer

The theorem says that you can raise the modulus to the power $n$ and multiply the angle by $n$.

####Exponential Form

Using Euler's Formula,

$$e^{i\theta} = \cos\theta + i\sin\theta$$

, a complex number can be written as:

$$z = re^{i\theta}$$

If we involve De Moivre's Theorem, it becomes even simpler!

$$(re^{i\theta})^n = r^ne^{in\theta}$$

This form is often the easiest to use in higher-level math.

(Video) De Moivre's Theorem

by Anusha Rahman

In this video by Anusha Rahman, you will learn about De Moivre's Theorem.