De Moivre's Theorem is a helpful tool for making it easier to work with complex numbers, especially when you raise them to a power. If you’re studying A-Level math, this theorem is something you'll come across a lot, especially with complex numbers.

So, what does De Moivre's Theorem say?

When a complex number is written in polar form like this:

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

where $r$ is the distance from the origin (called the modulus) and $\theta$ is the angle (called the argument), the theorem helps us find $z^n$ (which means raising $z$ to the power of n) very simply:

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

This makes it easier to do calculations involving complex numbers and shows how trigonometric functions connect with complex numbers.

If you try to calculate powers of complex numbers in a different way, like rectangular form (which looks like $a + bi$), it can take a long time. For example, to find $(1 + i)^3$, you would normally expand it like this:

$(1+i)^3 = (1+i)(1+i)(1+i).$

This involves a lot of extra steps, and it can get pretty messy. But if we change $1+i$ into polar form, we can make it much simpler.

First, let's find $r$ and $\theta$:

1. Modulus: $r = |1+i| = \sqrt{1^2 + 1^2} = \sqrt{2}.$
2. Argument: $\theta = \tan^{-1}(\frac{1}{1}) = \frac{\pi}{4}.$

Now we can write $1+i$ in polar form as:

$z = \sqrt{2} \left( \cos \frac{\pi}{4} + i \sin \frac{\pi}{4} \right).$

Now we can use De Moivre's Theorem to calculate $(1+i)^3$:

$z^3 = \left( \sqrt{2} \right)^3 \left( \cos(3 \cdot \frac{\pi}{4}) + i \sin(3 \cdot \frac{\pi}{4}) \right) = 2\sqrt{2} \left( \cos\frac{3\pi}{4} + i \sin\frac{3\pi}{4} \right).$

We know that $\cos \frac{3\pi}{4} = -\frac{\sqrt{2}}{2}$ and $\sin \frac{3\pi}{4} = \frac{\sqrt{2}}{2}$. Now we can finish our calculation:

$z^3 = 2\sqrt{2} \left( -\frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2} \right) = -2 + 2i.$

This shows how De Moivre's Theorem helps in calculating powers of complex numbers in a much easier way.

De Moivre's Theorem is also very useful for finding roots of complex numbers, which can be tricky with rectangular form. For example, if we want to find the cube roots of the complex number $z = 8 (\cos 0 + i \sin 0)$, we can use the theorem again to simplify our work. The formula for the roots looks like this:

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

where $k$ is 0, 1, 2, ..., up to $n-1$. Plugging in $r = 8$ and $\theta = 0$ for the cube roots, we get:

$z_k = 2 \left( \cos\left(\frac{2 k \pi}{3}\right) + i \sin\left(\frac{2 k \pi}{3}\right) \right).$

By calculating for each value of $k$, we can find the three unique cube roots of that complex number. This shows how De Moivre's Theorem neatly gives us all the roots we need.

In short, De Moivre's Theorem makes it easier to calculate powers and roots of complex numbers. It's a valuable tool in A-Level Mathematics and helps connect algebra and trigonometry, making complex ideas simpler to understand.