To find the roots of complex numbers using De Moivre's Theorem, you can follow these simple steps. This method is especially helpful in Year 13 A-Level Mathematics, where understanding complex numbers is really important.

Step 1: Change to Polar Form

First, you need to change the complex number from its usual form, like $a + bi$ (where $a$ is the real part and $b$ is the imaginary part), into polar form. The polar form looks like this:

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

In this formula:

• $r$ is how big the complex number is, calculated as $r = \sqrt{a^2 + b^2}$.
• $\theta$ is the angle, found using $\theta = \tan^{-1}(\frac{b}{a})$.

Step 2: Use De Moivre's Theorem

De Moivre's Theorem tells us that for any complex number in polar form:

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

If you want to find the $n$th roots of the complex number $z$, you can rearrange this to figure out $z^{1/n}$:

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

Here, $k$ can be any whole number from $0$ to $n-1$.

Step 3: Find the Modulus and Angle

Next, you need to find the $n$th root of the modulus:

$$r^{1/n} = \sqrt[n]{r}$$

Then, calculate the different angles for the roots using:

$$\theta_k = \frac{\theta + 2k\pi}{n} \quad \text{for } k = 0, 1, 2, \ldots, n-1$$

Step 4: Calculate the Roots

Now, plug in the values from the previous steps into the polar form:

$$z_k = r^{1/n} \left( \cos\left(\theta_k\right) + i \sin\left(\theta_k\right) \right)$$

Do this for each integer $k$ to find all $n$ roots of the complex number.

Step 5: Change Back to Rectangular Form (if needed)

Sometimes, it helps to change the polar coordinates back to the usual rectangular form. You can do this using:

$$x_k = r^{1/n} \cos\left(\theta_k\right)$$ $$y_k = r^{1/n} \sin\left(\theta_k\right)$$

This means each root can be written as:

$$z_k = x_k + iy_k$$

Example

Let’s say you want to find the cube roots of $-8$:

1. First, convert it to polar form: $-8 = 8(\cos(\pi) + i \sin(\pi))$.
2. Here, the modulus $r = 8$ and the angle $\theta = \pi$.
3. Find the roots: $r^{1/3} = 2$, and calculate the angles $\theta_k = \frac{\pi + 2k\pi}{3}$ for $k = 0, 1, 2$.
4. The resulting roots are $2(\cos(\theta_k) + i \sin(\theta_k))$ for $k = 0, 1, 2$, which gives you three cube roots.

These steps show how De Moivre's Theorem makes it easier to find the roots of complex numbers using polar form. This is an important skill for A-Level studies!