Complex numbers — practice questions

Given $u = -\sqrt{3} + i$. This is in the form $x+iy$ with $x = -\sqrt{3}$ and $y = 1$.

(i) Modulus and Argument
First, calculate the modulus:
$|u| = \sqrt{(-\sqrt{3})^2 + 1^2} = \sqrt{3+1} = \sqrt{4} = 2$. [M1, A1]

Next, find the argument $\theta$. Let's first find the related acute angle $\alpha$:
$\alpha = \tan^{-1} \left| \frac{y}{x} \right| = \tan^{-1} \left| \frac{1}{-\sqrt{3}} \right| = \tan^{-1} \left( \frac{1}{\sqrt{3}} \right) = \frac{\pi}{6}$. [M1]

Now, sketch an Argand diagram. The point $(-\sqrt{3}, 1)$ is in the second quadrant.
For the second quadrant, the principal argument is $\theta = \pi - \alpha$.
$\arg(u) = \theta = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$. [A1]

(ii) Exponential Form
The exponential form is $z = re^{i\theta}$. We have $r=2$ and $\theta = \frac{5\pi}{6}$.
So, $u = 2e^{i(5\pi/6)}$. [B1 ft]

To use De Moivre's theorem, we must first convert $z = 1 - i$ into polar or exponential form.

Step 1: Find modulus and argument of $z=1-i$.
$x=1, y=-1$.
Modulus: $|z| = \sqrt{1^2 + (-1)^2} = \sqrt{2}$. [M1]
Argument: The point $(1, -1)$ is in the fourth quadrant.
$\theta = \tan^{-1}\left(\frac{-1}{1}\right) = -\frac{\pi}{4}$. [M1]
So, $z = \sqrt{2} \left( \cos(-\frac{\pi}{4}) + i\sin(-\frac{\pi}{4}) \right)$.

Step 2: Apply De Moivre's Theorem.
We want to find $z^8$.
$z^8 = \left[ \sqrt{2} \left( \cos(-\frac{\pi}{4}) + i\sin(-\frac{\pi}{4}) \right) \right]^8$
$z^8 = (\sqrt{2})^8 \left( \cos(8 \times -\frac{\pi}{4}) + i\sin(8 \times -\frac{\pi}{4}) \right)$ [M1 for applying theorem]
$(\sqrt{2})^8 = (2^{1/2})^8 = 2^4 = 16$.
$8 \times -\frac{\pi}{4} = -2\pi$.
So, $z^8 = 16 (\cos(-2\pi) + i\sin(-2\pi))$.

Step 3: Convert back to Cartesian form.
We evaluate the trigonometric functions:
$\cos(-2\pi) = 1$
$\sin(-2\pi) = 0$
Therefore, $z^8 = 16(1 + i \cdot 0) = 16$. [A1]
The final answer is 16.