Question 8 i) a): Given that z1 = sqrt(3)+i and z2 = 1−3i, find (z1)/(z2) in the form a+bi, where a and b are real.

Okay, let's solve question 8 from the image you just sent.

---

Question 8 i) a): Given that $z_1 = \sqrt{3}+i$ and $z_2 = 1-3i$, find $\frac{z_1}{z_2}$ in the form $a+bi$, where $a$ and $b$ are real.

Step 1: Write down the given complex numbers.

$$z_1 = \sqrt{3}+i$$ $$z_2 = 1-3i$$

Step 2: Set up the division.

$$\frac{z_1}{z_2} = \frac{\sqrt{3}+i}{1-3i}$$

Step 3: Multiply the numerator and the denominator by the conjugate of the denominator. The conjugate of $1-3i$ is $1+3i$.

$$\frac{z_1}{z_2} = \frac{\sqrt{3}+i}{1-3i} \times \frac{1+3i}{1+3i}$$

Step 4: Multiply the numerators.

$$(\sqrt{3}+i)(1+3i) = \sqrt{3}(1) + \sqrt{3}(3i) + i(1) + i(3i)$$ $$= \sqrt{3} + 3\sqrt{3}i + i + 3i^2$$

Substitute $i^2 = -1$:

$$= \sqrt{3} + 3\sqrt{3}i + i - 3$$ $$= (\sqrt{3}-3) + (3\sqrt{3}+1)i$$

Step 5: Multiply the denominators.

$$(1-3i)(1+3i) = 1^2 - (3i)^2$$ $$= 1 - 9i^2$$

Substitute $i^2 = -1$:

$$= 1 - 9(-1)$$ $$= 1 + 9$$ $$= 10$$

Step 6: Combine the simplified numerator and denominator and express in the form $a+bi$.

$$\frac{z_1}{z_2} = \frac{(\sqrt{3}-3) + (3\sqrt{3}+1)i}{10}$$ $$\frac{z_1}{z_2} = \frac{\sqrt{3}-3}{10} + \frac{3\sqrt{3}+1}{10}i$$

The quotient $\frac{z_1}{z_2}$ is $\boxed{\frac{\sqrt{3}-3}{10} + \frac{3\sqrt{3}+1}{10}i}$.

---

Question 8 i) b): Given that $z_1 = \sqrt{3}+i$ and $z_2 = 1-3i$, find $\arg z_1^5$ and $|z_2^6|$.

Part 1: Find $\arg z_1^5$

Step 1: Convert $z_1$ to polar form $r_1(\cos\theta_1 + i\sin\theta_1)$.

$$z_1 = \sqrt{3}+i$$

The modulus $r_1 = |z_1| = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3+1} = \sqrt{4} = 2$.
The argument $\theta_1 = \arg z_1$. Since $z_1$ is in the first quadrant:

$$\tan\theta_1 = \frac{1}{\sqrt{3}}$$ $$\theta_1 = \frac{\pi}{6}$$

So, $z_1 = 2\left(\cos\left(\frac{\pi}{6}\right) + i\sin\left(\frac{\pi}{6}\right)\right)$.

Step 2: Use De Moivre's Theorem to find $\arg z_1^5$.
For $z_1^n = r_1^n(\cos(n\theta_1) + i\sin(n\theta_1))$, the argument is $n\theta_1$.

$$\arg z_1^5 = 5\theta_1 = 5 \times \frac{\pi}{6} = \frac{5\pi}{6}$$

The argument $\arg z_1^5$ is $\boxed{\frac{5\pi}{6}}$.

Part 2: Find $|z_2^6|$

Step 1: Find the modulus of $z_2$.

$$z_2 = 1-3i$$ $$|z_2| = \sqrt{1^2 + (-3)^2} = \sqrt{1+9} = \sqrt{10}$$

Step 2: Calculate $|z_2^6|$.
Using the property $|z^n| = |z|^n$:

$$|z_2^6| = (|z_2|)^6 = (\sqrt{10})^6$$ $$(\sqrt{10})^6 = (10^{1/2})^6 = 10^{(1/2) \times 6} = 10^3 = 1000$$

The modulus $|z_2^6|$ is $\boxed{1000}$.

---

Question 8 ii): Show that the roots of the equation $z^3 - 1 = 0$ lies at the vertices of an equilateral triangle.

Step 1: Find the roots of the equation $z^3 - 1 = 0$.

$$z^3 = 1$$

We need to find the cube roots of 1. In polar form, $1 = 1(\cos(0) + i\sin(0))$.
The $n$-th roots of a complex number $r(\cos\theta + i\sin\theta)$ are given by:

$$z_k = \sqrt[n]{r}\left(\cos\left(\frac{\theta + 2k\pi}{n}\right) + i\sin\left(\frac{\theta + 2k\pi}{n}\right)\right)$$

For $z^3=1$, we have $r=1$, $\theta=0$, and $n=3$. The values for $k$ are $0, 1, 2$.

For $k=0$:

$$$$