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 $k=0$: $$ z_0 = \sqrt[3]{1}\left(\cos\left(\frac{0 + 2(0