Given that $z_1 = -1 + i\sqrt{3}$ and $z_2 = 1 + i$: (a) Express $z_1z_2$ in polar form. (b) Express $\frac{z_1}{z_2}$ in exponential form.

Steps to Solve

1. Convert $z_1$ and $z_2$ to polar form

The polar form of a complex number $z = a + bi$ is $r(\cos\theta + i\sin\theta)$, where $r = \sqrt{a^2 + b^2}$ is the magnitude and $\theta = \arctan(\frac{b}{a})$ is the argument.

For $z_1 = -1 + i\sqrt{3}$: $r_1 = \sqrt{(-1)^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2$ $\theta_1 = \arctan(\frac{\sqrt{3}}{-1})$. Since the complex number is in the second quadrant, $\theta_1 = \frac{2\pi}{3}$ or $120^\circ$. So, $z_1 = 2(\cos(\frac{2\pi}{3}) + i\sin(\frac{2\pi}{3}))$.

For $z_2 = 1 + i$: $r_2 = \sqrt{1^2 + 1^2} = \sqrt{1 + 1} = \sqrt{2}$ $\theta_2 = \arctan(\frac{1}{1}) = \arctan(1) = \frac{\pi}{4}$ or $45^\circ$. So, $z_2 = \sqrt{2}(\cos(\frac{\pi}{4}) + i\sin(\frac{\pi}{4}))$.

2. Calculate $z_1z_2$ in polar form

To multiply complex numbers in polar form, multiply their magnitudes and add their arguments: $z_1z_2 = r_1r_2(\cos(\theta_1 + \theta_2) + i\sin(\theta_1 + \theta_2))$ $z_1z_2 = (2)(\sqrt{2})(\cos(\frac{2\pi}{3} + \frac{\pi}{4}) + i\sin(\frac{2\pi}{3} + \frac{\pi}{4}))$ $z_1z_2 = 2\sqrt{2}(\cos(\frac{8\pi + 3\pi}{12}) + i\sin(\frac{8\pi + 3\pi}{12}))$ $z_1z_2 = 2\sqrt{2}(\cos(\frac{11\pi}{12}) + i\sin(\frac{11\pi}{12}))$

3. Convert $z_1$ and $z_2$ to exponential form

The exponential form of a complex number $z = a + bi$ is $re^{i\theta}$, where $r$ is the magnitude and $\theta$ is the argument. $z_1 = 2e^{i(2\pi/3)}$ $z_2 = \sqrt{2}e^{i(\pi/4)}$

4. Calculate $\frac{z_1}{z_2}$ in exponential form

To divide complex numbers in exponential form, divide their magnitudes and subtract their arguments: $\frac{z_1}{z_2} = \frac{r_1}{r_2}e^{i(\theta_1 - \theta_2)}$ $\frac{z_1}{z_2} = \frac{2}{\sqrt{2}}e^{i(\frac{2\pi}{3} - \frac{\pi}{4})}$ $\frac{z_1}{z_2} = \sqrt{2}e^{i(\frac{8\pi - 3\pi}{12})}$ $\frac{z_1}{z_2} = \sqrt{2}e^{i\frac{5\pi}{12}}$