Let's solve each part step by step: --- ## (a) **Write \( z_1 = -\frac{1+i}{\sqrt{2}} \) and \( z_2 = \frac{1+i\sqrt{3}}{2} \) in polar form.** ### **For \( z_1 \):** 1. **Express in standard form:** \( z_1 = -\frac{1+i}{\sqrt{2}} = -\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}} \) 2. **Find the modulus \( r_1 \):** \[ r_1 = \sqrt{\left(-\frac{1}{\sqrt{2}}\right)^2 + \left(-\frac{1}{\sqrt{2}}\right)^2} = \sqrt{\frac{1}{2} + \frac{1}{2}} = \sqrt{1} = 1 \] 3. **Find the argument \( \theta_1 \):** \[ \tan\theta_1 = \frac{\text{Im}(z_1)}{\text{Re}(z_1)} = \frac{-\frac{1}{\sqrt{2}}}{-\frac{1}{\sqrt{2}}} = 1 \] So \( \theta_1 = \arctan(1) = \frac{\pi}{4} \). But both real and imaginary parts are negative, so \( z_1 \) is in the third quadrant: \[ \theta_1 = \pi + \frac{\pi}{4} = \frac{5\pi}{4} \] 4. **Polar form:** \[ z_1 = 1 \cdot \text{cis}\left(\frac{5\pi}{4}\right) \] or \[ z_1 = e^{i\frac{5\pi}{4}} \] --- ### **For \( z_2 \):** 1. **Express in standard form:** \( z_2 = \frac{1}{2} + \frac{\sqrt{3}}{2}i \) 2. **Find the modulus \( r_2 \):** \[ r_2 = \sqrt{\left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = \sqrt{1} = 1 \] 3. **Find the argument \( \theta_2 \):** \[ \tan\theta_2 = \frac{\sqrt{3}/2}{1/2} = \sqrt{3} \] So \( \theta_2 = \arctan(\sqrt{3}) = \frac{\pi}{3} \). 4. **Polar form:** \[ z_2 = 1 \cdot \text{cis}\left(\frac{\pi}{3}\right) \] or \[ z_2 = e^{i\frac{\pi}{3}} \] --- ## (b) **Find the exact value of \( z_1^{20} z_2^{-10} \) in Cartesian coordinates.** Recall: If \( z = r\,\text{cis}\,\theta \), then \( z^n = r^n\,\text{cis}(n\theta) \). - \( z_1 = \text{cis}\left(\frac{5\pi}{4}\right) \) - \( z_2 = \text{cis}\left(\frac{\pi}{3}\right) \) So: \[ z_1^{20} = \text{cis}\left(20 \cdot \frac{5\pi}{4}\right) = \text{cis}\left(25\pi\right) \] \[ z_2^{-10} = \text{cis}\left(-10 \cdot \frac{\pi}{3}\right) = \text{cis}\left(-\frac{10\pi}{3}\right) \] Now multiply: \[ z_1^{20} z_2^{-10} = \text{cis}(25\pi) \cdot \text{cis}\left(-\frac{10\pi}{3}\right) = \text{cis}\left(25\pi - \frac{10\pi}{3}\right) \] Calculate: \[ 25\pi - \frac{10\pi}{3} = \frac{75\pi}{3} - \frac{10\pi}{3} = \frac{65\pi}{3} \] Now, reduce modulo \( 2\pi \): Find \( \frac{65\pi}{3} \) modulo \( 2\pi \): Divide \( 65/3 \) by 2: \( \frac{65}{3} \div 2 = \frac{65}{6} \approx 10.833 \). So, \( 2\pi \) goes into \( \frac{65\pi}{3} \) 10 times (\( 10 \times 2\pi = 20\pi \)), with a remainder: \[ \frac{65\pi}{3} - 20\pi = \frac{65\pi-60\pi}{3} = \frac{5\pi}{3} \] So the angle is \( \frac{5\pi}{3} \). Thus: \[ z_1^{20}z_2^{-10} = \text{cis}\left(\frac{5\pi}{3}\right) \] Convert to Cartesian: \[ \cos\left(\frac{5\pi}{3}\right) = \frac{1}{2}, \quad \sin\left(\frac{5\pi}{3}\right) = -\frac{\sqrt{3}}{2} \] So, \[ \boxed{z_1^{20}z_2^{-10} = \frac{1}{2} - \frac{\sqrt{3}}{2}i} \] --- ## (c) **Write \( z \) in the form \( a + ib \) if:** ### (i) \( z = (2-i)(3+2i) \) Expand: \[ z = (2-i)(3+2i) = 2 \cdot 3 + 2 \cdot 2i - i \cdot 3 - i \cdot 2i = 6 + 4i - 3i - 2i^2 \] Recall \( i^2 = -1 \): \[ = 6 + 4i - 3i + 2 = (6+2) + (4i-3i) = 8 + i \] So, \[ \boxed{z = 8 + i} \] --- ### (ii) \( z = \frac{1+i}{3-4i} \) Multiply numerator and denominator by the conjugate of the denominator (\( 3+4i \)): \[ z = \frac{1+i}{3-4i} \cdot \frac{3+4i}{3+4i} = \frac{(1+i)(3+4i)}{(3-4i)(3+4i)} \] Numerator: \[ (1+i)(3+4i) = 1\cdot3 + 1\cdot4i + i\cdot3 + i\cdot4i = 3 + 4i + 3i + 4i^2 = 3 + 4i + 3i + 4(-1) = 3 + 7i - 4 = -1 + 7i \] Denominator: \[ (3-4i)(3+4i) = 3^2 - (4i)^2 = 9 - 16i^2 = 9 - 16(-1) = 9 + 16 = 25 \] So: \[ z = \frac{-1 + 7i}{25} = -\frac{1}{25} + \frac{7}{25}i \] So, \[ \boxed{z = -\frac{1}{25} + \frac{7}{25}i} \] --- ## **Summary of Answers** ### (a) - \( z_1 = \text{cis}\left(\frac{5\pi}{4}\right) \) - \( z_2 = \text{cis}\left(\frac{\pi}{3}\right) \) ### (b) - \( z_1^{20}z_2^{-10} = \frac{1}{2} - \frac{\sqrt{3}}{2}i \) ### (c) - (i) \( z = 8 + i \) - (ii) \( z = -\frac{1}{25} + \frac{7}{25}i \)