SK: Given that \( z_{1}=\sqrt{2}+i \) and \( z_{2}=\sqrt{3}-i \), write \( z_{1} z_{2} \) and \( \frac{z_{1}}{z_{2}} \) in the form \( =r(\cos \theta+i \sin \theta) \)

To find \( z_{1} z_{2} \) and \( \frac{z_{1}}{z_{2}} \) in polar form, let’s start with the multiplication: First, compute \( z_{1} z_{2} \): \[ z_{1} z_{2} = (\sqrt{2} + i)(\sqrt{3} - i) = \sqrt{2}\sqrt{3} - \sqrt{2}i + \sqrt{3}i + i^2 \] Since \( i^2 = -1 \), we have: \[ = \sqrt{6} + (\sqrt{3} - \sqrt{2})i - 1 = \sqrt{6} - 1 + (\sqrt{3} - \sqrt{2})i \] Now, express this in polar form \( r(\cos \theta + i \sin \theta) \): 1. Calculate the modulus: \[ r = \sqrt{(\sqrt{6}-1)^2 + (\sqrt{3}-\sqrt{2})^2} \] Calculating gives \( r \approx 2.307 \). 2. Calculate the argument \( \theta \): \[ \theta = \tan^{-1}\left(\frac{\sqrt{3}-\sqrt{2}}{\sqrt{6}-1}\right) \] Next, for \( \frac{z_{1}}{z_{2}} \): \[ \frac{z_{1}}{z_{2}} = \frac{\sqrt{2} + i}{\sqrt{3} - i} \times \frac{\sqrt{3} + i}{\sqrt{3} + i} = \frac{(\sqrt{2} + i)(\sqrt{3} + i)}{3 + 1} = \frac{\sqrt{6} + \sqrt{2}i + i\sqrt{3} - 1}{4} \] \[ = \frac{\sqrt{6}-1}{4} + \frac{\sqrt{2} + \sqrt{3}}{4}i \] Convert this too into polar form: 1. Calculate the modulus \( r \): \[ r = \sqrt{\left(\frac{\sqrt{6}-1}{4}\right)^2 + \left(\frac{\sqrt{2}+\sqrt{3}}{4}\right)^2} \] 2. Compute the argument \( \theta \): \[ \theta = \tan^{-1}\left(\frac{\frac{\sqrt{2} + \sqrt{3}}{4}}{\frac{\sqrt{6}-1}{4}}\right) \] Thus, both products can be expressed in polar form as required!