1.2. If \( w=1-i \sqrt{2} \) and \( z=1+i \), show that \( \left|\frac{w}{z}\right|=\frac{|w|}{|z|} \) 1.3. Express \( z=1+i \) in polar form.

To solve the problems, we will start with the first part, which is to show that \( \left|\frac{w}{z}\right|=\frac{|w|}{|z|} \). ### 1.2 Show that \( \left|\frac{w}{z}\right|=\frac{|w|}{|z|} \) Given: - \( w = 1 - i \sqrt{2} \) - \( z = 1 + i \) We need to find the magnitudes \( |w| \) and \( |z| \). **Step 1: Calculate \( |w| \)** The magnitude of a complex number \( a + bi \) is given by \( |a + bi| = \sqrt{a^2 + b^2} \). For \( w = 1 - i \sqrt{2} \): - \( a = 1 \) - \( b = -\sqrt{2} \) Thus, \[ |w| = \sqrt{1^2 + (-\sqrt{2})^2} = \sqrt{1 + 2} = \sqrt{3}. \] **Step 2: Calculate \( |z| \)** For \( z = 1 + i \): - \( a = 1 \) - \( b = 1 \) Thus, \[ |z| = \sqrt{1^2 + 1^2} = \sqrt{1 + 1} = \sqrt{2}. \] **Step 3: Calculate \( \left|\frac{w}{z}\right| \)** Using the property of magnitudes of complex numbers: \[ \left|\frac{w}{z}\right| = \frac{|w|}{|z|}. \] Now substituting the values we found: \[ \left|\frac{w}{z}\right| = \frac{|w|}{|z|} = \frac{\sqrt{3}}{\sqrt{2}}. \] Thus, we have shown that: \[ \left|\frac{w}{z}\right| = \frac{|w|}{|z|}. \] ### 1.3 Express \( z = 1 + i \) in polar form To express \( z \) in polar form, we need to find the magnitude \( r \) and the argument \( \theta \). **Step 1: Calculate the magnitude \( r \)** We already calculated \( |z| \): \[ r = |z| = \sqrt{2}. \] **Step 2: Calculate the argument \( \theta \)** The argument \( \theta \) is given by: \[ \theta = \tan^{-1}\left(\frac{b}{a}\right) = \tan^{-1}\left(\frac{1}{1}\right) = \tan^{-1}(1). \] Since \( z \) is in the first quadrant, we have: \[ \theta = \frac{\pi}{4}. \] **Step 3: Write in polar form** The polar form of a complex number is given by: \[ z = r(\cos \theta + i \sin \theta). \] Substituting the values we found: \[ z = \sqrt{2} \left( \cos\left(\frac{\pi}{4}\right) + i \sin\left(\frac{\pi}{4}\right) \right). \] Thus, the polar form of \( z = 1 + i \) is: \[ z = \sqrt{2} \left( \cos\left(\frac{\pi}{4}\right) + i \sin\left(\frac{\pi}{4}\right) \right). \]