91Ó°ÊÓ

Given z and w as complex numbers, we write them in rectangular form as: \(z = a + bi\) \(w = c + di\) where a, b, c, and d are real numbers.

Multiply the complex numbers z and w using the rectangular form: \(zw = (a+bi)(c+di)\) Apply the distributive property to get: \(zw = (ac + adi + bci + bdi^2)\) Since \(i^2 = -1\), the expression becomes: \(zw = (ac - bd) + (ad + bc)i\)

We will now find the magnitudes (absolute values) of the complex numbers z, w, and zw. For z: \(|z| = \sqrt{a^2+b^2}\) For w: \(|w| = \sqrt{c^2+d^2}\) For zw: \(|zw| = \sqrt{(ac-bd)^2 + (ad+bc)^2}\)

Our goal is to prove that \(|zw|=|z||w|\). Let's multiply the magnitudes of z and w. \(|z||w| = \sqrt{a^2+b^2}\sqrt{c^2+d^2}\) Now, let's square both sides of the equation: \((|zw|)^2 = (|z||w|)^2\) Expanding this equation, we get: \((ac-bd)^2 + (ad+bc)^2 = (a^2+b^2)(c^2+d^2)\) Now, expand both sides: \(a^2c^2 - 2acbd + b^2d^2 + a^2d^2 + 2abcd + b^2c^2 = a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2\) Observe that the terms \(a^2c^2\), \(b^2d^2\), \(a^2d^2\), and \(b^2c^2\) appear on both sides and can be canceled out: \(-2acbd + 2abcd = 0\) Divide by 2: \(-acbd + abcd = 0\) Factor out the common term "ac": \(ac(-bd + cd) = 0\) Either \(a = 0\), \(c = 0\), or \((-b+d) = 0\). In any of the cases, the equation holds, which means: \( |zw| = |z||w| \) Therefore, we have proven that for any two complex numbers z and w, the magnitude of their product is equal to the product of their magnitudes.