91Ó°ÊÓ

Start by finding \(|\alpha \beta|\). Let \(\alpha = a + bi\) and \(\beta = c + di\) where \(a, b, c, d\) are real numbers. Then, the product of \(\alpha\) and \(\beta\) can be found as follows: \[\alpha \beta = (a + bi)(c + di) = ac + adi + bci + bd(-1) = (ac - bd) + (ad + bc)i\] Now, we can find the absolute value of the product: \[ |\alpha \beta| = \sqrt{(ac - bd)^2 + (ad + bc)^2} \]

Now, find the absolute value of \(\alpha\) and \(\beta\) individually: \[ |\alpha| = \sqrt{a^2 + b^2} \] \[ |\beta| = \sqrt{c^2 + d^2} \] Then, find the product of their absolute values: \[ |\alpha||\beta| = \sqrt{a^2 + b^2} \sqrt{c^2 + d^2} \]

In this step, we will show that the results from Step 1 and Step 2 are equal. Start by expanding the product of the absolute values: \[ |\alpha||\beta| = \sqrt{a^2 + b^2} \sqrt{c^2 + d^2} = \sqrt{(a^2 + b^2)(c^2 + d^2)} \] Now, expand the expression inside the square root: \[ |\alpha||\beta| = \sqrt{a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2} \] Notice that \(a^2c^2 + a^2d^2 = a^2(c^2 + d^2)\) and \(b^2c^2 + b^2d^2 = b^2(c^2 + d^2)\). Therefore, the expression can be rewritten as: \[ |\alpha||\beta| = \sqrt{a^2(c^2 + d^2) + b^2(c^2 + d^2)} \] Now, factor out the common term \((c^2 + d^2)\): \[ |\alpha||\beta| = \sqrt{(a^2 + b^2)(c^2 + d^2)} \] Similarly, we can factor the expression for \(|\alpha \beta|\) which we found in Step 1: \[ |\alpha \beta| = \sqrt{(ac - bd)^2 +(ad + bc)^2}= \sqrt{(a^2 + b^2)(c^2 + d^2)} \] Finally, we can see that both expressions are equal: \[ |\alpha \beta| = |\alpha||\beta| \] Thus, we've proven that the absolute value (or modulus) of the product of two complex numbers is equal to the product of their individual absolute values.