91Ó°ÊÓ

Write \(w\) and \(z\) as complex numbers in the form of \(a+bi\): \( w = a + bi \) \( z = c + di \) where \(a, b, c\), and \(d\) are real numbers. Compute \(\bar{w}\) and \(\bar{z}\): \( \bar{w} = a - bi \) \( \bar{z} = c - di \)

Calculate \(\frac{w}{z}\): \( \frac{w}{z} = \frac{a + bi}{c + di} \) To eliminate the imaginary denominator, multiply both numerator and denominator by the complex conjugate of the denominator \((c - di)\): $$ \frac{w}{z} = \frac{(a + bi)(c - di)}{(c + di)(c - di)} $$ Now, calculate the complex conjugate \(\overline{\left(\frac{w}{z}\right)}\): $$ \overline{\left(\frac{w}{z}\right)} = \overline{\frac{(a + bi)(c - di)}{(c + di)(c - di)}} $$

Using the complex conjugates of \(w\) and \(z\), compute \(\frac{\bar{w}}{\bar{z}}\): \( \frac{\bar{w}}{\bar{z}} = \frac{a - bi}{c - di} \) Again, multiply both the numerator and denominator by the complex conjugate of the denominator \((c + di)\): $$ \frac{\bar{w}}{\bar{z}} = \frac{(a - bi)(c + di)}{(c - di)(c + di)} $$

Simplify the expressions for \(\overline{\left(\frac{w}{z}\right)}\) and \(\frac{\bar{w}}{\bar{z}}\), and compare if they are equal: For \(\overline{\left(\frac{w}{z}\right)}\): $$ \overline{\left(\frac{w}{z}\right)} = \overline{\frac{(a + bi)(c - di)}{(c + di)(c - di)}} = \overline{\frac{ac - adi + bci - bdi^2}{c^2 + 2cdi - d^2i^2}} $$ Since \(i^2 = -1\), we get: $$ \overline{\left(\frac{w}{z}\right)} = \overline{\frac{ac + bd + (bc - ad)i}{c^2 + d^2}} = \frac{ac + bd}{c^2 + d^2} - \frac{(bc - ad)i}{c^2 + d^2} $$ For \(\frac{\bar{w}}{\bar{z}}\): $$ \frac{\bar{w}}{\bar{z}} = \frac{(a - bi)(c + di)}{(c - di)(c + di)} = \frac{ac + adi - bci + bdi^2}{c^2 + 2cdi - d^2i^2} $$ Again, since \(i^2 = -1\), we have: $$ \frac{\bar{w}}{\bar{z}} = \frac{ac + bd - (ad + bc)i}{c^2 + d^2} = \frac{ac + bd}{c^2 + d^2} - \frac{(ad + bc)i}{c^2 + d^2} $$ Now, we can see that: $$ \overline{\left(\frac{w}{z}\right)} = \frac{\bar{w}}{\bar{z}} $$ Thus, the statement holds true, and we have completed the proof.