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Prove an identity for imaginary numbers of a particular form

Consider the imaginary

    \[ w = \frac{az+b}{cz+d} \qquad a,b,c,d \in \mathbb{R}, \quad z \in \mathbb{C}. \]

Prove that the following identity holds:

    \[ w - \overline{w} = \frac{(ad-bc)(z - \overline{z})}{|cz+d|^2}. \]

If the quantity ad-bc is positive, prove that the imaginary parts of the complex numbers w and z have the same sign.


Proof. For the first part of the proof we compute using the definition of w and the fact that \overline{\frac{az+b}{cz+d}} = \frac{a \overline{z} + b}{c \overline{z} + d}.

    \begin{align*}  w - \overline{w} &= \frac{az+b}{cz+d} - \frac{a\overline{z} + b}{c \overline{z} + d} \\[9pt]  &= \frac{(az+b)(c \overline{z} + d) - (a \overline{z} + b)(cz + d)}{(cz+d)(c\overline{z}+d)} \\[9pt]  &= \frac{ad(z - \overline{z}) + bc(\overline{z} - z)}{(cz+d)\overline{(cz+d)}} \\[9pt]  &= \frac{(ad-bc)(z - \overline{z})}{|cz+d|^2}. \end{align*}

This proves the first part of the theorem. To prove the second part, first we note that for any complex numbers w and z we have

    \[ w - \overline{w} = 2 \operatorname{Im}(w) \quad \text{and} \quad z - \overline{z} = 2 \operatorname{Im}(z). \]

Therefore, from the identity we established in the first part we have,

    \[  2 \operatorname{Im}(w) &= \frac{ad-bc}{|cz+d|^2} \cdot 2 \operatorname{Im}(z). \]

So, if ad -bc > 0 then the quantity

    \[ \frac{ad-bc}{|cz+d|^2} > 0 \]

(since the denominator is always positive as well). Therefore, \operatorname{Im}(w) and \operatorname{Im}(z) have the same sign. \qquad \blacksquare

One comment

  1. Anonymous says:

    One of the rules for complex conjugates given is that the ratio of two complex numbers is equal to the ratio of their respective conjugates. Using your definition of the conjugate of w would result in it being the same as w. So the conjugate of w is wrong here. If you multiply both sides by the denominator, then use the fact that the product of two complex numbers is equal to the product of their respective conjugates, you get the conjugate of w being the numerator unchanged, but the denominator is the conjugate of the initial denominator. This would mean the problem is wrong though.

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