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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

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