Consider the imaginary
![\[ w = \frac{az+b}{cz+d} \qquad a,b,c,d \in \mathbb{R}, \quad z \in \mathbb{C}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-79367adb6bb51c8f3c1f93d069a7cb76_l3.png "Rendered by QuickLaTeX.com")
Prove that the following identity holds:
![\[ w - \overline{w} = \frac{(ad-bc)(z - \overline{z})}{|cz+d|^2}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-afa7279a4ba0c7f67e97ee07cfcf6ca7_l3.png "Rendered by QuickLaTeX.com")
If the quantity 
is positive, prove that the imaginary parts of the complex numbers 
and 
have the same sign.
Proof. For the first part of the proof we compute using the definition of and the fact that 
.
![\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*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-efdae9e3c26afaa9f26134817da1c102_l3.png "Rendered by QuickLaTeX.com")
This proves the first part of the theorem. To prove the second part, first we note that for any complex numbers and we have
![\[ w - \overline{w} = 2 \operatorname{Im}(w) \quad \text{and} \quad z - \overline{z} = 2 \operatorname{Im}(z). \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-f4f89e4bf5586cdeaf203042c3192ec9_l3.png "Rendered by QuickLaTeX.com")
Therefore, from the identity we established in the first part we have,
![\[ 2 \operatorname{Im}(w) &= \frac{ad-bc}{|cz+d|^2} \cdot 2 \operatorname{Im}(z). \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-ee1a46b4602d2bffc9072b141f22ae12_l3.png "Rendered by QuickLaTeX.com")
So, if 
then the quantity
![\[ \frac{ad-bc}{|cz+d|^2} > 0 \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-6fc0d1842e28dc78086bd2d45486b982_l3.png "Rendered by QuickLaTeX.com")
(since the denominator is always positive as well). Therefore, 
and 
have the same sign
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.
I hope this is a joke. Everything in the problem and this solution by RoRi is fine. Your claim that the product of two complex numbers is equal to the product of their respective conjugates is wrong. The product of two complex numbers is equal to the *conjugate* of the product of their respective conjugates.