I am wondering why the following claim in Ahlfors' book Complex Analysis' page 79 is true:
Suppose $w$ is a complex number or $\infty$ such that $$\frac{aw + b}{cw + d} = \frac{\overline{aw + b}}{\overline{cw + d}}.$$ where $ad-bc \neq 0$. By cross multiplication we get that $$(a \bar c - c \bar a)|w^2| + (a \bar d - c \bar b)w + (b \bar c - d \bar a)\bar w + b \bar d - d \bar b = 0.$$ If $a \bar c - c \bar a = 0$ then this is the equation of a straight line, for under this condition the coefficient $a \bar d - c \bar b$ cannot also be zero.
Why is it true that the coefficient $a \bar d - c \bar b$ cannot also be zero? If that were zero, we would have $b \bar d - d \bar b = 0$ also, but I am not sure how that would lead to a contradiction.
Edit after accepting @geetha290krm's answer: To suppliment his answer, if $a = 0$ and $a \bar d - c \bar b = 0$, then $c \bar b = 0$ so that either $b = 0$ or $c = 0$. But then $ad - bc = 0 - 0 = 0$, a contradiction. Similar reasoning applies to the case if $c = 0$.
Suppose $a \bar d - c \bar b = 0$ and $a \bar c - c \bar a = 0$. If $\lambda =\frac {\bar a} a$ then $\lambda =\frac {\bar c} c$ from the second equation. So the first equation gives $\lambda a \bar d -\lambda c \bar b = 0$ or $\bar a \bar d -\bar b \bar c=0$ so $ad-bc=0$, a contradiction.
[The special case when $a$ or $c$ vanishes is left to you].