If a, b are complex numbers then the maximum value of $\dfrac{a\bar b+\bar ab}{|ab|}$ If a, b are complex numbers then the maximum value of $\dfrac{a\bar b+\bar ab}{|ab|}$ is
(A) 2
(B) 1
(C) the expression may not always be a real number and hence maximum does not make sense
(D) none of the above.
My Steps:
Let $a=x+iy$ and $b=p+iq$.
Then $$\begin{align}
\dfrac{a\bar b+\bar ab}{|ab|} &= \dfrac{(x+iy)(p-iq)+(x-iy)(p+iq)}{|(x+iy)(p+iq)|}\\
&= \dfrac{2xp+2yq}{(xp-yq)^2+(xq+yp)^2}
\end{align}$$
If I take $y=q=0$ and $x=p=1/n$, the maximum value can be taken to $\infty$.
So, is the answer option D ?
Please advise.

Hint:
$$ \dfrac{a\bar{b}+\bar{a}b}{|ab|}=\dfrac{2\Re(\bar{a}b)}{|\bar{a}b|}\leq 2$$
