Maximum value of $\frac{\alpha\overline{\beta}+\overline{\alpha}\beta}{|\alpha\beta|}$ Maximum value of  $$\frac{\alpha\overline\beta+\overline\alpha\beta}{|\alpha\beta|}$$ is
1) 2
2) 1
3) none of the above.
Considering $\alpha=x+iy$ and $\beta=m+in$ , on evaluating the expression I got $\frac{2.(xm+ny)}{\sqrt{(x^{2}+y^{2})(m^{2}+n^{2})}}$ which is $\leq \frac{2.(xm+ny)}{\sqrt{4xymn}}$. Least value of this is 2. So can we call it as maximum value ?

To show that $2$ is possible to obtain, let $\alpha = \beta = 1$.
For all other values of $\alpha, \beta$, the triangle inequality gives 
$$
|\alpha\bar \beta + \bar \alpha\beta| \leq |\alpha\bar\beta|+|\bar\alpha\beta|
$$
Can you finish from here?
Alternatively, let $\gamma = \alpha\bar\beta$. In that case, your fraction may be written as $\frac{\gamma + \bar \gamma}{|\gamma|}$. Now it should be easier to apply $\gamma = x + iy$ than it was when you had both $\alpha$ and $\beta$ to work with.
