plz have a look at the aatached picture
thank you
$\displaystyle sinh(z) = \frac{e^{z} - e^{-z}}{2} $
$\displaystyle = \frac{e^{x+iy} - e^{-x-iy}}{2} $
$\displaystyle = \frac{e^{x}.e^{iy} - e^{-x}.e^{-iy}}{2} $
$\displaystyle = \frac{e^{x}.e^{iy} - e^{-x}.\frac{1}{e^{iy}}}{2} $
$\displaystyle = \frac{e^{x}.(cos(y)+isin(y)) - \frac{1}{e^{x}.(cos(y)+isin(y))}}{2} $
$\displaystyle = \frac{e^{x}.(cos(y)+isin(y)) - \frac{1}{e^{x}.(cos(y)+isin(y))}}{2} $
Here you have a complex number which is composed of subtraction of two complex numbers.
Remember that if $\displaystyle w$ and $\displaystyle z$ are complex numbers then:
$\displaystyle \overline{(v - w)} = \overline{v} - \overline{w}$
In this case $\displaystyle v = e^{x}.cos(y)+ie^{x}sin(y) $ $\displaystyle w = \frac{1}{e^{x}.cos(y)+ie^xsin(y)} $
And you're trying to find $\displaystyle \frac{1}{2} \overline{(v - w)}$
Get $\displaystyle w$ in a nicer form by multiplying top and bottom by the conjugate of the denominator first.
$\displaystyle w = \frac{1}{e^{x}.cos(y)+ie^xsin(y)} = \frac{e^{x}.cos(y)-ie^xsin(y)}{e^{2x}.cos^2(y)+ie^{2x}sin^2(y)} = \frac{cos(y)-isin(y)}{e^{x}}$
Hence
$\displaystyle \frac{1}{2}( v - w) = \frac{1}{2}(e^{x}.cos(y)+ie^{x}sin(y) - (\frac{cos(y)-isin(y)}{e^{x}})$
Should be relatively easy to find the conjugate now.