Originally Posted by

**demode** In the following, I don't understand how the expression on the LHS equals to the one on the RHS:

$\displaystyle Im \left( \frac{(-e^{\pi}-1)(1-i)}{(1+i)(1-i)} \right) =\frac{e^{\pi}+1}{2}$

("Im" means "imaginary")

Of course if on LHS we cancel out the (1-i) term on the numerator with the one on denominator we get

$\displaystyle Im \left( \frac{(-e^{\pi}-1)}{(1+i)} \right)$

And if $\displaystyle i=1$, then the denominator becomes 2. But I can't work out the rest. Any help is appreciated.