hi, this is a stupid question but is the real part of the function
$\displaystyle f(z) = e^{iz} $
$\displaystyle cos z $ ?
No, it's not.
Remember that $\displaystyle z$ is a complex number.
In other words, $\displaystyle z = x + iy$.
You need to write $\displaystyle f(z) = e^{iz}$ as $\displaystyle u + iv$, where $\displaystyle u, v$ are real functions of $\displaystyle x$ and $\displaystyle y$.
So $\displaystyle f(z) = e^{iz}$
$\displaystyle = e^{i(x + iy)}$
$\displaystyle = e^{ix + i^2y}$
$\displaystyle = e^{-y + ix}$
$\displaystyle = e^{-y}e^{ix}$
$\displaystyle = e^{-y}(\cos{x} + i\sin{x})$ by Euler's Rule
$\displaystyle = e^{-y}\cos{x} + i\,e^{-y}\sin{x}$.
$\displaystyle = u + iv$.
So the real part of $\displaystyle e^{iz}$ is $\displaystyle e^{-y}\cos{x}$ and the imaginary part of $\displaystyle e^{iz}$ is $\displaystyle e^{-y}\sin{x}$.