hi, this is a stupid question but is the real part of the function

$\displaystyle f(z) = e^{iz} $

$\displaystyle cos z $ ?

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- Mar 24th 2010, 10:20 PMAdre5real part of a function
hi, this is a stupid question but is the real part of the function

$\displaystyle f(z) = e^{iz} $

$\displaystyle cos z $ ? - Mar 24th 2010, 10:26 PMProve It
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 $\displaystyle e^{-y}\sin{x}$. - Mar 24th 2010, 10:44 PMAdre5
ohh thank you!! you made it so clear for me (Happy)