Not sure if I have done this correctly, seems too straight forward, any help is very appreciated.

Find the real and imaginary parts of $f(z) = \cos(z)$.

$\cos(z) = \cos(x+iy) = \cos x\cos(iy) − \sin x\sin(iy) = \cos x\cosh y − i\sin x\sinh y$

Is that correct?

  • $\begingroup$ Looks good! So the real and imaginary parts are (...). $\endgroup$ – StackTD Apr 15 '16 at 15:45
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    $\begingroup$ see: google.com/… $\endgroup$ – imranfat Apr 15 '16 at 15:46

By definition, $$ \cos z=\frac{e^{iz}+e^{-iz}}{2},\qquad \sin z=\frac{e^{iz}-e^{-iz}}{2i} $$ In particular, for real $y$, $$ \cos(iy)=\frac{e^{-y}+e^{y}}{2}=\cosh y $$ and $$ \sin(iy)=\frac{e^{-y}-e^{y}}{2i}=i\frac{e^{y}-e^{-y}}{2}=i\sinh y $$

So, yes, you're correct.


Using the exponential definition of the cosine,

$$2\cos(z)=e^{iz}+e^{-iz}=e^{-y+ix}+e^{y-ix}\\ =e^{-y}(\cos(x)+i\sin(x))+e^{y}(\cos(x)-i\sin(x))\\ =(e^y+e^{-y})\cos(x)-i(e^y-e^{-y})\sin(x)).$$

  • $\begingroup$ Actually, the question doesn't state clearly what kind of answer is expected, so $\Re(\cos(z))$ and $\Im(\cos(z))$ are equally good ! $\endgroup$ – Yves Daoust Apr 15 '16 at 16:01

It is simple, but tis the beauty of the trig/exponential functions! You're $correct$!


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