interesting complex equation

**tombrownington** · September 25th 2008, 08:43 AM

Here is an interesting problem I'm having some difficulty with:
Find all the complexes z that satisfy the equation:
cos(z) = cosh(z).
The answer should be z=k(pi)(1+i) & z=k(pi)(1-i), but I don't know how to show this.
Any help, folks?
Thanks.

**bobak** · September 25th 2008, 09:08 AM

Salam

you must use the two following identities.

$\cosh z = \cos i z$

$\cos A = \cos B \ \ \ \rightarrow A = 2 \pi n \pm B$

Can you take it form here, if you need help proving those identities just ask.

Bobak

**tombrownington** · September 25th 2008, 09:34 AM

hi bobak,
thanks a lot for the help. only i thought cosA = cosB => A = 2n(pi) +/- B holds true for A & B real numbers. how does this also work for complexes. is cos a periodic function for complexes.
can you please clarify?
thanks.

**bobak** · September 25th 2008, 01:43 PM

Originally Posted by tombrownington

hi bobak,
thanks a lot for the help. only i thought cosA = cosB => A = 2n(pi) +/- B holds true for A & B real numbers. how does this also work for complexes. is cos a periodic function for complexes.
can you please clarify?
thanks.

consider the function $\cos z = \frac{e^{iz} + e^{-iz} }{2}$ adding $2 \pi$ to z will make no difference as $\cos (z +2 \pi) = \frac{e^{i(z +2 \pi)} + e^{-i(z +2 \pi)} }{2} \ = \ \frac{e^{iz}e^{i 2\pi} + e^{-iz}e^{-i 2\pi} }{2}$ however $e^{i 2\pi} = 1$ so $\cos z = \cos (z + 2 \pi)$ .

Bobak

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