complex trig Q

**Danneedshelp** · September 24th 2010, 09:39 AM

Hi, I am not seeing the following step:

We are given $cos(z)=\frac{e^{iz}+e^{-iz}}{2}=\frac{1}{2}$ . Therefore,

$e^{2iz}-e^{iz}+1=0$ , and sp by the quadratic formula....

I don't see how the got $cos(z)=\frac{e^{iz}+e^{-iz}}{2}=\frac{1}{2}$ into the form $e^{2iz}-e^{iz}+1=0$ . I can see the mult. the 2 and subtracted $e^{iz}$ over, but I don't see how the got the $e^{2iz}$ term.

Thanks

**Traveller** · September 24th 2010, 10:23 AM

In the original equation, multiply both sides by 2, bring the 1 to the LHS, and multiply both sides by $e^{iz}$ . Can you see what happens?

**Danneedshelp** · September 24th 2010, 10:53 AM

$e^{iz}+e^{-iz}-1=0$
$e^{iz}e^{iz}+e^{iz}e^{-iz}-e^{iz}=0$
$e^{iz+iz}+e^{-iz+iz}-e^{iz}=0$
$e^{2iz}+e^{0}-e^{iz}=0$
$e^{2iz}-e^{iz}+1=0$

...bad post, sorry.

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