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The example in my book says this:

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equation 1 is:

$$\cos{z} = \frac{1}{2} ( e^{iz} + e^{-iz} )$$

Where is $e^{2iz} - 10e^{iz} + 1 = 0 $ coming from? Can someone show me how they are doing this solution?

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    $\begingroup$ Can you multiply with $e^{iz}$? $\endgroup$ – Fakemistake Oct 25 '20 at 13:08
  • $\begingroup$ Is this some homework? $\endgroup$ – DavidW Oct 26 '20 at 0:06
  • $\begingroup$ Yes it is david $\endgroup$ – Jwan622 Oct 27 '20 at 3:57
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You have:

$$ \cos z = \dfrac{1}{2} \left( e^{iz} + e^{-iz} \right) \tag{1} $$

Multiplying both sides by $ e^{iz} $ as said (and replacing $ \cos z $ with $5$):

$$ 5 e^{iz} = \dfrac{e^{iz}}{2} \left( e^{iz} + e^{-iz} \right) \\ \implies 10 e^{iz} = e^{2iz} + 1 $$

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$\frac12(e^{iz}+\frac1{e^{iz}})=5\implies e^{2iz}-10e^{iz}+1=0.$

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  • $\begingroup$ Why the downvote tho $\endgroup$ – Shubham Johri Oct 28 '20 at 11:38
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From $\cos z = 5$, we have $2 \cos z = 10$, so that $e^{2iz} + 1 = 10e^{iz}$, or $e^{2iz} - 10e^{iz} +1 = 0$.

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