# Complex trig function example. If $\cos z=5$, why does $e^{2iz}-10e^{iz}+1=0$?

The example in my book says this:

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?

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

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 $$5$$):

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

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

• Why the downvote tho – Shubham Johri Oct 28 at 11:38

From $$\cos z = 5$$, we have $$2 \cos z = 10$$, so that $$e^{2iz} + 1 = 10e^{iz}$$, or $$e^{2iz} - 10e^{iz} +1 = 0$$.