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Prove that cos z = 0 if and only if z = π/2 + kπ, where k is an integer.
The solution is :
cos z = (eiz+e-iz)/2 = 0
<=>eiz+e-iz)=0
<=>e2iz=-1
<=>2iz=πi +2kπi
<=>z=π/2 + kπ
But i dont understand how does these 2 steps work
<=>e2iz=-1
<=>2iz=πi +2kπi
how can a exp function suddenly become πi +2kπi
Thanks
The solution is :
cos z = (eiz+e-iz)/2 = 0
<=>eiz+e-iz)=0
<=>e2iz=-1
<=>2iz=πi +2kπi
<=>z=π/2 + kπ
But i dont understand how does these 2 steps work
<=>e2iz=-1
<=>2iz=πi +2kπi
how can a exp function suddenly become πi +2kπi
Thanks