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I want to show that $f(z) = \frac{\cos z}{e^z}$ is analytic. I've used the definition of $\cos z = \frac{e^{iz} + e^{-iz}}{2}$ to get $f(z) = \frac 1 2 (e^{(-1 + i)z} + e^{(-1 - i)z})$.

Since $e^z$ is analytic for any complex $z$, then this is just the sum of two analytic functions, so $f(z)$ must be analytic.

Does this proof suffice?

  • How do you know that the sum or product of analytic functions is analytic? What are you relying on? – Mark Viola Sep 15 '22 at 03:43
  • @MarkViola We can use the limit definition to prove that the derivative of $(f + g)$ exists at any point in the domain, so their addition must also be analytic. –  Sep 15 '22 at 03:49
  • Ratio of two analytic functions is analytic if the denominator has no zeros. – Kavi Rama Murthy Sep 15 '22 at 05:02
  • Another way would be to write $\cos(z)/e^z=e^{-z}\cos(z)$ and apply the Cauchy Riemann equations. – Mark Viola Sep 15 '22 at 13:06
  • Another way is to represent the function $f(z) = \frac 1 2 (e^{(-1 + i)z} + e^{(-1 - i)z})$ by the power series, which is in this case convergent on the entire space. By definition such function is analytic. See https://math.stackexchange.com/questions/573984/difference-between-analytic-and-holomorphic-function – Ryszard Szwarc Sep 15 '22 at 14:04

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Hint for such excercises: Try writing your function with $x,y.$ That way you can use Cauchy Riemann equations. In general, when you are asked if a function is holomorphic or analytic, you need to check Cauchy-Riemann equations, or apply Morera's theorem to prove it.

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