# Deducing that $\cosh$ and $\sinh$ are entire and calculating the complex derivatives $\sinh'$ and $\cosh'$

I know that $\cosh(z)=\cos(iz)$ and $\sinh(z)=-i\sin(iz)$ my question is how can I deduce that $\cosh$ and $\sinh$ are entire? I know that $\cos(x)$ has an infinite radius of convergence and the same goes for $\sin(x)$. If I were to substitute $\cos(iz)$ into the power series for $\cos(x)$ would I be able to show that the radius of convergence is also $\infty$ which would prove that $\cosh(z)$ is entire? To find the derivatives I assume I need to use the Cauchy Riemann equations? Or would I just be able to differentiate the power series?

• By definition $\sinh z=\frac12(e^z-e^{-z})$. – Nosrati Apr 12 '17 at 20:02
• does this imply that it is therefore entire? and if so why? – user395952 Apr 12 '17 at 20:03
• Sum of two entire functions is entire.... – Nosrati Apr 12 '17 at 20:04

Those three facts allow you to conclude that $\sinh$ and $\cosh$ are entire based on their definition in terms of $e^x$, since $e^x$ is an entire function.
• @Gibberish well, you can use the equations you've given + the chain rule, so $\frac{d}{dx}\cosh(z)=-i\sin{iz}=\sinh{z}$ or you can use the definition in terms of $e$ – Stella Biderman Apr 13 '17 at 4:16