0
$\begingroup$

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?

$\endgroup$
  • 3
    $\begingroup$ By definition $\sinh z=\frac12(e^z-e^{-z})$. $\endgroup$ – Nosrati Apr 12 '17 at 20:02
  • $\begingroup$ does this imply that it is therefore entire? and if so why? $\endgroup$ – user395952 Apr 12 '17 at 20:03
  • 1
    $\begingroup$ Sum of two entire functions is entire.... $\endgroup$ – Nosrati Apr 12 '17 at 20:04
0
$\begingroup$

As mentioned in the comments, the best approach is to use the definitions of the two functions. The product rule tells us that the product of two entire functions is entire. Similarly, the sum rule tells us that the sum of two entire functions is entire and the chain rule tells us that the composition of two entire functions is entire. These three rules say not only what forms the functions derivatives take, but also the fact that they are differentiable (up to some caveats that don't apply here)

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.

$\endgroup$
  • $\begingroup$ Ahh okay that makes sense, how would one go about finding the complex derivatives of cosh and sinh? $\endgroup$ – user395952 Apr 13 '17 at 4:00
  • $\begingroup$ @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 $e$ $\endgroup$ – Stella Biderman Apr 13 '17 at 4:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy