I'm trying to understand analytic continuation and I noticed on wolfram that it
allows the natural extension of the definition trigonometric,
exponential, logarithmic, power, and hyperbolic functions from the
real line $\mathbb{R}$ to the entire complex plane $\mathbb{C}$
So how does it extend, say, $f(x) = \sin(x)$, $x \in \mathbb{R}$ to the complex plane? What are the steps that have to be taken to extend this function (and others) to the complex plane?
Best Answer
There are various ways to extend a function $f:\>J\to{\mathbb R}$ defined on some open interval $J\subset{\mathbb R}$ to a holomorphic function $\tilde f:\>\Omega\to{\mathbb C}$, if such an extension is at all possible. But whatever method you use, you will always end up with the same function $\tilde f$.
(i) If $f$ is given as a rational expression $f(x)={p(x)\over q(x)}$ with $p$ and $q$ polynomials, $q(x)\ne0$ on $J$, then the extension is simply $$\tilde f(z):={p(z)\over q(z)}\ ,$$ and is defined on all of ${\mathbb C}$, minus the complex zeros of $q$.
(ii) If $f$ is given by some power series with positive radius of convergence $\rho$: $$f(x)=\sum_{k=0}^\infty c_k(x-a)^k\qquad(|x-a|<\rho)$$ then you may again put $$\tilde f(z):=\sum_{k=0}^\infty c_k(z-a)^k\qquad(|z-a|<\rho)$$ and can be sure that $\tilde f$ is holomorphic at least in the disk $D_\rho(a)$.
(iii) If $f$ is the solution of some IVP $y'=g(x,y)$, $y(x_0)=y_0$, and $g$ is "real analytic" (example: $y'=\lambda y$) then this ODE can be viewed at also in a complex sense (i.e., $x$ and $y$ are complex variables). The IVP then defines a holomorphic function $\tilde f$ in the neighborhood of $(x_0,y_0)$, which is the extension of $f$ to the complex domain.
An example: Euler's formula gives $$\sin t={e^{it}-e^{-it}\over 2i}\ .$$ We can then say right away that $$\sin z:={e^{iz}-e^{-iz}\over 2i}$$ is the complex extension of $\sin$.