Let's define $\exp(z)$ in the following form: $$\exp(z) = \sum \limits_{n=0}^{\infty} \frac{z^n}{n!}, z \in \mathbb{C} \tag{1}$$ We know that the definition above stands (I mean that the series converges).
We are to prove that $z\mapsto \exp(z)$ is a holomorphic function over $\mathbb{C}$.


Firstly I would like to prove that $\exp \in H(\mathbb{C})$.
I know that $f(z)=z$ is a holomorphic function (I've proven it using Cauchy-Riemann's equations).
It is obvious that $\forall_{n \in \mathbb{N}}$ $f_n(z) = \frac{z}{n}$ is holomorphic (is it?).
If the statement above stands it would be good to say that an infinite sum of holomorphic functions is holomorphic. I'm not sure it it's true however.


Secondly let's calculate $\frac{\mbox{d}}{\mbox{dz}}\exp(z)$.
Because we know that $z\mapsto \exp(z)$ is holomorphic (to my mind it basically means that I can calculate this derivative using methods from real analysis, am I correct?) and the series converges we can calculate it in the following form (using $(1)$): $$\frac{\mbox{d}}{\mbox{dz}}\exp(z) =\frac{\mbox{d}}{\mbox{dz}} \sum \limits_{n=0}^{\infty} \frac{z^n}{n!} = \sum \limits_{n=0}^{\infty} \frac{\mbox{d}}{\mbox{dz}}\frac{z^n}{n!} = \sum \limits_{n=1}^{\infty} n \frac{z^{n-1}}{n!} = \sum \limits_{n=1}^{\infty} \frac{z^{n-1}}{(n-1)!} = \sum \limits_{n=0}^{\infty} \frac{z^{n}}{n!} = \exp(z)$$Is it calculated well?

up vote 2 down vote accepted

You must know that every analytic function (i.e. which can be written as the sum of a power series, like $\exp$) is holomorphic.

Moreover, if you recall the theorem saying that it's possible to exchange $\displaystyle \sum_{n=0}^{+\infty}$ and $\dfrac{d}{dz}$ in the case of power series, then your computation is correct.

In any case, be careful to check first the radius of convergence of the power series. Here, using ratio test:

$$\dfrac{\dfrac{z^{n+1}}{(n+1)!}}{\dfrac{z^{n}}{n!}} = \dfrac{z}{n+1}$$

goes to $0$ when $n$ go to $+\infty$, thus it's convergent on $\Bbb C$.

  • Thank you! So using the theorem that every analytical function is holomorphic we get the thesis immediately from the definition of $\exp(z)$? – Hendrra May 11 at 20:31
  • 1
    That's exactly my point. – paf May 11 at 20:32
  • Thank you! :) And yes, you are right with the radius of convergence. But in this case it was written in my task that the radius in infinite thus in is convergent on $\mathbb{C}$. – Hendrra May 11 at 20:33
  • One more question - the holomorphism of $\sin$ and $\cos$ is obvious because they are sum of $\exp(iz)$ and $\exp(-iz)$ which are holomorphic? – Hendrra May 11 at 20:38
  • 1
    Linear combination but yes, it's correct. – paf May 11 at 20:39

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