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Using proof by contradiction, show that $e^z$ is the only entire function that satisfies the conditions $f'(z) = f(z)$ and $f(0) = 1$

I'm stuck on finding the contradiction:

Suppose that $f(z)$ is an entire function, is not equal to $e^z$, and satisfies the given conditions

Let $f(z) = u(x,y) + iv(x,y)$

Then $f'(z) = u_x(x,y) + iv_x(x,y)$

And because $f'(z) = f(z)$, we see that $u = u_x$

However, I do not know if this is the best way to approach the problem, and if it is, what I can derive from this.

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  • $\begingroup$ I think an easy way could considering power series, since $f$ is holomorphic it has a power series centered at $z=0$, try to use this fact and conclude $f \equiv e^z$ $\endgroup$ – JoseSquare Feb 27 at 17:17
  • $\begingroup$ @JoseSquare so far, we have only defined holomorphic as one which is differentiable at all points on the set it is contained in. How would the power series lead to a contradiction? $\endgroup$ – bigsbylp Feb 27 at 17:59
  • $\begingroup$ Ups, well I let the answer for anyone who finds it helpful, the answer provide by Bobo it's what you are looking for then. $\endgroup$ – JoseSquare Feb 27 at 18:03
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Suppose that there is another function other than $e^z$ which satisfies this condition, which we will call $f$. Let $g(z) = e^{-z}f(z)$. This function is holomorphic because it is the product of two holomorphic functions. Furthermore, we have $g'(z) = e^{-z}(f'(z) - f(z)) = 0$. Thus, $g(z) = C$ for some constant $C \in \mathbb{C}$. We then have $C = e^{-z}f(z)$. Hence, $f(z) = Ce^z$. Apply the initial condition to get a contradiction to our assumption.

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Let $f(z) = \sum_{n=0}^{\infty} a_n z^n$, and so $f'(z) = \sum_{n=1}^{\infty} n \,a_n z^{n-1}=\sum_{n=0}^{\infty} (n+1)a_{n+1} z^n$. Now must be $a_n = (n+1) a_{n+1}$ for all $n\in \Bbb{N}$, and because $f(0)=1$, then $a_0 = 1$, and so as $a_{n+1} = \frac{a_n}{n+1}$ then $a_1 = 1$, $a_2 = \frac 1 2$ , $a_3= \frac 1 6, \ldots, a_n = \frac{1}{n!}$. So $f$ has the same development as $e^z$ and so must be $f(z)=e^z$

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