# Prove the Following Using Proof By Contradiction

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.

• 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 $f \equiv e^z$ – JoseSquare Feb 27 at 17:17
• @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? – bigsbylp Feb 27 at 17:59
• 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. – JoseSquare Feb 27 at 18:03

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.
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 $$f(z)=e^z$$