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Prove with the definition of differentiability that $\exp(z)$ is differentiable in $\mathbb C$ and $(\exp(z))' = \exp(z)$ for all $z \in \mathbb C.$

I tried:

\begin{align*} \frac{\exp(z+h) - \exp(z)}{h} &= \frac{\sum_{k = 0}^\infty \frac{(x+h)^k}{k!} - \sum_{k = 0}^\infty x^k/k!}{h} \\ &= \frac{1}{h} \sum_{k = 0}^\infty \frac{(x+h)^k - x^k}{k!} \\ &= \frac{1}{h} \sum_{k = 0}^\infty \frac{\sum_{j = 0}^k \binom{k}{j} x^{k-j}h^j - x^k}{k!} \\ &= \frac{1}{h} \sum_{k = 0}^\infty \sum_{j = 0}^k\frac{ \frac{k!}{(k-j)! j!} x^{k-j}h^j - x^k}{k!} = \frac{1}{h} \sum_{k = 0}^\infty \sum_{j = 0}^k\frac{x^{k-j}h^j - x^k}{(k-j)! j!}. \end{align*}

I don't know how to go on from here.

numerion
  • 673

1 Answers1

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If you have at your disposal the fact that

$e^{z_1 + z_2} = e^{z_1} e^{z_2} \tag{1}$

for all $z_1, z_2 \in \Bbb C$, then this calculation becomes a lot easier; for then we may assert that

$e^{z + h} = e^z e^h, \tag{2}$

whence

$e^{z + h} - e^z = e^z e^h - e^z = e^z(e^h - 1), \tag{3}$

so that

$(e^{z + h} - e^z) / h = e^z((e^h -1) / h). \tag{4}$

Now when we take the limit as $h \to 0$, we only need to look at $((e^h - 1) / h)$, and the power series for $e^h$ is much easier to apply; indeed we have

$(e^h - 1) / h = \sum_{n = 1}^{n = \infty} (h^{n - 1} / n!) = 1 + \sum_{n = 2}^{n = \infty} (h^{n - 1} / n!), \tag{5}$

and it is easy to see that

$\sum_{n =2}^{n = \infty}(h^{n - 1} / n!) \to 0 \; \text{as} \; h \to 0; \tag{6}$

thus we have

$(e^z)' = e^z. \tag{7}$

QED!!!

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!

Robert Lewis
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