Show that $e^z$ is continuous on $\mathbb{C}$

Question

I know that $e^z$ is continuous on $\mathbb{R}$, but how would I show this rigorously on $\mathbb{C}$ using the $\epsilon - \delta$ definition of continuity?

I know how to begin:

If $|z - z_0| < \delta$ then we want $|f(z) - f(z_0)| < \epsilon$.

To work backwards, I know we want to basically play around with $|f(z) - f(z_0)| = |e^z - e^{z_0}|$ and then pick $\delta$ to have some relationship with $\epsilon$ so that we get the inequality.

However, I am having a hard time figuring out how to proceed with expanding $|e^z - e^{z_0}|$ in a way that gets me to a point where I can get $|z - z_0|$ to appear somewhere.

Answer 1

Idea: $$ |e^z - e^{z_0}|\le |e^x - e^{x_0}||\cos y + i\sin y| + e^{x_0}|(\cos y + i\sin y) - (\cos y_0 + i\sin y_0)|\le\cdots $$ Can you continue?

Answer 2

1) $a+bi \rightarrow A+Bi $ iff $a\rightarrow A$ and $b\rightarrow B$

2) $$ |e^{a+bi} -e^{A+Bi}| =| e^{A+Bi}| |e^{a-A}e^{(b-B)i} - 1|$$

$$ |e^{a-A}e^{(b-B)i} - 1| \leq |e^{a-A}\cos\ (b-B) -1 | +|e^{a-A}\sin\ (b-B)| $$

$e^x,\ \cos\ x,\ \sin\ x$ are continuous so that the proof is followed

Answer 3

Note that $$e^w-e^z=e^z(e^{w-z} - 1)$$ and if $h=w-z=x+iy$ then $$e^h-1=e^x\cos y-1+ie^x\sin y=(e^x-1)\cos y+\cos y-1+ie^x\sin y$$ Using the fact that $|\sin x|\leq |x|$ for $|x|<1$ we get $|1-\cos y|\leq |y^2|/2<|y|$ if $|y|<1$ and then $$|e^h-1|\leq |e^x-1|+|y|+e^{|x|} |y|$$ Next we can use the inequality $$1+t\leq e^t\leq \frac{1}{1-t}, 0<t<1$$ to show that if $|x|<1/2,|y|<1/2$ then $$|e^x-1|<2|x|,e^{|x|}|y|<2|y|$$ and therefore if $|x|<1/2,|y|<1/2$ then $$|e^h-1|<3(|x|+|y|)$$ Now let $M=|e^z|$ and $\epsilon>0$ be given. And let's choose $\delta=\min (1/2,\epsilon/6M)$ then for all $w$ with $|h|=|w-z|<\delta$ we have $$|e^w-e^z|=|e^z||e^h-1|=M|e^h-1|<3M(|x|+|y|)<6M\delta\leq \epsilon$$ because $|x|\leq |h|<\delta, |y|\leq |h|<\delta$. This proves the continuity of $f(z) =e^z$ at an arbitrary point $z\in\mathbb {C} $.

---

The proof is much simpler if we assume the power series for $e^z$ for then we have $$|e^h-1|\leq \sum_{k=1}^{\infty} \frac{|h|^k} {k!} \leq \sum_{k=1}^{\infty}|h|^{k}=\frac{|h|}{1-|h|}<2|h|$$ if $|h|<1/2$.

Answer 4

This is more of a philosophical answer; the two previous answers already explain how to choose a $\delta$ given some $\epsilon$, which is enough for the proof. However, I think that it is very easy to get caught in a circular argument here: In order to give a precise definition of $e^z$, you need to use power series, and for these you prove continuity (inside the radius of convergence) more or less at the same time as well-definedness (i.e. convergence inside the radius of convergence).

In the comments, you said that you define $e^{x+iy} = e^x(\cos(x)+i\sin(x)$. This only makes sense if you have already defined $\cos$ and $\sin$. The only way I know how to do this is to define them via their Taylor series $\cos x = \sum_{k=0}^\infty (-1)^k\frac{x^{2k}}{(2k!)}$ and $\sin x = \sum_{k=0}^\infty (-1)^k\frac{x^{2k+1}}{(2k+1)!}$. Then, one may ask how you know that these two functions are continuous; the usual way is to use the theorem that power series are the uniform limit of the polynomials one gets by truncating the series to a finite sum on any closed ball of radius smaller than the radius of convergence, that the uniform limit of continuous functions is continuous, and that polynomials are continuous (proof by induction starting from constant functions and the identity function $x\mapsto x$, for which one can check continuity using arbitrary $\delta$ and $\delta = \epsilon$, respectively). In particular, proving their continuity is at least as hard as proving that sums and products of continuous functions are continuous, so you might as well assume that. Alternatively, you can define $e^z = \sum_{k=0}^\infty \frac{z^k}{k!}$ by its complex Taylor series, and the same argument shows that it is continuous. Note that in order to prove that $e^{z\cdot z'} = e^z e^{z'}$ you need to use absolute convergence of the power series on the right to rearrange its summands, which is more or less as difficult as showing its continuity.

In general, mathematical definitions come in two flavours: Some impose very few conditions on the object they define, so it is easy to come up with lots of examples. Others impose very strong conditions, so if you want to prove something about the objects they define, you have a lot of properties you can assume, so it is easy to prove lots of theorems. The epsilon-delta definition sits inbetween: There are at least two examples where you can check the condition in your head (constant functions and the identity function $x\mapsto x$), and many operations (sum, product, composition, uniformly converging limit...) yield continuous functions when applied to continuous functions. All of these theorems basically work by taking a point $x$ and a number $\epsilon > 0$, breaking it up into some pieces $\epsilon_1+\cdots+\epsilon_n = \epsilon$, choosing a $\delta_i$ for $(x,\epsilon_i)$ by using continuity of the fucntion $f_i$, and then taking $\delta = \min_i\delta_i$ or something. Theoretically, this means that you can start with $(z,\epsilon)$ and come out with some $\delta > 0$ which will prove that $e^z$ is continuous, but there is little actual value in providing an explicit $\delta$ - the important point is that one exists, and in almost all applications of continuity you will replace it with a suitably chosen smaller number, anyway.

Answer 5

Let $z_0$ be given and $|z-z_0|<\delta$ therefore$$|e^z-e^{z_0}|=|e^{z_0}|\cdot|e^{z-z_0}-1|=|e^{z_0}|\cdot|e^{z-z_0}-1|$$let $z=z_0+re^{i\theta}$ where $r<\delta$ and $\delta$ being sufficiently small therefore$$|e^{z-z_0}-1|{=|e^{re^{i\theta}}-1|\\=\sqrt{e^{2r\cos \theta}+1-2e^{r\cos \theta}\cos (r\sin\theta)}\\=\sqrt{(e^{r\cos \theta}-1)^2+4e^{r\cos \theta}\sin^2 (r\sin\theta)}\\\le \sqrt{(e^\delta-1)^2+4e^\delta \sin^2\delta}}$$so we have$$|e^z-e^{z_0}|\le|e^{z_0}|\sqrt{(e^\delta-1)^2+4e^\delta \sin^2\delta}$$since we can make both the terms $(e^\delta-1)^2$ and $4e^\delta \sin^2\delta$ sufficiently small so we can with $|e^z-e^{z_0}|$ and this completes our proof.