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I need to prove the following that

$e^{\bar{z}} = \bar{e^{z}}$

with $e^z := \Sigma_{k = 0}^{k = \infty} \frac{z^k}{k!}$

The problem that I am having is first we know $e^z$ defined that way always converge, but what I don't understand is we have to see that $\bar{z^k}$ put into the formula above converges to same number as if we do the summation and then converge to a number, and conjugate that. I don't know how to do that..

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First, we have

$$\overline{\lim_{N\to \infty}S_N}=\lim_{N\to \infty}\overline{S_N}$$

Second, we have

$$\overline{S_N}=\overline{\sum_{k=0}^Nf_k(z)}=\sum_{k=0}^N\overline{f_k(z)}$$

Third, we have

$$\overline{f_k(z)}=\overline{\left(\frac{z^k}{k!}\right)}=\frac{\overline{z^k}}{k!}$$

Finally, we have

$$\overline{z^k}=\bar z^k$$

and we are done, as this last equality follows inductively from the fact that $\overline{z_1z_2}=\bar z_1\bar z_2$!

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Say $z=x+iy$. Then $$e^{\bar{z}}=e^xe^{-iy}=e^x\cos(y)-ie^x\sin(y)$$ $$\bar{e^{z}}=\bar{e^x\cos(y)+ie^x\sin(y)}=e^x\cos(y)-ie^x\sin(y).$$

To do it your way: conjugation is continuous (it is an isometry), so it respects limits of sequences.

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  • $\begingroup$ how do we know that conjugation is continuous ? $\endgroup$ – Dude Sep 24 '15 at 20:58
  • $\begingroup$ As I said, it is an isometry: $|\bar{z}-\bar{w}|=|z-w|$ $\endgroup$ – Sonner Sep 24 '15 at 23:14
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We know that $$\sum_{k=0}^\infty\frac{z^k}{k!}$$ converges for all $z\in\Bbb C$. Let's let $a_k=1/k!$, and $f(z)=e^z$.

Now,

$$\begin{align}\overline{f(z)} & =\overline{\sum_{k=0}^\infty a_kz^k}\\ & = \sum_{k=0}^\infty a_k\overline z^k,\quad\text{because }a_k\in\Bbb R\\ & = f(\overline z). \end{align}$$

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  • $\begingroup$ how do we know that conjugation is continuous ? $\endgroup$ – Dude Sep 24 '15 at 20:58
  • $\begingroup$ Well, let $\phi:\Bbb R^2\to\Bbb R^2$ where $\phi(x,y)=(\phi_1(x,y),\phi_2(x,y)$ and $\phi_1(x,y)=x$ and $\phi_2(x,y)=-y$. Since $\phi_1,\phi_2$ are clearly continuous, $\phi$ is continuous. Also writing $(x,y)$ as $x+iy$, we see that $\phi(x+iy)=x-iy$. $\endgroup$ – Tim Raczkowski Sep 24 '15 at 21:04
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Let $x^*$ denote the complex conjugate of $ x$ ) .We have $$e^z= x_n(z)+\sum_{k=0}^{k=n} z^n/n!$$ and $$e^{z^*}=x_n(z^*)+\sum_{k=0}^{k=n}(z^*)^n/n!$$ where $(x_n(z))_{n \in N}$ and $(x_n(z^*))_{n \in N}$ are sequences converging to $0$ . Therefore $$(e^z)^*-e^{z^*}=(x_n(z))^*-x_n(z^*)$$.In the above equation the LHS is independent of $n$ and the RHS converges to $0$ as $n \to \infty$. Therefore the LHS is $0$.This method will apply for any convergent power series whose co-efficients are all real numbers.

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