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How to find the derivative of $e^{z}$ , where $z$ is a complex number. Any proof ?

I tried it by using the definition. But it seems difficult.

Angelo Mark
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    Hint: Series expansion – user251257 Oct 16 '15 at 01:21
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    There are a lot of definitions of $e^z$. Which are you using? – Chappers Oct 16 '15 at 01:30
  • Here $z=x+iy$ . So how can I use series expansion ? Thank you. – Angelo Mark Oct 16 '15 at 01:33
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    @AmilaAbesingha definition of $e^z$, not what is $z$ ... – user251257 Oct 16 '15 at 10:59
  • @AmilaAbesingha Was there a specific type of Answer you're looking for on this? I posted clear and simple proofs below, and I quickly got a bunch of troll down-votes. Were THEY looking for some specific type of answer? – Jerry Guern Oct 16 '15 at 15:22
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    @JerryGuern, your proofs are incorrect because they use operations which are not valid in general and their use has not been justified for this particular problem. That is why they were downvoted. – Antonio Vargas Oct 16 '15 at 15:25
  • @AntonioVargas So why is the Question itself getting down-voted if not straight trolling? – Jerry Guern Oct 16 '15 at 16:01
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    @JerryGuern, the question is likely being downvoted because it lacks context necessary to give a useful answer (i.e. what definition of $e^z$ is being used). It also does not give enough information about where the asker is stuck (i.e. "I tried it by using the definition. But it seems difficult."), which could really help answerers frame their answers in ways that would be instructive for the asker. I am very familiar with this website and I assure you that, based on my knowledge of the community norms here, nothing is suspicious or "trolling" about the votes on this question or its answers. – Antonio Vargas Oct 16 '15 at 16:28
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    @JerryGuern ...A third reason that users might downvote this question is that it shows a lack of research effort. There are lots of resources on the internet (not to mention every calculus textbook) that explain how the derivative of the exponential function can be calculated. – Antonio Vargas Oct 16 '15 at 16:34
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    I should have said "every complex analysis textbook" in my last comment. – Antonio Vargas Oct 16 '15 at 16:46
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    It's reasonable to assume that "by definition" means trying to compute $$\lim_{\mathbb{C} \ni h \to 0} \frac{e^{z+h}-e^z}{h},$$ which, while doable, is fairly tedious. – mrf Oct 17 '15 at 15:00

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Assuming that $e^z$ is defined as $e^z = e^{x+iy} = e^x(\cos y + i \sin y)$, you can check that Cauchy-Riemann's equations are satisfied:

$$ u = e^x \cos y, \qquad v = e^x \sin y $$ so $$ u'_x = e^x \cos y = v'_y \qquad u'_y = -e^x \sin y = -v'_x. $$ Hence (since $u$ and $v$ are clearly $C^1$), $e^z$ is complex differentiable everywhere, and (also from Cauchy-Riemann) the derivative is given by $$ u'_x + iv'_x = e^x\cos y + i e^x\sin y = e^z $$ as expected.

mrf
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