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Today I've encountered this proof of Euler's Formula. The proof basically says that $e^{iz}$ and $(\cos z + i \sin z)$ are both solutions of the differential equation $f'(z) = i f(z)$ and do not differ by a constant.

Everything is clear to me, however I have a doubt: isn't it necessary to demonstrate that if:

$$ f'(x) = k f(x) \\ g'(x) = k g(x) $$

Then $f' = g'$? If so, how can I proceed to build a proof?

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What was given in that link is that: when $f'=g'$ we know that $f=g+C$ wherein $C$ is any constant. Now, if we consider an initial condition [for example if $z=0$ then $e^{iz}=e^{i\times 0}=1$ and $\cos(0)+i\sin(0)=1$]; then we get $C=0$.

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    $\begingroup$ Fine, but my question is: how do we know that $f' = g'$? $\endgroup$ – hey hey Feb 21 '13 at 11:47
  • $\begingroup$ If $f=\exp(iz),~g=\cos(z)+i\sin(z)$ then $f'=i\exp(iz)=i\times[\cos(z)+i\sin(z)]=i\cos(z)-\sin(z)=g'$ $\endgroup$ – mrs Feb 21 '13 at 11:55
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    $\begingroup$ But where is the proof for $i \exp(iz)= i\times[\cos(z)+i\sin(z)]$? $\endgroup$ – hey hey Feb 21 '13 at 11:59
  • $\begingroup$ @heyhey: This exactly what you noted above. In fact according to the link both are solutions of the ODE and then they satisfy the same OE as you asked above. I mean $f'=if$ $\endgroup$ – mrs Feb 21 '13 at 12:04
  • $\begingroup$ I get your point. Probably my question isn't much clear. I'll try to rephrase it: how do we know that the solutions to the ODE are unique? (Or, to be more precise, they differ by a constant?) $\endgroup$ – hey hey Feb 21 '13 at 12:20
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To show $f=g$ you can use the Picard-Lindelöf theorem.

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  • $\begingroup$ In fact you did it here. +1 $\endgroup$ – mrs Feb 21 '13 at 13:28

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