# Identity of two differential equations

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?

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 $C=0$.
• Fine, but my question is: how do we know that $f' = g'$? – hey hey Feb 21 '13 at 11:47
• If $f=\exp(iz),~g=\cos(z)+i\sin(z)$ then $f'=i\exp(iz)=i\times[\cos(z)+i\sin(z)]=i\cos(z)-\sin(z)=g'$ – mrs Feb 21 '13 at 11:55
• But where is the proof for $i \exp(iz)= i\times[\cos(z)+i\sin(z)]$? – hey hey Feb 21 '13 at 11:59
• @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 $f'=if$ – mrs Feb 21 '13 at 12:04
To show $f=g$ you can use the Picard-Lindelöf theorem.