Why does Euler's formula have to be $e^{ix} = \cos(x) + i\sin(x)$

Question

In part one of this youtube video the uploader goes on to explain the calculus proof for Euler's Formula.

The Formula $$e^{ix} = \cos(x) + i\sin(x)$$ Differentiate $$ie^{ix} = f'(x) + i g'(x)$$ Multiply original formula by $i$ $$ie^{ix} = if(x) - g(x)$$ Equate the differentiation and the multiplied version $$f'(x) + ig'(x) = if(x) - g(x)$$ Equate real and imaginary (and cancel the i) $$f'(x) = -g(x) \qquad g'(x) = f(x)$$

Then he goes on to explain $f(x) = \cos(x)$ and $g(x) = \sin(x)$. My question is why can't $f(x) = \sin(x)$ and $g(x) = -\cos(x)$? Can further proof be added to this proof to eliminate $f(x) = \sin(x)$ and $g(x) = -\cos(x)$?

Else the equality would be false at $x=0$, or you wouldn't have a holomorphic function to do complex analysis with, etc. and that all gets messy. — Simply Beautiful Art, Mar 11 '17 at 15:22
Possibly related: How to prove Euler's formula: $e^{it}=\cos t +i\sin t$? — Simply Beautiful Art, Mar 11 '17 at 15:23
@SimplyBeautifulArt It is related but not the same as I already have a proof but want some to explain the end result of it — Dan, Mar 11 '17 at 15:27
Yes, of course. Hence I did not have it as possible duplicate :-) — Simply Beautiful Art, Mar 11 '17 at 15:30

score 7 · Accepted Answer · answered Mar 11 '17 at 15:27

7

Because we know the initial condition $e^{i0}=1$ holds. As with most differential equations, there's an family of answers that you need to use the initial condition to find the correct one for.

answered Mar 11 '17 at 15:27

Stella Biderman

31,155

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Stella (+1) for the answer. I reject the so-called proof of Euler's formula inasmuch as it uses a result from real analysis (i.e., $\frac{de^x}{dx}=e^x$) and presumes that it holds for the yet-to-be-defined object $e^{ix}$. – Mark Viola Mar 11 '17 at 15:36
@Dr.MV Agreed. My preferred way is to define $e^{ix}$ by the Taylor polynomial and then use complex analysis to prove that the power series preserves the usual properties of the real analysis function $e^x$. Euler's Equation is then simply obtained from comparing power series. – Stella Biderman Mar 11 '17 at 18:07
Yes, that's a good standard approach. – Mark Viola Mar 11 '17 at 18:12

Serouj Ghazarian · Answer 2 · 2019-01-16T05:20:22.690

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Let z=cos(x)+isin(x) When you 8derive it (in terms of x), z'=-sin(x)+icos(x) And since we are in the complex world, -1=i×i =>z'=i×isin(x)+icos(x)=i(cos(x)+isin(x))=iz This pretty much makes it a first order differential equation. After dividing by z and multiplying by dx (z'=dz/dx) both on both sides and integrating both sides, you get: ln(z)=ix+c ln(cos(x)+isin(x))=ix+c If we set x=0, cos(0)=1,sin(0)=0 and ln(1)=0 => ln(1)=c=>c=0 => ln(cos(x)+isin(x))=ix e^ix=cos(x)+isin(x). Hope this helps.

edited Jan 16 '19 at 05:20

answered Jan 16 '19 at 05:07

Serouj Ghazarian

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Hi and welcome to the Math.SE. Producing beautiful and beautifully formatted text is one of the points that makes an answer good: if you want to learn how to do this, please have a look at MathJax basic tutorial and quick reference. – Daniele Tampieri Jan 16 '19 at 05:37

Why does Euler's formula have to be $e^{ix} = \cos(x) + i\sin(x)$

2 Answers2