Let $y = e^{ix}$. Then $y'' = -y$, which is the equation for simple harmonic motion. The general solution is
$$y = A\cos(x) + B\sin(x) \tag1$$ where
$$y(0) = e^0 = 1 = A\cos(0) + B\sin(0) = A \tag2$$ so $A = 1$, and $$y'(0) = ie^0 = i = -A\sin(0) + B\cos(0) =B \tag3$$ so $B = i$, and that's pretty much it.
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- 1$\begingroup$ Hi, welcome to Math SE. I edited your question to use our preferred MathJax format. $\endgroup$J.G.– J.G.2025-11-16 12:15:10 +00:00Commented Nov 16 at 12:15
- 1$\begingroup$ You‘d better use MathJax. And you’d better say what you’re talking about. Little things like which definitions you’re using, and what exactly you want to prove. $\endgroup$wasn't me– wasn't me2025-11-16 12:21:51 +00:00Commented Nov 16 at 12:21
- $\begingroup$ If you imply anything known about $ \exp, \sin , \cos$ and second order ODEs , there is nothing to be shown. Your example is at best a confirmation for a special case of $x$, eg $x=(1+2i)^{-1/5}$ for the inverse functions of integrals of $1/x, 1/\sqrt{1-x^2} $ $\endgroup$Roland F– Roland F2025-11-16 12:29:41 +00:00Commented Nov 16 at 12:29
- 1$\begingroup$ Here you are differentiating the complex exponential before having defined it, so the argument is circular. In effect, you are defining $x \mapsto e^{ix}$ to be a solution of the ODE $y^{\prime\prime} = -y$ and then checking that it does indeed solve that equation. $\endgroup$Enforce– Enforce2025-11-16 13:06:29 +00:00Commented Nov 16 at 13:06
- $\begingroup$ What is your definition of $e^{ix}?$ $\endgroup$Adam Rubinson– Adam Rubinson2025-11-16 13:43:31 +00:00Commented Nov 16 at 13:43
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3 From what you have said, I can see at least one glaring issue, you assumed the solution to be $ y = e^{ix} $. But one can also see that $ y = e^{ix} + c $ is also a solution to the simple harmonic oscillator where $ c $ is an arbitrary constant. So, following the same procedure would lead you to believe that $ y(0)=1+c=A $. Then your process would lead that $ e^{ix} = B \sin(x) + (1+c) \cos(x) $. So, I do not think this qualifies as a valid proof.
- $\begingroup$ $$(\cos x +c)^{\prime \prime} = -\cos x \ne -(\cos x +c)$$ $\endgroup$Roland F– Roland F2025-11-16 12:37:26 +00:00Commented Nov 16 at 12:37
- $\begingroup$ Thanks for the correction, I am editing it right now. $\endgroup$Hamidul Miad– Hamidul Miad2025-11-16 12:38:52 +00:00Commented Nov 16 at 12:38
- 1$\begingroup$ $y = e^{ix} + c$ is still not a solution when $c\neq 0.$ $\endgroup$md2perpe– md2perpe2025-11-16 15:06:48 +00:00Commented Nov 16 at 15:06