# Complex part of the solution for physical values

What's a physical meaning of, for example, complex part of the solution for coordinate change of the anharmonic oscillator? Why after substitute (for diff. equation solve) for real x we can earn $x = Re(z) + iIm(z)$? It's because of substitute?

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Can you give some more context for your question? What does "coordinate change of the anharmonic oscillator" mean? –  Ryan Thorngren Sep 7 '12 at 3:43
For example, solve for equation $a \ddot x + b \dot x + c x = f(t)$. –  user8817 Sep 7 '12 at 3:49
Okay, got it. I'll write something up in the answers. –  Ryan Thorngren Sep 7 '12 at 3:50
@Maxim_Ovchinnikov: This is not called "anharmonic", it's called "damped". Anharmonic is nonlinear, and the complex business doesn't work for this. –  Ron Maimon Sep 7 '12 at 15:28
Oh, really. Thank you. –  user8817 Sep 8 '12 at 14:15

When we wish to solve a differential equation like your example, a good guess for when $f(t)=0$ is something of the form $x(t) = e^{i \alpha t}$. This just gives us an algebraic equation for $\alpha$, which we can just solve. I interpret your question to mean: but what is this solution? Why is it complex? Indeed, position should certainly be just a real number. The resolution is that a general solution to the differential equation will not be of the form $x(t)=e^{i\alpha t}$ for $\alpha$ solving our algebraic equation. In general there will be a couple solutions to the algebraic equation for $\alpha$--just two in your example--say $\alpha_1$ and $\alpha_2$. Then a general solution to your diffeq with $f(t)=0$ is of the form $x(t) = A e^{i \alpha_1 t} + B e^{i \alpha_2 t}$ for some complex constants $A$ and $B$. We then need boundary conditions to choose $A$ and $B$ properly. The crux of the biscuit is we can choose $A$ and $B$ so that an $x(t)$ of this form is real for all times $t$ (this assumes $a$, $b$, and $c$ are real in the original equation).
So using the complex guess $x(t) = e^{i\alpha t}$ is just a computational trick which makes things easy because the complex numbers have nice properties with algebraic equations (they are algebraically closed). We could have done the whole thing without $e$, instead just using sines and cosines and it would have worked out too.
Oh, I understand. When we use substitute $x = Ae^{i\varphi t}$, we add a complex part to the real part of $x$. So after finding the solution we need to discard a complex part. But when we use substitution $x = Ae^{\varphi t}$, we need to apply a condition of $x = x*$ to the solution. –  user8817 Sep 7 '12 at 15:10