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When we have a linear homogeneous ODE, suppose $M\; \ddot{x} + K \;x =0$

Where $x$ is a real function of some parameter. We can analyze, for a complex function of the same parameter $M\; \ddot{z} + K \;z =0$

As the coefficients are real, if $z_0$ is a solution then so is $\bar{z}_0$. From linearity we construct the real solutions ($x$'s) by taking the linear combinations, $\large \frac{z_0+\bar{z}_0}{2}$, $\large\frac{z_0-\bar{z}_0}{2i}$.

Can someone explain to me why the second expression in the last paragraph ($\large\frac{z_0-\bar{z}_0}{2i}$) is valid, as up until now, the ODE was linear over the Reals, and if $M$ or $K$ were chosen to be complex, then $\bar{z}_0$ would not be a solution.

My question is, why is it valid to create a linear combination with complex coefficients?

Addition:

From the comments I get that I havent been able to explain my doubt, so I will try more.

  1. The solutions $x_i$ of the ODE form a linear vector space over the field of reals, i.e if $x_1$ and $x_2$ are solutions then so is $ax_1+bx_2$ for some $a,b\in\mathbb{R}$

  2. The solutions of $M\; \ddot{z} + K \;z =0$ form a linear vector space over the field of complex numbers, i.e for $z_1$ and $z_2$ to be solutions of the ODE, $\alpha z_1 + \beta z_2$ where $\alpha,\beta\in \mathbb{C}$ is also a solution.

  3. It is apparent that we can choose $\alpha,\beta = -i/2$ for $z_i$ and $\bar{z}_i$ to create some $x_i$

Everything is fine in this sequence. Except that I am having trouble understanding how step 3 relates to step 1.


Thanks to anon for answering in the comment.

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    Funny you said you had no doubt $\frac{z_0 - \overline{z_0}}2$ was real when this was exactly the problem =P I guess anon's comment should be an answer.2011-08-09

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The two solutions $\frac{z+\overline{z}}{2}$ and $\frac{z-\overline{z}}{2i}$ are both real and linearly independent, so all real solutions must be linear combinations of them with coefficients over the field of reals. The $1/2i$ part is a cancelling factor that is intrinsic to the second given solution in order to make it real, it is not considered part of the real coefficients. One might as well call the two given solutions $u$ and $v$ and then write all real solutions as combinations $au+bv,$ and $a,b\in\mathbb{R}$ so one doesn't have to explicitly see an $i$ in the expression (or $z$ or $\overline{z}$, both of which are complex, for that matter).