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Say we got two complex roots *p*+*q*i and *p*–*q*i. Using the solution for two different roots we get

Using the fact that

we get that

Now we want both *C*+*D* and (*C–**D*)*i* to be real, We can choose *C* and *D* to be complex conjugates, say *C*=*U*+*Vi* this gives us

In that way, the constants may be any value. Let us write *C* instead of 2*U* and *D* instead of -2*V*, this will finally give us the solution

**An example**

Let us check against a special case if this works. Say the equation has pure complex roots then we must have started with an equation of the form

that will have the characteristic equation

If b is positive we get the solutions

This would give us

as a solution to the differential equation. This is something we already found when we solved the equation

where we found the solution to be

that, with a bit of algebra, can be found to be equivalent to the solution we found now. We can use the formula

and we let *t* be *x* and we can let the constant in front of *t* be *ω*. This would give us

so *q*=* ω*, *C*=*A*cos*θ* and *D*=*–**A*sin*θ*.

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