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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 really 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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