For a homogeneous second order differential equation, why does a complex conjugate pair solution (m+in and m-in) to the auxiliary equation result in the complementary function y(x)=e^(mx)(Acos(nx)+Bisin(nx)), where i represents √(-1).

For a second order differential equation, our auxiliary equation, am²+bm+c=0, has two roots. Let's denote these roots as A and B. Note that A and B can be the same if there is a repeated root, but we're not considering that case here, and that A and B are not the same constants here as the ones used in the question.Now, we know that the solution to our differential equation will have the following form: y(x)=Ce^Ax+De^Bx. As we know our roots are a complex conjugate pair, we can rewrite A and B as (m+in) and (m-in), where m and n are real constants: y(x)=Ce^(m+in)x+De^(m-in)x=> y(x)=Ce^mx+inx+De^mx-inx. By laws of exponents, we can rewrite y(x) as follows: y(x)=Ce^mxe^inx+De^mxe^-inx. By factoring out e^mx, we have the following: y(x)=e^mx(Ce^inx+De^-inx). Now, by Euler's Identity, we can rewrite our e^inx and e^-inx terms. Recalling that e^ikX=cos(kX)+isin(kX), we have: y(x)=e^mx(C[cos(nx)+isin(nx)]+D[cos(-nx)+isin(-nx)]). By distributing the constants C and D into the brackets, and recalling that cos(-X)=cos(X) (as cos(x) is an even function) and sin(-X)=-sin(X) (as sin(x) is an odd function), we find: y(x)=e^mx(Ccos(nx)+Cisin(nx)+Dcos(nx)-Disin(nx)) => y(x)=e^mx((C+D)cos(nx)+i(C-D)sin(nx)). Now, if we let C+D=A and C-D=B, where A and B are real constants, we have the following result: y(x)=e^mx(Acos(nx)+Bisin(nx)), as required. Note that the constants A and B here represent the constants in the question rather than the A and B used earlier in our answer.