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I have searched all the websites but couldn't get the answer...everywhere I am seeing linear combinations with real coefficients..but in case of exponential solution of second order differential equation of SHM there are complex coefficients present in linear combination.. The general solution of 2nd order D.E of S.H.M is...

$$x(t)=Ae^{i\omega t} +A^\ast e^{-i\omega t}.$$

Here, A represents a complex number and $A^\ast$ represents complex conjugate of $A$.

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I am not used to the fact that linear combinations are possible using complex coefficients

Correct me if I am wrong somewhere in presenting my doubt and please show me is this possible to make linear combinations with complex coefficients with examples...

surrutiaquir
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Manish Sharma
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    Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community Nov 08 '21 at 05:49
  • Possible duplicates: https://physics.stackexchange.com/q/77156/2451 , https://physics.stackexchange.com/q/316135/2451 and links therein. – Qmechanic Nov 08 '21 at 06:09

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