For a simple harmonic oscillator,

$$x(t) = A \cos(\omega t)$$

We can also write $x(t)$ as: $$x(t) = C_1 e^{i\omega t} + C_2 e^{-i\omega t}$$

Why is it necessary that the coefficients $C_1$ and $C_2$ be complex conjugate of each other? If they are not, then we still get real values of $x(t)$ as $(C_2+ C_1) \cos(\omega t)$.

So why the condition for complex conjugation?

  • $\begingroup$ $\mathbf C_1$ and $\mathbf C_2$ are both real so $\mathbf {x(t)}$ is real as well. $\endgroup$
    – Ruchi
    Jan 14, 2021 at 8:51
  • $\begingroup$ This is still only the real part of the above expression. So if you want $x(t)$ to be automatically real, you do need them to be complex conjugates, right? You manually threw away the $\sin$ piece. $\endgroup$
    – kaylimekay
    Jan 14, 2021 at 8:55
  • $\begingroup$ This is so dumb of me ugh $\endgroup$
    – Ruchi
    Jan 14, 2021 at 8:58
  • $\begingroup$ No worries..... $\endgroup$
    – kaylimekay
    Jan 14, 2021 at 8:59
  • $\begingroup$ Related : Need help understanding an equation of motion for a pendulum. $\endgroup$
    – Frobenius
    Jan 14, 2021 at 17:30

3 Answers 3


There is a problem with your algebra, in general
$$C_1 e^{i\omega t} + C_2 e^{-i\omega t} \neq (C_2+C_1)\cos(\omega t)$$ Recall $$\cos(\omega t) = \frac{1}{2}(e^{i\omega t}+e^{-i\omega t})$$ both imaginary exponentials must have the same coefficient in front in order to give a cosine.

So remember the space of solutions of a differential equation of order two, such as the case of a harmonic oscillator has two dimensions (or parameters). However this space depending on the application can be written as the span of $e^{\pm i\omega t}$ or $\sin$ and $\cos$ as you might have encountered. However it is only by using the boundary conditions of your problem that you can determine both coefficients.

With just your first line as given, the most general requirement for the two real parameters $C_1$ and $C_2$ to give you just a $\cos$ is if they are the same, otherwise you still get a term proportional to $\sin$


I am not sure if that condition is necessary. As you have pointed out,

$$ \mathcal{Re}\, (C_1 e^{i\omega t} + C_2 e^{-i\omega t}) = (C_1 + C_2) \cos{\omega t} = A \cos{\omega t},$$ where $A \equiv C_1 + C_2$. No need whatsoever for $C_1$ and $C_2$ to be linked in any way.

  • $\begingroup$ Thus they are linked by $A=C_1+C_2$. $\endgroup$
    – Roger
    Jan 14, 2021 at 8:52

Why is it necessary that the coefficients C1 and C2 be complex conjugate

you want that $~x'(t)~$ be equal to $~x(t)$


$$x(t)=A\,\cos(\omega\,t)\tag 1$$

$$x'(t)=C_1\,e^{i\,\omega\,t}+C_2\,e^{-i\,\omega\,t}\tag 2$$

and the Euler equation

$$e^{i\varphi}=\cos(\varphi)+i\,\sin(\varphi)\tag 3$$

$x'(t)~$ will be only real if $~C_2$ is complex conjugate to $~C_1$

with $C_1=a+i\,b~,C_2=a-i\,b$ and Eq. (3) you obtain

$$x'(t)=\left( a+ib \right) {{\rm e}^{i\omega\,t}}+ \left( a-ib \right) { {\rm e}^{-i\omega\,t}} =2\,a\cos \left( \omega\,t \right) -2\,b\sin \left( \omega\,t \right)$$

thus $~x'(t)=x(t)~$ if $~b=0~$and $2a=A$

$$x'(t)=\left( A/2 \right) {{\rm e}^{i\omega\,t}}+ \left( A/2 \right) { {\rm e}^{-i\omega\,t}} =x(t)$$


general case


$$x'(t)=\left( a+ib \right) {{\rm e}^{i\omega\,t}}+ \left( a-ib \right) { {\rm e}^{-i\omega\,t}}$$

with $~a=A/2~,b=-B/2~$ $~x'(t)~$ will be equal to $~x(t)$


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