When dealing with Laplace’s equation in spherical coordinates (convention : azimuthal angle = $\varphi$) and using separation of variables, we come to the equation: $$ \frac{1}{\Phi} \frac{\partial^2{\Phi}}{\partial\varphi^2}= - m^2 $$ Now, the general solution of this equation is $\Phi(\varphi)=c_1 \cos(m\varphi)+c_2 \sin(m\varphi)$ (as given readily for example by $WolframAlpha$).
Browsing the Wikipedia page “Spherical harmonics”, I read $\Phi$ “ is a linear combination of the complex exponentials $e^{\pm{i m\varphi}}$ ”. But the simplicity of the expression confuses me. If I apply Euler’s formula, I get: $$ \Phi(\varphi)=c_1 \frac {e^{ +i m\varphi } + e^{ -i m\varphi}}{2} + c_2\frac {-i e^{ +i m\varphi } +i e^{ -i m\varphi}}{2}$$
$$ = e^{+{im\varphi}} \left( \frac{c_1}{2} -i \frac{c_2}{2} \right) + e^{-{im\varphi}} \left( \frac{c_1}{2} +i \frac{c_2}{2} \right) $$ So, the so-called “linear combination” is not a general linear combination, but is actually a very specific linear combination with a complex constant and its conjugate: $$ \Phi(\varphi)= c \ e^{+{im\varphi}} + \bar{c} \ e^{-{im\varphi}} $$ Am I right, or did I make some some silly nonsense?
You are correct under the assumption that $c_1$ and $c_2$ are purely real. Yet, the coefficients $c_1$ and $c_2$ are not restricted to the real numbers. And hence, $c_1+ic_2$ is not in general the complex conjugate of $c_1-ic_2$.
However, if the solution $\Phi(\phi)$ is restricted to be a real-valued function, then any solution written as $Ae^{im\phi}+Be^{-im\phi}$ must be real for all $\phi$. This would require that $A+B$ be purely real and $A-B$ be purely imaginary. And this, in turn, implies that $A=B^*$, which is what you observed under the tacit assumption that $c_1$ and $c_2$ are purely real.