Why $\lambda_n=sgn(n)\pi i \sqrt{n^2+\alpha}$?
I have this:
$\varphi_{xx}-(\alpha+\lambda^2)\varphi=0$ and $\varphi(0)=\varphi(1)=0$ then $\varphi(x)=c\sin(\sqrt{-(\alpha+\lambda^2)}x)+d\cos(\sqrt{-(\alpha+\lambda^2)}x)$, and $d=0$ then $\varphi(x)=c\sin(\sqrt{-(\alpha+\lambda^2)}x)$ and $0=\sin(\sqrt{-(\alpha+\lambda^2)}) \Leftrightarrow \sqrt{-(\alpha+\lambda^2)}=n\pi$
$\sqrt{-(\alpha+\lambda^2)}=n\pi\Rightarrow \lambda_n=sgn(n)\pi i \sqrt{n^2+\alpha}$?
Thanks in advance