For which value is this function a characteristic function of a random variable?

Question

Let $\varphi(t)=(k+pe^{it})^4$, where $p,k\in\mathbb{R}$. I want to know for which values of $k$, the function $\varphi$ is the characteristic function of a random variable $X$.

Of course, if $X\sim\mathrm{Binom}(4,p)$, the function $\varphi$ is the characteristic function of $X$ as long as $p\in(0,1)$ and $k=1-p$. But, a priori, there could be other values of $k$ for which $\varphi$ is the characteristic function of another random variable with a different distribution. I computed the first and second derivatives of $\varphi$ if I could get something, and I also tried to apply the Fourier inversion theorem, but I don't know how to proceed.

Any help will be appreciated, thank you very much.

Answer 1

From $\phi(0)=1$, the value of $k$ has to be one of the four possibilities: $\pm1-p$ and $\pm i-p$.

For $k=\pm i-p$, the $\phi(-t)=\overline{\phi(t)}$ condition fails.

For $k=-1-p$, the $\left\vert \phi(t) \right\vert\leq1$ condition fails.

So $k=1-p$ is the only choice.