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

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.

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.
• By pulling out $(-1)^{4}$ we may suppose $p>0$. We show that $p \le 1$ by taking $e^{it}=-1$ in $|\phi (t)| \le 1$. Commented Aug 8 at 11:42