# Is $f(\alpha, \beta) = \exp\left\{ -j \ \alpha \ d^{\beta} \right\}$ convex?

Is the following function convex? \begin{align} f(\alpha, \beta) = \exp\left( -j \cdot \alpha \cdot d^{\beta} \right), \end{align} where $$j = \sqrt{-1}$$, $$\alpha \geq 0$$, $$\beta \geq 0$$, and $$d \in \mathbb{R}$$.

If yes, how to show it?

If the function was a single variable dependent and twice differentiable, then one can show that the $$f^{\prime\prime} \geq 0$$ or show that $$f(\lambda x_1 + (1-\lambda)x_2) \leq \lambda f(x_1) + (1-\lambda) f(x_2)$$ is true.

But here a function of two variables $$\alpha$$ and $$\beta$$ is difficult for me. Can you please help me? Thank you so much in advance.

• How is convexity defined for complex valued functions? Also, please do not abuse the square root function. – LinAlg Nov 4 '18 at 14:05
• Well, since $j=\sqrt{-1}$, we have that $f(\alpha,\beta)=\exp(-j\alpha d^\beta)=\cos(-\alpha d^\beta)+ i\sin(-\alpha d^\beta)$. So $f$ is a function from $\mathbb R^2$ to $\mathbb C$. I don't know the definition of convex function when they have image in $\mathbb C$. – André Porto Nov 4 '18 at 14:09