Is $W(S_1,S_2)$ convex? Let $F=\mathbb{C}^4$  be endowed with the norm $\|\cdot\|_2$. Let the operators $$
S_1=\left(\begin{array}{cccc}0&1&0&0\\0&0&0&0\\0&0&0&0\\0&0&0&0\end{array}\right)\;\mbox{and}\;\;S_2=\left(\begin{array}{cccc}0&0&0&0\\0&0&0&0\\0&0&1&0\\0&0&0&1\end{array}\right)
.$$
The numerical range of $(S_1,S_2)$ is given by
$$W(S_1,S_2) =\left\{\left(\overline{a}b,|c|^2+|d|^2\right);\;(a,b,c,d) \in \mathbb{C}^4\;\;\hbox{and}\;|a|^2+|b|^2+|c|^2+|d|^2=1\right\}.$$

Is $W(S_1,S_2)$ convex?

We can see that $(0,1),\,(\frac{1}{2},0)\in W(S_1,S_2)$. I hope to find a point in the segment joining $(0,1)$ and $(\frac{1}{2},0)$ which is not in $W(S_1,S_2)$.

Note that the imaginary component of the second coordinate vanishes, and the set can be identified with 
$$\{(r\cos\theta,r\sin\theta,s)∈\mathbb{R}^3;\;4r^2+s^2≤1\}.$$
Up to scaling, this is the Euclidean unit ball, and in particular convex
