Let $\mathcal{B}(F)$ the algebra of all bounded linear operators on a complex Hilbert space $F$. Let $(S\_1,S\_2)\in \mathcal{B}(F)^2$. We define
$$W(S\_1,S\_2)=\{(\langle S\_1 y\; ,\;y\rangle,\langle S\_2 y ,\;y\rangle):y \in F,\;\;\|y\|=1\}.$$


Consider the matrices
$$
S\_1 = \left[\begin{array}{ccc} 0&1&0 \\ 0& 0&0 \\0&0&0 \end{array}\right] \ \ \textrm{and} \ \ S\_2 = \left[\begin{array}{ccc} 0&0&0 \\ 0& 0&0 \\0&1&0 \end{array}\right].
$$
Clearly, $S\_1S\_2=S\_2S\_1$. Moreover, we get
$$
W(S\_1,S\_2)=\{(b\overline{a},b\overline{c});\;(a,b,c) \in \mathbb{C}^3\;\;\hbox{and}\;|a|^2+|b|^2+|c|^2=1\}.
$$



> 
> Is $W(S\_1, S\_2)$ a convex subset of $\mathbb{C}^2$ in this case? 
> 
> 
> 



example of convexity
In your specific case, $W(S\_1,S\_2)$ is the polydisk of radius $\frac12$. For
$$|b\bar a|^2+|b\bar c|^2=|b|^2(1-|b|^2)\le\frac14$$
and conversely if $|w|^2+|z|^2\le\frac14$, one takes $b=\sqrt t\in(0,1]$ a root of $t(1-t)=|w|^2+|z|^2$, then $\bar a=w/t$, $\bar c=z/t$.


