and $c\in H^+(M_n(\mathcal A))$. Denote $$B^+_n(c,r)=\left\{a\in H^+(M_n(\mathcal A))\colon\left\|(\Im a)^{-1/2}(a-c)(\Im c)^{-1/2}\right\|\leq r
\right\}.$$ Then $B_n^+(c,r)$ is a norm-closed norm-bounded convex subset of $H^+(M_n(\mathcal A))$ with nonempty interior, which is bounded away from the topological boundary in the norm topology of $H^+(M_n(\mathcal A))$. Moreover, it is noncommutative. More precisely, $$\label{estimBall}
\|a\|\leq\|\Re c\|+\|\Im c\|\left(\frac{r^2+2+r\sqrt{r^2+4}}{2}+r\sqrt{\frac{r^2+2+r\sqrt{r^2+4}}{2}}\right),\ a\in B^+_n(c,r),$$ and $$\label{estimBdry}
\Im a\ge\frac{1}{2+r^2}\Im c,\quad a\in B^+_n(c,r).$$

The set $B^+_n(c,r)$ is norm-bounded: $a\in B^+_n(c,r)$ if and only if $(\Im a)^{-\frac12}(a-c)
(\Im c)^{-1}(a-c)^*(\Im a)^{-\frac12}\leq r^2\cdot1$, relation which implies $(a-c)(\Im c)^{-1}
(a-c)^*\leq r^2\|\Im a\|\cdot1$, which in its own turn implies $\left\|[(a-c)(\Im c)^{-\frac12}]
[(a-c)(\Im c)^{-\frac12}]^*\right\|\leq r^2\|\Im a\|$. Recalling that in any C${}^*$-algebra the adjoint (star) operation is isometric and that $\|x^*x\|=\|x\|^2$, this implies that $\left\|[(a-c)(\Im c)^{-\frac12}]^*
[(a-c)(\Im c)^{-\frac12}]\right\|\leq r^2\|\Im a\|$, which again implies $(\Im c)^{-\frac12}(a-c)^*(a-c)(\Im c)^{-\frac12}\leq r^2\|\Im a\|\cdot1$. Thus, repeating once again the above computations, we obtain $$\|a-c\|^2\leq r^2\|\Im a\|\|\Im c\|,\quad a\in B^+_n(c,r).$$ Recall that $\Im x=(x-x^*)/2i$, so $\|\Im x\|\leq(\|x\|+\|x^*\|)/2=\|x\|$. Similarly, $\|\Re x\|\leq
\|x\|.$ Applying this to $x=a-c$, we obtain $$\left(\|\Im a\|-\|\Im c\|\right)^2\leq\|\Im (a-c)\|^2\leq\|a-c\|^2\leq r^2\|\Im a\|\|\Im c\|
,\quad a\in B^+_n(c,r).$$ Direct computation shows that this relation imposes $$\label{estimIm}
\frac{\|\Im c\|}{2}\left(r^2+2-r\sqrt{r^2+4}\right)\leq\|\Im a\|\leq
\frac{\|\Im c\|}{2}\left(r^2+2+r\sqrt{r^2+4}\right),$$ for all $a\in B^+_n(c,r).$ Similarly,$\|\Re(a-c)\|^2\leq\|a-c\|^2\leq  r^2\|\Im a\|\|\Im c\|$ implies $$\label{estimRe}
0\leq\|\Re a\|\le\|\Re c\|+r\|\Im c\|\sqrt{\frac{r^2+2+r\sqrt{r^2+4}}{2}},\quad a\in B^+_n(c,r).$$ Adding 