[TIKZ]
\begin{scope}
\draw (0,0) circle(3);
\end{scope}
\node (1) at (0,0) {c};
\draw (0,0) node[anchor=south] {.};
\coordinate[label=left: $z_2$] (E) at (-2,-2.236);
\coordinate[label=left: $z_1$] (A) at (-2,2.236);
\coordinate[label=above: $z$] (B) at (-1,2.828);
\coordinate[label=above: $z_3$] (C) at (1.2,2.75);
\coordinate[label=below: $z$] (D) at (2,-2.236);
\draw (E) -- (A);
\draw (E) -- (B);
\draw (E) -- (C);
\draw (E) -- (D);
\draw (A) -- (B);
\draw (B) -- (C);
\draw (C) -- (D);
\node (1) at (-1.8,2) {$\theta_1$};
\node (2) at (-0.8,2.6) {$\theta_2$};
\node (3) at (1.8,-2.0) {$\theta_2$};
[/TIKZ]
Let $\dfrac{z_3-z_1}{z_2-z_1}=r_1\cis(\theta_1)$, where $0<\theta_1<180^{\circ}$.
If $z$ is on or below the line through $z_2$ and $z_3$, then $\dfrac{z-z_2}{z-z_3}=r_2\cis(\theta_2)$, where $0<\theta_2<180^{\circ}$. Because $r_1 \cis(\theta_2)\cdot r_2 \cis(\theta_2)=r_1\cdot r_2\cdot \cis(\theta_1+\theta_2)$ is real, it follows that $\theta_1+\theta_2=180^{\circ}$, meaning that $z_1,\,z_2,\,z_3$ and $z$ lie on a circle.
On the other hand, if $z$ is above the line through $z_2$ and $z_3$, then $\dfrac{z-z_2}{z-z_3}=r_2\cis(-\theta_2)$, where $0<\theta_2<180^{\circ}$. Because $r_1 \cis(\theta_1)\cdot r_2 \cis(\theta_2)=r_1\cdot r_2\cdot \cis(\theta_1-\theta_2)$ is real, it follows that $\theta_1=\theta_2$, meaning that $z_1,\,z_2,\,z_3$ and $z$ lie on a circle.
In either case, $z$ must lie on the circumcircle of $\triangle z_1 z_2 z_3$ whose center is the intersection of the perpendicular bisectors of $\overline{z_1z_2}$ and $\overline{z_1z_3}$, namely, the lines $y=\dfrac{39+83}{2}=61$ and $16(y-91)=-60(x-48)$.
Thus the center of the circle is $c=56+61i$. The imaginary part of $z$ is maximal when $z$ is at the top of the circle, and the real part of $z$ is 56.