Let $M:\mathbb{C}\to \mathbb{C}$ be a linear transformation and let $p(z)=(z-z_1)(z-z_2)(z-z_3)$. Define $$p_M(z)=(z-M(z_1))(z-M(z_2))(z-M(z_3)).$$ Then if $z$ is a root of $p^{\prime }(z)$, $M(z)$ is a root of $p_M^{\prime }(z)$. In other words, we can transform the initial triangle $\mathrm{\Delta }{z}_1{z}_2{z}_3$ via any linear transformation and its roots will follow appropriately.

Proof. Let $M(z)=az+b$. Note M'(z) = a. See that $$\begin{array}{rl}{p}_M(M(z))& =(M(z)-M(z_1))(M(z)-M(z_2))(M(z)-M(z_3))\\ & =a^3(z-z_1)(z-z_2)(z-z_3)\\ & =a^{3}p(z).\end{array}$$ Now differentiate: $$a{p}_M^{\prime }(M(z))=a^3{p}^{\prime }(z).$$ So in particular, since $a\ne 0$, $M(z)$ is a root of $p_M^{\prime }$ if and only if $z$ is a root of $p^{\prime }(z)$. $◻$