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 \(p'(z)\), \(M(z)\) is a root of \(p_M'(z)\). In other words, we can transform the initial triangle \(\Delta z_1z_2z_3\) via any linear transformation and its roots will follow appropriately.

Proof. Let \(M(z) = az+b\). Note M'(z) = a. See that \begin{align*} 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^3p(z). \end{align*} Now differentiate: $$ a p_M'(M(z)) = a^3 p'(z). $$ So in particular, since \(a \neq 0\), \(M(z)\) is a root of \(p_M'\) if and only if \(z\) is a root of \(p'(z)\). \(\Box\)