Roots of a quadratic function with complex coefficient

Question

Consider the quadratic function $$ x^2 + \alpha x +1 = 0 $$ where $\alpha = \gamma + \delta i$ for $\gamma, \delta \in [-100, 100]$. It is claimed that for any root of the equation $\hat{x}$, we have $$ |\hat{x}| \leq C \delta^{-1} $$ for some universal constant $C$. I'm not familiar with quadratic equations with complex coefficients, so I'd appreciate a hint or sketch of the solution.

Answer 1

The product of the two roots is $1$. If both roots satisfied that inequality, we would have $$ 1 \le C^2 \delta^{-2}$$ so $|C| \ge \delta$. Thus there can't be such a constant.

EDIT: The roots (given by the usual quadratic formula) are continuous functions of $\alpha$, so if $\alpha$ is bounded the roots are also bounded.

Answer 2

You claim is equivalent to that if $\hat z=δ\hat x$ is a root of $$ z^2+δ(γ+δi)z+δ^2=0 $$ then $\hat z$ is inside some bounded set. That is true if you bound $δ$ and $γ$ (by $100$), as a simple root bound gives that $$ δ|\hat x|=|\hat z|\le \max(1,|δ(γ+δi)|+δ^2)\le C=(1+\sqrt2)100^2 $$