Question

Let $z_1$ and $z_2$ be complex numbers such that $z{{}_1}^2-4z_2=16+20i.$ Suppose the roots $\alpha \&\beta \text{of}{x}^2+z_{1}x+z_2+m=0$ for some complex number $m$ satisfy $|\alpha -\beta |=2\sqrt{7}.$

The complex number $m$ lies on

• A. A square with side $7$ and centre $(–4,5)$
• B. A circle with radius $7$ and centre $(4,5)$
• C. A square with side $7$ and centre $(4,5)$
• D. A circle with radius $7$ and centre $(–4,5)$

Answer 1

The correct option is B A circle with radius $7$ and centre $(4,5)$

We have,
$\alpha +\beta =-z,\text{while}\alpha \beta =z_2+m$

Now,
$(\alpha -\beta {)}^2=(\alpha +\beta {)}^2-4\alpha \beta$
$=z{{}_1}^2-4zi-4m$
$=16+20i-4m$
As $|\alpha -\beta |=2\sqrt{7}\text{we have}|4+5i-m|=7$

Thus point $M$ representing the complex number m lies within a circle of radius $7,$ centered at $C(4,5).$

$OC=\sqrt{4^2+5^2}=\sqrt{41}$
maximum value of $|m|=7+\sqrt{41}$
minimum value of $|m|=7-\sqrt{41}$