If for z = α + iβ, |z + 2| = z + 4 ( 1 + i ) , then α + β and αβ...

Given,

$z = \alpha + i\beta \text{ and } \left|z + 2\right| = z + 4\left(1 + i\right)$

Now putting the value of $z = \alpha + i\beta$ in \left|z + 2\right| = z + 4\left(1 + i\right)$we get,</p><p>\left|z + 2\right| = z + 4\left(1 + i\right)$

$⇒ \sqrt{{\left(\alpha + 2\right)}^{2} + {\beta}^{2}} = \left(\alpha + 4\right) + i\left(\beta + 4\right)$

Now on comparing real and imaginary part, we get

$\sqrt{( \alpha + 2{)}^{2} + {\beta}^{2}} = \alpha + 4 . . . ( i )$ and $\beta + 4 = 0 ⇒ \beta = - 4 . . . ( ii )$

Now solving,

$\sqrt{( \alpha + 2{)}^{2} + 16} = \alpha + 4$

$⇒ {\alpha}^{2} + 4\alpha + 20 = {\alpha}^{2} + 8\alpha + 16$

$⇒ \alpha = 1$

So, $\alpha + \beta = - 3, \alpha\beta = - 4$

We know that quadratic equation is given by,

${x}^{2} - \left(\text{sum of roots}\right)x + \left(\text{product of roots}\right) = 0$

So, equation with roots $- 3$ and $- 4$ will be,

${x}^{2} + 7x + 12 = 0$