Let S=z C:z^2+4z+16=0.

Question

Let $S={z\in C:{z}^{2}+4z+16=0}$. Then $\sum _{z\in S}|z+\sqrt{3}i{|}^{2}$ is equal to :

Accepted Answer

Solve the quadratic equation first:

${z}^{2}+4z+16=0$

Using the quadratic formula,

$z=\frac{-4\pm \sqrt{16-64}}{2}=\frac{-4\pm \sqrt{-48}}{2}=\frac{-4\pm 4\sqrt{3}i}{2}=-2\pm 2\sqrt{3}i$

So the set is

$S={-2+2\sqrt{3}i,\, -2-2\sqrt{3}i}$

Now we need to find

$\sum _{z\in S}|z+\sqrt{3}i{|}^{2}$

That is,

$|-2+2\sqrt{3}i+\sqrt{3}i{|}^{2}+|-2-2\sqrt{3}i+\sqrt{3}i{|}^{2}$

Simplify each term:

For $z=-2+2\sqrt{3}i$,

$z+\sqrt{3}i=-2+3\sqrt{3}i$

Hence,

$|-2+3\sqrt{3}i{|}^{2}=(-2{)}^{2}+(3\sqrt{3}{)}^{2}=4+27=31$

For $z=-2-2\sqrt{3}i$,

$z+\sqrt{3}i=-2-\sqrt{3}i$

Hence,

$|-2-\sqrt{3}i{|}^{2}=(-2{)}^{2}+(-\sqrt{3}{)}^{2}=4+3=7$

Therefore,

$\sum _{z\in S}|z+\sqrt{3}i{|}^{2}=31+7=38$

So the correct answer is:

$38$

Option D

Solutions · 1