Interpret the given equation

Let $z=\alpha+i\beta.$ Then $z+2=(\alpha+2)+i\beta,$ so $|z+2|=\sqrt{(\alpha+2)^2+\beta^2},$ which is a real number.

The question gives $|z+2|=z+4(1+i).$

Since $z+4(1+i)=\alpha+i\beta+4+4i=(\alpha+4)+i(\beta+4),$ and the left side is real, the imaginary part of the right side must be zero.

Equate imaginary parts

$\beta+4=0 \implies \beta=-4.$

Equate real parts

Now, $|z+2|=\alpha+4.$

Also,

Solve for $\alpha$

Squaring both sides: $(\alpha+2)^2+16=(\alpha+4)^2.$

Expand: $\alpha^2+4\alpha+4+16=\alpha^2+8\alpha+16.$

$4\alpha+20=8\alpha+16$ $4=4\alpha$ $\alpha=1.$

So, $\alpha=1,\quad \beta=-4.$

Find $\alpha+\beta$ and $\alpha\beta$

$\alpha+\beta=1+(-4)=-3,$ $\alpha\beta=1\cdot(-4)=-4.$

Form the quadratic whose roots are $-3$ and $-4$

If roots are $r_1=-3$ and $r_2=-4$, then the equation is $x^2-(r_1+r_2)x+r_1r_2=0.$

Here,

Check options

• A: $x^2+2x-3=0$ → roots not $-3,-4$
• B: $x^2+3x-4=0$ → roots not $-3,-4$
• C: $x^2+x-12=0$ → roots not $-3,-4$
• D: $x^2+7x+12=0$ → roots are $-3,-4$

So the correct option is D.