Given,
z=\alpha +i\beta &|z+2|=z+4(1+i)
Now putting the value of z=α+iβ in ∣z+2∣=z+4(1+i) we get,
∣z+2∣=z+4(1+i)
⇒(α+2)2+β2=(α+4)+i(β+4)
Now on comparing real and imaginary part, we get
(α+2)2+β2=α+4...(i) and β+4=0⇒β=−4...(ii)
Now solving,
(α+2)2+16=α+4
⇒α2+4α+20=α2+8α+16
⇒α=1
So, α+β=−3,αβ=−4
We know that quadratic equation is given by,
x2−(sum of roots)x+(product of roots)=0
So, equation with roots −3 and −4 will be,
x2+7x+12=0