Equality of Complex Numbers

Given
$$z=a+ib.$$

The condition $$\arg(z-2)=\arg(z+4i)$$ implies that $$z-2$$ and $$z+4i$$ have the same direction.

Therefore,

$$z-2=\lambda(z+4i)$$ for some positive real number $$\lambda.$$

Substituting

$$z=a+ib,$$

$$(a-2)+ib=\lambda\bigl(a+i(b+4)\bigr).$$

Comparing real and imaginary parts,

$$a-2=\lambda a$$

$$b=\lambda(b+4).$$

Also,

$$|z-2|=2|z+4i|.$$

Since

$$z-2=\lambda(z+4i),$$

we have

$$|z-2|=\lambda|z+4i|.$$

Hence,

$$\lambda=2.$$

Substituting in the earlier equations,

$$a-2=2a$$

$$a=-2,$$ and $$b=2(b+4)$$

$$b=-8.$$

Therefore,

$$a+b=-2-8=-10.$$