First Principles AI

Solution

Let $z=x+iy$ .

z-2-3i=(x-2)+i(y-3),\quad z+2+3i=(x+2)+i(y+3)

Given:

\arg(z-2-3i)-\arg(z+2+3i)=2\pi

So:

\arg\left(\frac{z-2-3i}{z+2+3i}\right)=2\pi

That means the quotient lies on the positive real axis (argument $0$ modulo $2\pi$ ). Hence its imaginary part is $0$ .

Let:

w=\frac{(x-2)+i(y-3)}{(x+2)+i(y+3)}

Rationalize:

\begin{aligned} w&=\frac{[(x-2)+i(y-3)]\,[(x+2)-i(y+3)]}{(x+2)^2+(y+3)^2} \end{aligned}

Imaginary part of the numerator is:

(y-3)(x+2)-(x-2)(y+3)=4y-6x

For $w$ to be real:

4y-6x=0\Rightarrow 2y=3x

\boxed{2y=3x}