Theory of Intelligence • First-Principles AI & STEM Education

Solution

Given $z=x+iy$ and:

\arg\left(\frac{z-1+2i}{z+1-2i}\right)=\frac{9\pi}{4}

Since $\frac{9\pi}{4}=2\pi+\frac{\pi}{4}$ , this is equivalent to:

\arg\left(\frac{z-1+2i}{z+1-2i}\right)=\frac{\pi}{4}

Let:

w=\frac{z-1+2i}{z+1-2i}

Then $\arg(w)=\frac{\pi}{4}$ , so $\tan\arg(w)=1$ , hence (same sign case):

\operatorname{Im}(w)=\operatorname{Re}(w)

Now:

\begin{aligned} \frac{z-1+2i}{z+1-2i} &=\frac{(x-1)+i(y+2)}{(x+1)+i(y-2)}\cdot\frac{(x+1)-i(y-2)}{(x+1)-i(y-2)}\\ &=\frac{[(x-1)+i(y+2)][(x+1)-i(y-2)]}{(x+1)^2+(y-2)^2} \end{aligned}

Compute the numerator:

(x-1)(x+1)+(y+2)(y-2)=x^2-1+y^2-4=x^2+y^2-5

(y+2)(x+1)-(x-1)(y-2)=4x+2y

So:

w=\frac{(x^2+y^2-5)+i(4x+2y)}{(x+1)^2+(y-2)^2}

Since the denominator is real and positive (except where undefined), $\arg(w)$ depends on the numerator.

Given $\operatorname{Im}(w)=\operatorname{Re}(w)$ :

4x+2y=x^2+y^2-5

Rearrange:

\boxed{x^2+y^2+4x+2y-5=0}