First Principles AI

Solution

Given:

\operatorname{Arg}\left(\frac{z-2+i}{-2-2i}\right)=\frac{2\pi}{3},\quad z=x+iy

Write:

z-2+i=(x-2)+i(y+1)

Also, for $-2-2i$ (Quadrant III):

\operatorname{Arg}(-2-2i)=\frac{5\pi}{4}

Using $\operatorname{Arg}\left(\frac{A}{B}\right)=\operatorname{Arg}(A)-\operatorname{Arg}(B)$ (mod $2\pi$ ):

\operatorname{Arg}(z-2+i)-\frac{5\pi}{4}=\frac{2\pi}{3}

So:

\operatorname{Arg}(z-2+i)=\frac{2\pi}{3}+\frac{5\pi}{4}=\frac{23\pi}{12}

Hence:

\tan\frac{23\pi}{12}=\frac{y+1}{x-2}

Now:

\tan\frac{23\pi}{12}=\tan\left(2\pi-\frac{\pi}{12}\right)=-\tan\frac{\pi}{12}=-(2-\sqrt{3})=\sqrt{3}-2

Therefore:

\frac{y+1}{x-2}=\sqrt{3}-2

\boxed{y=(\sqrt{3}-2)x-2\sqrt{3}+3}