If `arg(z-1)=arg(z+3i)`, then find `(x-1):y`, where `z=x+iy`

Explanation

Given `arg(Z−1)=arg(Z+3i)`

`arg(x+iy−1)=arg(x+iy+3i)`

`arg(x−1+iy)=arg(x+i(y+3))`

`tan^(−1)(y/(x−1)​)=tan^(−1)((y+3)/x​)`

`y/(x−1)​=(y+3)/x​`

`xy=xy−y+3x−3`

`y=3(x−1)`

`1/3​=(x−1)/y​`

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