Find the Cartesian equation of the locus of `arg((z-4)/(z+4))=pi/4`?

Set `z=x+yi`, so `arg((z-4)/(z+4))=pi/4`

`arctan((z-4)/(z+4))=pi/4`

`arctan((x+yi-4)/(x+yi+4))=pi/4`

`arctan((y-4)/x)-arctan((y+4)/x)=pi/4`

After taking tangent both sides,

`[(y-4)/x-(y+4)/x]/[1+(y-4)/x*(y+4)/x]=tan(pi/4)`

`(-8/x)/[(x^2+y^2-16)/x^2]=1`

`-8/x=(x^2+y^2-16)/x^2`

`x^2+y^2-16=-8x`

`x^2+8x+y^2-16=0`

`x^2+8x+16+y^2=32`

`(x+4)^2+y^2=32`

This equation describes a circle with its centre `M (-4,0)` and its radius `r=4sqrt2`.

We know that

`z-4 = r_1 e^(i phi_1)` with `phi_1 = arctan(y/(x-4))`
`z+4 = r_2 e^(i phi_2)` with `phi_2 = arctan(y/(x+4))`

then

`"arg"((z-4)/(z+4)) = "arg"(r_1/r_2e^(i(phi_1-phi_2))) = phi_1-phi_2 = pi/4` or

`arctan(y/(x-4))-arctan(y/(x+4))=pi/4`

now

`tan(arctan(y/(x-4))-arctan(y/(x+4))) = tan(pi/4) = 1` or

`(8 y)/(x^2 + y^2-16) = 1` or

`x^2+y^2-8y-16=0` or

`x^2+(y-4)^2 = 32`

NOTE

From `tan(phi_1-phi_2) = (tanphi_1-tanphi_2)/(1+tanphi_1 tanphi_2)`

making

`phi_1 = arctan(y/(x-4))` and `phi_2 = arctan(y/(x+4))`

and knowing that `tan(arctan xi))= xi` we have

`((y/(x-4))-(y/(x+4)))/(1+(y/(x-4))(y/(x+4))) = (8 y)/(x^2 + y^2-16)`