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)#