argument question (1 Viewer)

[stag_j]

ok i have this problem that i'm trying to figure out
(z-z')^2 + 8a(z+z') = 16a^2
show that |Arg z| <= pi/4

i've graphed it, and can see that this is true - for any value of 'a' there is one point where the argument is +/- pi/4, namely the point (2a,2a) but how do i prove this nicely?
any ideas?

 

[MyLuv]

1st-prove it a parabola (let z=x+iy) y^2=4a(x-a);
2nd-differentiate the parabola equation -->equation of tangent through O :y-y0=2a/y0*(x-x0)
sub x=0,y=0 u got 2 tangent with make with Ox angle of +-(pi)/4 .
3rd-The parabola lie inside angle made by 2 tangents
ie larg(z)l<= pi/4