So: Given z = a + bi = r*cis(t), find w = c +di so that |z| = |w| = |z+w|.
Here I will use lowercase "p" to represent "pi" and a lowercase "t" to represent "theta".
I wasted a lot of time messing around with a^2,b^2,c^2,d^2,(a+b)^2,(c+d)^2 until I finally realized
that z,w,z+w have to form an equilateral triangle to satisfy the given conditions.
So the angles are all p/3.
Sorry for the out-focus jpg.
Here are the steps as I see it.
1) Pick any real pair a,b where not both of a,b are 0 (trivial case) and then z = a + bi.
2) Express z in polar form: z = r*cos(t) + ri*sin(t)
Find r = sqrt(a^2 + b^2)
You can find t by using: cos(t) = b/r and sin(t) = a/r.
3) As seen in the diagram the angle for w is t+2p/3
Set w = r*cos(t+2p/3) + ri*sin(t+2p/3) and you have it.
4) Now having the angles for z and w and some trig identities and you can easily show that the
angle for (z + w) is (t+p/3) as seen in the diagram.
cos(t) + cos(t+2p/3) = cos(t+p/3)
sin(t) + sin(t+2p/3) = sin(t+p/3)
so that z+w = r*cos(t+p/3) + ir*sin(t+p/3)
Another oddity that comes out of this is that for z = a+ bi and w = c + di we get
ac +bd = -|z|^2/2 = -|w|^2/2 under the constraint that |z| = |w| = |z+w|.
Thanks for the problem - Ian