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
I am so sorry but I got thrown off when I saw r*cis(t) and read that at "racist". What is wrong with me.
I think the younger generation needs to take a break.
@Ian Fowler I've seen what happens on Twitter and it is NOT pretty. Someone said it feels like the Salem witch trials all over again, people blindly executing (cancelling) other people just because of something absolutely tiny.
@Setsuna Kujo What you say is so true. Anyway, my response was jokingly done. No harm. I hope the solution was helpful. Ian