Ian FowlerDec 14, 2019 · Edited: Dec 15, 2019So: 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

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 triangleto 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|^2/2under the constraint that |z| = |w| = |z+w|.Thanks for the problem - Ian