hey all you math whizzes, chew on this!!!?

Question

we have three complexes z1, z2, & z3 such that:

z1+z2+z3=1
z1*z2*z3=1

find the three complexes z1, z2, and z3;

chew on that!!!

in case you were wondering: this system is solvable. you do not need 3 equations.

Answer 1

First, solve the second equation for z₁:

z₁=1/(z₂z₃)

Substitute it into the first equation:

1/(z₂z₃) + z₂ + z₃ = 1

Multiply by z₂z₃:

1 + z₃z₂² + z₃²z₂ = z₃z₂

Subtract z₃z₂:

z₃z₂² + (z₃²-z₃)z₂ + 1 = 0

Solve for z₂ using the quadratic formula:

z₂ = (z₃-z₃² ± √(z₃⁴-2z₃³+z₃²-4z₃))/(2z₃)

Substituting in the first equation:

z₁ = 1 - z₃ - (z₃-z₃² ± √(z₃⁴-2z₃³+z₃²-4z₃))/(2z₃)

This gives us values for z₁ and z₂ in terms of z₃. z₃ itself may be chosen arbitrarily. For instance, choosing z₃=1, we have:

z₂ = (1-1² ± √(1⁴-2(1)³+1²-4(1)))/(2(1))
z₂ = (0 ± √(1-2+1-4))/2
z₂ = (2i)/2 = i

And z₁ = 1 - 1 - i = -i

So one possible solution is (z₁, z₂, z₃) = (-i, i, 1)

Answer 2

You need one more equation to define z1, z2 and z3 uniquely.

If z1 is given, then we get the system
... z2 + z3 = 1 - z1
... z2 * z3 = 1/z1

We substitute z3 = 1 - z1 - z2 and get
... z2 * (1 - z1 - z2) = 1/z1,
... z2^2 - (1 - z1)z2 + 1/z1 = 0,

The discriminant is
... D = (1 - z1)^2 - 4/z1,
so that this equation can be solved whenever
... (1 - z1)^2 > 4/z1.
... (1 - z1)^2 z1 > 4 OR z1<0
... z1 >~ 2.3145 OR z1<0

The solutions are
... z2, z3 = {1 - z1 +- sqrt(D)} / 2.

For instance, if z1 = 4, then
... D = (1 - 4)^2 - 4/4 = 8,
... z2, z3 = (1 - 4 +- V8)/2 = -3/2 +- V2.

Indeed,
... 4 + (-3/2 + V2) + (3/2 - V2) = 1,
... 4 * (-3/2 + V2) * (3/2 - V2) = 1.

But as I said, there are many more solutions, depending on what value you pick for z1.

Answer 3

Don't you need 3 equations to determine 3 unknowns?

There could be any number of solutions. This is one of them.
z1 = 1, z2 = i, z3 = -i
If you let z1 = 1, then
z2 + z3 = 0 or z2 = -z3
z2*z3 = 1 = -z3^2
Thus z2 and z3 are +/- i.

Another solution is
z1 = 2, z2 = (-1-i)/2, z3 = (-1+i)/2

And the list goes on.

You could thumbs me down if you want, you know I'm right.

Answer 4

z1=1
z2=1
z3=1