Complex Numbers Brilliance

Finally, after waiting a very long time, I found a really good question with complex numbers. Usually those questions are really lame, and they are an exact copy of each other. However, today I found a really good one (like ones in super math tips), and the solution someone gave there is pure brilliance.

z is a complex number with modulus 1. Solve:
5z⁴ - 11z³ + 16z² - 11z + 5 = 0

Because of its special coefficient pattern, some people factored it. But doing so takes the fun, and not everyone can spot this pattern. However, someone used the modulus property to create a truely amazing solution.

A complex number can be represented in two ways:
z = a + bi
z = r(cos θ + isin θ)

r is the modulus, and it is also given by √(a² + b²).

Let w be the conjugate of z (which means w = a - bi). Since a² + b² = z * w, z * w = 1. Square that, and you get:
z²w² = 1

Multiply the entire equation by w² (w is non-zero because z is non-zero):
5z⁴w² - 11z³w² + 16z²w² - 11zw² + 5w² = 0

Using what we showed before, this equation turns into:
5z² - 11z + 16 - 11w + 5w² = 0

Now it's just playing with numbers (remember that i² = -1 by definition):
5(z² + w²) = 11(z + w) - 16

Since z = a + bi, w = a - bi:
5((a + bi)² + (a - bi)²) = 11(a + bi + a - bi) - 16
5(a² + 2abi - b² + a² - 2abi - b²) = 11(2a) - 16
5(2a² - 2b²) = 22a - 16
10(a² - b²) = 22a - 16

Since √(a² + ²) = 1, we can conclude that:
a² + b² = 1
b² = 1 - a²

Plugging back into the equation:
10(a² - 1 + a²) = 22a - 16
10(2a² - 1) = 22a - 16
20a² - 10 = 22a - 16
20a² - 22a + 6 = 0
10a² - 11a + 3 = 0
(2a - 1)(5a - 3) = 0
a = 1/2, 3/5

Now it's really easy to find b (notice that each a has two b's):
When a = 1/2:
(1/2)² + b² = 1
b² = 3/4
b = ±√3/2

When a = 3/5:
(3/5)² + b² = 1
b² = 16/25
b = ±4/5

The four solutions of the equation are:
z₁ = 1/2 + i√3/2
z₂ = 1/2 - i√3/2
z₃ = 3/5 + i4/5
z₄ = 3/5 - i4/5

Hope you liked it. I know I did.

Have a great, non-complex weekend,
Nadav

nadavs