Problem
Let \(\bar{z}\) denote the complex conjugate of a complex number z and let i = √−1. In the set of complex numbers, find the number of distinct roots of the equation \(\bar{z}\) − z² = i(\(\bar{z}\) + z²).
Solution
Let’s discuss three different methods to solve the given equation.
Method 1
On rearranging the given equation, we’ll get (\(\bar{z}\) − z²)/(\(\bar{z}\) + z²) = i.
This gives z²/\(\bar{z}\) = (1 – i)/(1 + i). The RHS simplifies to -i. So, we get z² = -i\(\bar{z}\).
Now, \(\bar{z}\) can be written as |z|2/z, so the equation becomes z3 = -i|z|2.
To proceed, we can take the absolute value of both sides, which gives us |z|3 = |z|2.
On solving this for |z|, we get |z| = 0, and |z| = 1 (rejecting |z| = -1).
When |z| = 0, z must be 0. So, that’s one root.
When |z| = 1, the our equation changes to z3 = -i. This can also be written as z3 = i3.
This gives us three more roots: i, iω, iω2, where ω is the complex cube root of 1. So, a total of 4 distinct roots.
In case you’re unaware of or forgot about ω, we can look at another method to solve this equation.
And that’s by factorizing: z3 – i3 = 0 ⇒ (z – i)(z2 + i2 + zi) = 0.
One root is clearly i. And if you solve the quadratic equation, you’ll get two more roots. I’ll skip that part.
Method 2
Let’s continue from the simplified equation that we got in the previous method: z² = -i\(\bar{z}\)
Now, to find the number of roots, we can substitute z = x + iy, and find all possible pairs (x, y).
Let’s do that! We’ll get (x + iy)2 = -i(x – iy), which gives us x2 – y2 + 2ixy = -y – ix.
Now, we’ll equate the real and imaginary parts to get x2 – y2 = -y and 2xy = -x.
The second equation gives x = 0 or y = -1/2. Let’s substitute these values in the first to get the corresponding values of y and x respectively.
When x = 0, we get -y2 = -y, which gives y = 0 or 1. So, we get two roots: (0, 0) and (0, 1).
When y = -1/2, we get x2 – (-1/2)2 = -(-1/2). This gives x2 = 3/4, or x = ±√3/2.
So, we get two more roots: (√3/2, -1/2) and (-√3/2, -1/2).
Therefore the given equation has 4 distinct roots.
Method 3
Let’s assume z = |z|eiθ. Then the given equation changes to (|z|e-iθ – |z|2e2iθ) = i(|z|e-iθ + |z|2e2iθ).
This simplies to |z|(1 – |z|e3iθ) = i|z|((1 + |z|e3iθ).
Now, one equation that we get from here is |z| = 0, which means z = 0. And that’s one of the roots.
When |z| is non-zero, we get (1 – |z|e3iθ) = i((1 + |z|e3iθ).
And this gives us |z|e3iθ = (1 – i)/(1 + i) = -i.
Now, we can equate the real and imaginary parts of both sides to get cos(3θ) = 0 and sin(3θ) = -1/|z|
Now, wherever cos(3θ) = 0, sin(3θ) must be 1 or -1. If sin(3θ) equals 1, |z| comes out to be -1, which isn’t possible.
So, sin(3θ) can only be -1, which gives |z| = 1.
Finally, we’ll find the values of θ (in [0, 2π)) that satisfy the two equations (i.e. cos(3θ) = 0 and sin(3θ) = -1), and find the number of distinct roots.
This is straightforward – the required values are given by 3θ = (4n – 1)π/2, or θ = (4n – 1)π/6.
Putting n = 1, 2, and 3 gives us θ = π/2, 7π/6, and 11π/6 – three more roots.
So, the given equation has a total of 4 distinct roots.
Comments
A moderate problem – just the right amount of challenge. Method 1 wins over the other two, as it’s shorter. Also, we could’ve stopped at the point where we got z3 = -i. We didn’t really have to find the roots to conclude that there are three distinct roots. The nth root of a complex number (except 0) will always have n distinct values. So, we can be sure that the equation z3 = -i will have three distinct roots.
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