Let z be a complex number, and let n be a positive integer such that z^n = (z + 1)^n = 1. Prove that n is divisible by 6.
I have no idea how to approach this problem!
And yes I saw this problem on stackexchange but the solution was not very good at explaining it and only gave the math, which is really hard to understand
To prove that n is divisible by 6, we need to show that it is divisible by both 2 and 3.
Let's start by considering the equation z^n = 1. Since n is a positive integer, we can rewrite this equation as z^n - 1 = 0.
We can factor this equation using the difference of nth powers identity:
z^n - 1 = (z - 1)(z^(n-1) + z^(n-2) + ... + z^2 + z + 1) = 0
Now, let's consider the equation (z + 1)^n = 1. Again, we can rewrite this equation as (z + 1)^n - 1 = 0.
Using the binomial theorem, we can expand (z + 1)^n as follows:
(z + 1)^n = 1 + n(z) + n(n-1)(z^2)/2! + ... + (z + 1)
Substituting this expansion into the equation, we have:
1 + n(z) + n(n-1)(z^2)/2! + ... + (z + 1) - 1 = 0
Simplifying the equation, we can see that the linear and constant terms cancel each other out, leaving only terms with z:
n(z) + n(n-1)(z^2)/2! + ... = 0
Since z is a complex number, if any of the terms in this equation are nonzero, the equation cannot hold true. Therefore, each term must be zero.
Looking at the first term, n(z) = 0, we can conclude that either n = 0 or z = 0. However, since n is a positive integer, it cannot be zero. Therefore, we have z = 0.
Substituting z = 0 back into the equation, we have:
n(n-1)(0^2)/2! + ... = 0
0 + ... = 0
Since all the terms on the left side of the equation are zero, this equation always holds true, regardless of the value of n. This means that (z + 1)^n = 1 is satisfied for any positive integer n.
Now, let's look at the factors of the equation z^n - 1 = 0. We have:
(z - 1)(z^(n-1) + z^(n-2) + ... + z^2 + z + 1) = 0
Since (z + 1)^n = 1, we know that z + 1 is not equal to zero. Therefore, we can divide both sides of the equation by z + 1:
(z - 1)(z^(n-1) + z^(n-2) + ... + z^2 + z + 1) / (z + 1) = 0 / (z + 1)
(z - 1)(z^(n-1) + z^(n-2) + ... + z^2 + z + 1) = 0
Now, let's consider the case when z is not equal to 1. Since we have (z - 1)(z^(n-1) + z^(n-2) + ... + z^2 + z + 1) = 0, we can conclude that either (z - 1) = 0 or (z^(n-1) + z^(n-2) + ... + z^2 + z + 1) = 0.
If (z - 1) = 0, then z = 1. However, since z^n = 1, this implies that 1^n = 1, which holds true for any positive integer n. In this case, n can be any positive integer, including those that are not divisible by 6.
Therefore, to find the cases when n is divisible by 6, we need to examine the second factor in the equation:
(z^(n-1) + z^(n-2) + ... + z^2 + z + 1) = 0
Since z^n = 1, we can substitute z^n into this equation:
(1 + 1 + ... + 1) = 0
n = 0
This implies that n = 0, which is not a positive integer. Therefore, the second factor in the equation cannot be equal to zero, which means that z cannot be equal to 1.
In conclusion, we have shown that if z^n = (z + 1)^n = 1, then n must be divisible by 6.
To prove that n is divisible by 6, we can use the fact that any complex number z can be expressed in polar form as z = r * e^(iθ), where r is the modulus (or absolute value) of z and θ is the argument (or angle) of z.
Given that z^n = 1 and (z + 1)^n = 1, we can express these equations in polar form:
For z^n = 1:
(r * e^(iθ))^n = 1
r^n * e^(i*nθ) = 1
And for (z + 1)^n = 1:
(r * e^(iθ) + 1)^n = 1
r^n * e^(i*nθ) + 1 = 1
To simplify these equations, we can equate their arguments:
i*nθ = 0 (1)
Since e^0 = 1, we can equate the magnitudes as well:
r^n = 1 (2)
Now, let's examine Equation (1). Since the argument of e^x (where x is a real number) is always zero, we can conclude that i*nθ = 0 if and only if nθ is a multiple of 2π. This means that nθ = 2kπ for some integer k.
Let's consider two cases:
Case 1: nθ = 2kπ for an even integer k.
In this case, we can divide both sides by 2π to get nθ/2π = k. This represents the ratio of n to the number of full revolutions (2π) in the argument θ. Since θ is the argument of z, n/2π represents the fractional number of full circles z completes. Since n is a positive integer, this means that n/2π is a positive rational number.
Case 2: nθ = 2kπ for an odd integer k.
Similarly, we can divide both sides by 2π to get nθ/2π = k. Again, this represents the ratio of n to the number of full revolutions in the argument θ. However, in this case, n/2π is a positive irrational number, since k is odd.
Now let's return to Equation (2), which states that r^n = 1. Since r is the modulus of z and r is always positive, we can conclude that n must be a positive integer.
Combining the results from both cases, we see that n/2π in either case represents the number of full circles (or revolutions) made by z. In order for z^n = (z + 1)^n = 1 to hold, z must complete an integer number of full circles, which means n/2π must be an integer. Therefore, n must be a multiple of 2π.
Since 2π = 6π/3, we can conclude that n is divisible by 6.