At 3:46 you have (z² + 1)(z² + z + 1) = z² In fact this is not hard to factor. Rather than expanding the left hand side completely, we can note that z² + z + 1 = (z² + 1) + z and rewrite this as (z² + 1)² + (z² + 1)z = z² At the left hand side (z² + 1)² is the square of z² + 1 and (z² + 1)z = 2·(z² + 1)·(¹⁄₂z) is twice the product of z² + 1 and ¹⁄₂z so we can complete the square at the left hand side by adding (¹⁄₂z)² = ¹⁄₄z² to both sides, which gives (z² + 1)² + 2·(z² + 1)·(¹⁄₂z) + (¹⁄₂z)² = z² + ¹⁄₄z² and this can be written as (z² + 1 + ¹⁄₂z)² = ⁵⁄₄z² or (z² + 1 + ¹⁄₂z)² = (¹⁄₂√5·z)² since ⁵⁄₄ is the square of ¹⁄₂√5. We now have a square on both sides and subtracting (¹⁄₂√5·z)² from both sides this gives (z² + 1 + ¹⁄₂z)² − (¹⁄₂√5·z)² = 0 and applying the difference of two squares identity a² − b² = (a − b)(a + b) to the left hand side this can be written as (z² + (¹⁄₂ − ¹⁄₂√5)z + 1)(z² + (¹⁄₂ + ¹⁄₂√5)z + 1) = 0 and our quartic is factored into two quadratics with real coefficients. The discriminant of each of the two quadratics is negative, so this equation has no real solutions. Note, however, that it is easy to show without factoring or actually solving the equation that z⁴ + z³ + z² + z + 1 = 0 has no real solutions, since we can rewrite this as (z² + ¹⁄₂z)² − ¹⁄₄z² + z² + z + 1 = 0 or (z² + ¹⁄₂z)² = −³⁄₄z² − z − 1 The discriminant of the quadratic at the right hand side is (−1)² − 4·(−³⁄₄)·(−1) = 1 − 3 = −2 so this quadratic has no real zeros and must therefore be negative for any real z since it is negative for z = 0. But the left hand side is nonnegative for any real z. Consequently, no real z can satisfy this equation. Since the quartic equation has real coefficients, the conjugate of any complex solution is also a solution, so the four solutions are two pairs of complex conjugates. This implies that, apart from the order of the factors, this monic quartic can only have a _single_ factorization into two monic quadratics with _real_ coefficients. This is because a quadratic factor with real coefficients can only be obtained by pairing two linear factors with conjugate complex zeros. Note that the two pairs of conjugate complex zeros of the quartic polynomial z⁴ + z³ + z² + z + 1 must be different from one another, because if they were identical this quartic would be the square of a quadratic polynomial, which it is not. At 10:22 you observe that the roots of this quartic equation are the fifths roots of unity except for 1 itself, so we have zₖ = exp(2kπi/5), k = 1..4 Note that the complex exponential function is periodic with a period 2πi. So, for k = 3 we have z₃ = exp(6πi/5) = exp(−4πi/5) and therefore z₂·z₃ = exp(4πi/5)·exp(−4πi/5) = 1 and z₂ + z₃ = exp(4πi/5) + exp(−4πi/5) = 2·cos(4π/5) Similarly, for k = 4 we have z₄ = exp(8πi/5) = exp(−2πi/5) and therefore z₁·z₄ = exp(2πi/5)·exp(−2πi/5) = 1 and z₁ + z₄ = exp(2πi/5) + exp(−2πi/5) = 2·cos(2π/5) Consequently, we have (z − z₁)(z − z₄) = z² − (z₁ + z₄)·z + z₁·z₄ = z² − 2·cos(2π/5)·z + 1 and (z − z₂)(z − z₃) = z² − (z₂ + z₃)·z + z₂·z₃ = z² − 2·cos(4π/5)·z + 1 so we can factor z⁴ + z³ + z² + z + 1 = 0 as (z² − 2·cos(2π/5)·z + 1)(z² − 2·cos(4π/5)·z + 1) = 0 Comparing this with the factorization (z² + (¹⁄₂ − ¹⁄₂√5)z + 1)(z² + (¹⁄₂ + ¹⁄₂√5)z + 1) = 0 and noting that cos(2π/5) > 0 and cos(4π/5) < 0 and that, apart from the order of the factors, this monic quartic has only a single factorization into two monic quadratics with real coefficients, we can conclude that cos(2π/5) = −¹⁄₄ + ¹⁄₄√5 and cos(4π/5) = −¹⁄₄ − ¹⁄₄√5 This means that cos(2π/5) and cos(4π/5) are the roots of the quadratic equation 4x² + 2x − 1 = 0 This same result can also be found without using complex numbers. For α = 2kπ/5 where k is any integer we have cos 3α = cos(6kπ/5) = cos(−4kπ/5) = cos(4kπ/5) = cos 2α. Therefore, we have 4·cos³α − 3·cos α = 2·cos²α − 1 whenever α = 2kπ/5 where k is any integer. This implies that, for any integer k, x = cos(2kπ/5) is a root of the equation 4x³ − 3x = 2x² − 1 or 4x³ − 2x² − 3x + 1 = 0 Accordingly, for k = 0, 1, 2, this equation has the roots x₁ = cos 0 = 1, x₂ = cos(2π/5), x₃ = cos(4π/5). Of course, other integer values of k do not give us any more roots because a cubic equation has no more than three different roots and because the cosine is an even periodic function with a period 2π. So, e.g. k = 3 gives the same value as k = −2 which is the same value we get for k = 2. Likewise, k = 4 gives the same value as k = −1 and k = 1 and k = 5 again gives the same value as k = 0. Since x₁ = cos 0 = 1 is a root of the cubic we can factor out x − 1 to get (x − 1)(4x² + 2x − 1) = 0 which means that x₂ = cos(2π/5) and x₃ = cos(4π/5) are indeed the roots of the quadratic equation 4x² + 2x − 1 = 0 and solving this equation algebraically and noting that cos(2π/5) > 0 and cos(4π/5) < 0 we again find that cos(2π/5) = −¹⁄₄ + ¹⁄₄√5 and cos(4π/5) = −¹⁄₄ − ¹⁄₄√5.
I find that multiplying by (z-1) gives you z^5-1=0 much easier for most students to remember (not forgetting to considering z=1 case). from there Standard root of unity e(2 k pi i t / 5)= 1 giving you the 5th roots of unity (remembering again the initial consideration for case z=1). Not sure why going to the 1/x and 1/x^2 step. Feels much more convoluted and difficult for students to reproduce under stress.
@@black_eagle" ...otherwise they'd be impossible to solve ;)" TIL that real world math problems are impossible to solve. Just kidding. I know you didnt really mean what you said. I used math in my job for decades, mostly algebra and geometry or finding the derivative. I used it for : break even analysis, interpreting prints, various stupid human tricks with compression/tension of various substrates. That last one covers so much from bend allowances to optimal torque values for a bolt pattern. I really do get what you mean though. If you look at an algebra book's answers to the problems, it is quite evident that they often follow a formula you learn from a more advanced course, and the rest started from the answer and work back from there. Story problems are a different beast altogether, and perhaps that is why you don't see them used much at all. Not good ones at least.
At 3:46 you have
(z² + 1)(z² + z + 1) = z²
In fact this is not hard to factor. Rather than expanding the left hand side completely, we can note that z² + z + 1 = (z² + 1) + z and rewrite this as
(z² + 1)² + (z² + 1)z = z²
At the left hand side (z² + 1)² is the square of z² + 1 and (z² + 1)z = 2·(z² + 1)·(¹⁄₂z) is twice the product of z² + 1 and ¹⁄₂z so we can complete the square at the left hand side by adding (¹⁄₂z)² = ¹⁄₄z² to both sides, which gives
(z² + 1)² + 2·(z² + 1)·(¹⁄₂z) + (¹⁄₂z)² = z² + ¹⁄₄z²
and this can be written as
(z² + 1 + ¹⁄₂z)² = ⁵⁄₄z²
or
(z² + 1 + ¹⁄₂z)² = (¹⁄₂√5·z)²
since ⁵⁄₄ is the square of ¹⁄₂√5. We now have a square on both sides and subtracting (¹⁄₂√5·z)² from both sides this gives
(z² + 1 + ¹⁄₂z)² − (¹⁄₂√5·z)² = 0
and applying the difference of two squares identity a² − b² = (a − b)(a + b) to the left hand side this can be written as
(z² + (¹⁄₂ − ¹⁄₂√5)z + 1)(z² + (¹⁄₂ + ¹⁄₂√5)z + 1) = 0
and our quartic is factored into two quadratics with real coefficients.
The discriminant of each of the two quadratics is negative, so this equation has no real solutions. Note, however, that it is easy to show without factoring or actually solving the equation that
z⁴ + z³ + z² + z + 1 = 0
has no real solutions, since we can rewrite this as
(z² + ¹⁄₂z)² − ¹⁄₄z² + z² + z + 1 = 0
or
(z² + ¹⁄₂z)² = −³⁄₄z² − z − 1
The discriminant of the quadratic at the right hand side is (−1)² − 4·(−³⁄₄)·(−1) = 1 − 3 = −2 so this quadratic has no real zeros and must therefore be negative for any real z since it is negative for z = 0. But the left hand side is nonnegative for any real z. Consequently, no real z can satisfy this equation.
Since the quartic equation has real coefficients, the conjugate of any complex solution is also a solution, so the four solutions are two pairs of complex conjugates. This implies that, apart from the order of the factors, this monic quartic can only have a _single_ factorization into two monic quadratics with _real_ coefficients. This is because a quadratic factor with real coefficients can only be obtained by pairing two linear factors with conjugate complex zeros. Note that the two pairs of conjugate complex zeros of the quartic polynomial z⁴ + z³ + z² + z + 1 must be different from one another, because if they were identical this quartic would be the square of a quadratic polynomial, which it is not.
At 10:22 you observe that the roots of this quartic equation are the fifths roots of unity except for 1 itself, so we have
zₖ = exp(2kπi/5), k = 1..4
Note that the complex exponential function is periodic with a period 2πi. So, for k = 3 we have z₃ = exp(6πi/5) = exp(−4πi/5) and therefore
z₂·z₃ = exp(4πi/5)·exp(−4πi/5) = 1
and
z₂ + z₃ = exp(4πi/5) + exp(−4πi/5) = 2·cos(4π/5)
Similarly, for k = 4 we have z₄ = exp(8πi/5) = exp(−2πi/5) and therefore
z₁·z₄ = exp(2πi/5)·exp(−2πi/5) = 1
and
z₁ + z₄ = exp(2πi/5) + exp(−2πi/5) = 2·cos(2π/5)
Consequently, we have
(z − z₁)(z − z₄) = z² − (z₁ + z₄)·z + z₁·z₄ = z² − 2·cos(2π/5)·z + 1
and
(z − z₂)(z − z₃) = z² − (z₂ + z₃)·z + z₂·z₃ = z² − 2·cos(4π/5)·z + 1
so we can factor
z⁴ + z³ + z² + z + 1 = 0
as
(z² − 2·cos(2π/5)·z + 1)(z² − 2·cos(4π/5)·z + 1) = 0
Comparing this with the factorization
(z² + (¹⁄₂ − ¹⁄₂√5)z + 1)(z² + (¹⁄₂ + ¹⁄₂√5)z + 1) = 0
and noting that cos(2π/5) > 0 and cos(4π/5) < 0 and that, apart from the order of the factors, this monic quartic has only a single factorization into two monic quadratics with real coefficients, we can conclude that
cos(2π/5) = −¹⁄₄ + ¹⁄₄√5
and
cos(4π/5) = −¹⁄₄ − ¹⁄₄√5
This means that cos(2π/5) and cos(4π/5) are the roots of the quadratic equation
4x² + 2x − 1 = 0
This same result can also be found without using complex numbers. For α = 2kπ/5 where k is any integer we have cos 3α = cos(6kπ/5) = cos(−4kπ/5) = cos(4kπ/5) = cos 2α. Therefore, we have 4·cos³α − 3·cos α = 2·cos²α − 1 whenever α = 2kπ/5 where k is any integer. This implies that, for any integer k, x = cos(2kπ/5) is a root of the equation
4x³ − 3x = 2x² − 1
or
4x³ − 2x² − 3x + 1 = 0
Accordingly, for k = 0, 1, 2, this equation has the roots x₁ = cos 0 = 1, x₂ = cos(2π/5), x₃ = cos(4π/5). Of course, other integer values of k do not give us any more roots because a cubic equation has no more than three different roots and because the cosine is an even periodic function with a period 2π. So, e.g. k = 3 gives the same value as k = −2 which is the same value we get for k = 2. Likewise, k = 4 gives the same value as k = −1 and k = 1 and k = 5 again gives the same value as k = 0. Since x₁ = cos 0 = 1 is a root of the cubic we can factor out x − 1 to get
(x − 1)(4x² + 2x − 1) = 0
which means that x₂ = cos(2π/5) and x₃ = cos(4π/5) are indeed the roots of the quadratic equation
4x² + 2x − 1 = 0
and solving this equation algebraically and noting that cos(2π/5) > 0 and cos(4π/5) < 0 we again find that cos(2π/5) = −¹⁄₄ + ¹⁄₄√5 and cos(4π/5) = −¹⁄₄ − ¹⁄₄√5.
Nice. z^2 +1 is the equation at the core of the Mandelbrot set.
That was elegant, super nice
Thank you! 😊
I find that multiplying by (z-1) gives you z^5-1=0 much easier for most students to remember (not forgetting to considering z=1 case). from there Standard root of unity e(2 k pi i t / 5)= 1 giving you the 5th roots of unity (remembering again the initial consideration for case z=1).
Not sure why going to the 1/x and 1/x^2 step. Feels much more convoluted and difficult for students to reproduce under stress.
Did you start with the fifth roots of unity and remove 1 to construct this problem?
Yes - I guess that was contrived.
@@scottleung9587 Most math problems are contrived, otherwise they'd be impossible to solve ;)
Good thinking! 😁
@@black_eagle" ...otherwise they'd be impossible to solve ;)"
TIL that real world math problems are impossible to solve. Just kidding. I know you didnt really mean what you said.
I used math in my job for decades, mostly algebra and geometry or finding the derivative. I used it for : break even analysis, interpreting prints, various stupid human tricks with compression/tension of various substrates. That last one covers so much from bend allowances to optimal torque values for a bolt pattern.
I really do get what you mean though. If you look at an algebra book's answers to the problems, it is quite evident that they often follow a formula you learn from a more advanced course, and the rest started from the answer and work back from there. Story problems are a different beast altogether, and perhaps that is why you don't see them used much at all. Not good ones at least.