Cube Root of A Complex Number | Problem 455

[@f5673-t1h]

The meta approach would be to realize that the exam writers would not give you something too tedious, so a and b must be integers. We have that a(a^2 - 3b^2) + ib(3a^2-b^2) = 52 + 47i, so b is a divsor of 47, which is prime. It probably won't be 47, since the cube would be too big, so b must be 1. Solving for a quickly gives a = 4 (in both components, and you can also verify it afterwards), so 4+i is one solution, and then you multiply by the two other third roots of 1 to get the other solutions.

[@NadiehFan]

How about this. At 7:30 you have a system (1) a³ − 3ab² = 52 (2) 3a²b − b³ = 47 and at 9:33 you found that | a + bi | = √17. But this implies (3) a² + b² = 17 From (3) we get b² = 17 − a² which we can substitute in (1) to get a³ − 3a(17 − a²) = 52 so (4) 4a³ − 51a − 52 = 0 Using the rational root theorem and by rewriting (4) as a(4a² − 51) − 52 = 0 and noting that for a positive solution we must have 4a² − 51 > 0 and therefore a > 3 we can find that a = 4 is a solution of (4) since (5) 4·4³ − 51·4 − 52 = 0 and subtracting (5) from (4) we have 4(a³ − 4³) − 51(a − 4) = 0 4(a − 4)(a² + 4a + 16) − 51(a − 4) = 0 (a − 4)(4a² + 16a + 64) − 51(a − 4) = 0 (a − 4)(4a² + 16a + 13) = 0 a − 4 = 0 ⋁ (2a + 4)² = 3 a = 4 ⋁ a = −2 + ½√3 ⋁ a = −2 − ½√3 So, we have three values for a, but evidently for the principal cube root we must have a > 0 because the principal value of the argument of a nonzero complex number is on the interval (−π, π] so the principal value of the argument of the principal cube root of a nonzero complex number is on the interval (−⅓π, ⅓π] and therefore in the right half of the complex plane. For a = 4 we get b = 1 ⋁ b = −1 but only (a, b) = (4, 1) satisfies (2) so ∛(52 + 47i) = 4 + i The other two complex numbers whose cube is 52 + 47i can obviously be found from a = −2 + ½√3 ⋁ a = −2 − ½√3 by substituting these values for a in (3) to get the corresponding values for b and checking in each case which of the two values of b satisfies (2). However, we can also simply multiply 4 + i by each of the complex cube roots of unity ω₁ = −½ + i·½√3 and ω₂ = −½ − i·½√3 to get (−½ + i·½√3)(4 + i) = (−2 − ½√3) + i·(− ½ + 2√3) (−½ − i·½√3)(4 + i) = (−2 + ½√3) + i·(− ½ − 2√3) as the other two complex numbers whose cube is 52 + 47i. As you can see, the real parts of these complex numbers are indeed the two negative solutions of equation (4).

[@WindyNight114]

Your videos are so cool. I really appreciate it. I took complex analysis back in college and I forgot so much. Thank you!

[@aplusbi]

Thanks for the kind words! 😄

[@jorgeheyaime8810]

Hello from Dominican Republic

[@aplusbi]

Hi there! 😀

[@0over0]

hello!

[@moeberry8226]

The real way to obtain all 3 complex cube roots of that expression is to write it in polar form of the inside and then take the cube root so the radius is NOT 17root17 but rather just root17. And the angle will be reduced by a factor of 1/3. And then using integer values 0, 1 and 2 you will get all 3 cube roots. It’s no different then finding the cube roots of unity of z^3=57+42i.

[@scottleung9587]

I also got z=4+i.

[@key_board_x]

x = ³√(52 + 47i) x³ = 52 + 47i → we assume that: y³ = 52 - 47i ------------------------------------------------------sum x³ + y³ = 104 ← equation (1) x³y³ = (52 + 47i).(52 - 47i) x³y³ = (52)² - (47i)² x³y³ = 2704 - 2209i² → where: i² = - 1 x³y³ = 4913 x³y³ = 17³ xy = 17 (x + y)³ = (x + y)².(x + y) (x + y)³ = (x² + 2xy + y²).(x + y) (x + y)³ = x³ + x²y + 2x²y + 2xy² + xy² + y³ (x + y)³ = x³ + y³ + 3x²y + 3xy² (x + y)³ = x³ + y³ + (3x²y + 3xy²) (x + y)³ = x³ + y³ + 3xy.(x + y) → recall (2): xy = 17 (x + y)³ = x³ + y³ + 51.(x + y) → recall (1): x³ + y³ = 104 (x + y)³ = 104 + 51.(x + y) → let: u = x + y u³ = 104 + 51u u³ - 51u - 104 = 0 u = 8 → recall: u = x + y x + y = 8 → recall (2): xy = 17 → y = 17/x x + (17/x) = 8 (x² + 17)/x = 8 x² + 17 = 8x x² - 8x + 17 = 0 Δ = (- 8)² - (4 * 17) = 64 - 68 = - 4 = 4i² x = (8 ± 2i)/2 x = 4 ± i Let's check: x = 4 + i x³ = x².x x³ = (4 + i)².(4 + i) x³ = (16 + 8i + i²).(4 + i) → where: i² = - 1 x³ = (15 + 8i).(4 + i) x³ = 60 + 15i + 32i + 8i x³ = 60 + 47i - 8 x³ = 52 + 47i ← ok Let's check: x = 4 - i x³ = x².x x³ = (4 - i)².(4 - i) x³ = (16 - 8i + i²).(4 - i) → where: i² = - 1 x³ = (15 - 8i).(4 - i) x³ = 60 - 15i - 32i + 8i² x³ = 60 - 47i - 8 x³ = 52 - 47i ← rejected Solution = { (4 + i) }

[@aplusbi]

Wow! This is amazing 😍

[@0over0]

wow.

[@key_board_x]

My answer is not complete, because there are 3 roots. Restart x = ³√(52 + 47i) x³ = 52 + 47i x³ - (52 + 47i) = 0 → recall the solution: ³√(52 + 47i) = (4 + i) x³ - (4 + i)³ = 0 → recall: a³ - b³ = (a - b).(a² + ab + b²) [x - (4 + i)].[x² + x.(4 + i) + (4 + i)²] = 0 First case: [x - (4 + i)] = 0 x - (4 + i) = 0 x = 4 + I ← first root Second case: [x² + x.(4 + i) + (4 + i)²] = 0 x² + x.(4 + i) + (4 + i)² = 0 Δ = (4 + i)² - 4.(4 + i)² Δ = - 3.(4 + i)² Δ = 3i².(4 + i)² x = [- (4 + i) ± i.(4 + i).√3]/2 First case: x = [- (4 + i) + i.(4 + i).√3]/2 x = [- 4 - i + (4i + i²).√3]/2 x = [- 4 - i + (4i - 1).√3]/2 x = [- 4 - i + 4i√3 - √3]/2 x = [- 4 - √3 - i + 4i√3]/2 x = [- (4 + √3) - i.(1 - 4√3)]/2 ← second root Second case: x = [- (4 + i) - i.(4 + i).√3]/2 x = [- 4 - i - (4i + i²).√3]/2 x = [- 4 - i - (4i - 1).√3]/2 x = [- 4 - i - 4i√3 + √3]/2 x = [- 4 + √3 - i - 4i√3]/2 x = [- (4 - √3) - i.(1 + 4√3)]/2 ← third root

[@SidneiMV]

(a + bi)³ = 52 + 47i a³ + 3i²ab² = 52 => a³ - 3ab² = 52 3ia²b + i³b³ = 47i => 3a²b - b³ = 47 a³ + b³ - 3ab(a + b) = 5 a³ - b³ + 3ab(a - b) = 99

[@giuseppemalaguti435]

Se non sbaglio (52^2+47^2)(1/6)e^iarctg47/3*52)=4,1231e^i16,766

[@jeanmariebergeron4759]

The angle i ie^i(arctg(47/52))/3