A Radical Equation | Problem 422

[@iabervon]

I used the second method, but I replaced the constant with c until I'd done all of the manipulation. It turns out that the c² terms cancel out in the quadratic formula, so there's a bunch of calculation you can skip if you put it off.

[@scottleung9587]

I used a third method, in which I replaced w with c+di and did the arithmetic from there.

[@Don-Ensley]

problem z - √z = 1 - 3i Let u = √z u² - u - (1 - 3i) = 0 By quadratic formula, u = { 1 ± √[ (1+4( 1 - 3i ) ] } / 2 = (1/2) ± (1/2) √(5 - 12 i) = (1/2) ± (1/2) √13 (5 / 13 - i 12 / 13) ¹ᐟ² = (1/2) ± (1/2)√13 [ ( cos tan⁻¹(12/5) - i sin tan⁻¹(12/5) ) ] ¹ᐟ² By DeMoivre's theorem, u = (1/2) ± (1/2)√13 [(cos ½ tan⁻¹12/5 - i sin ½ tan⁻¹12/5)] By half angle formulas: cos ½ tan⁻¹12/5 = √[(1+5/13)/2] = √(18/26) = 3/√13 sin ½ tan⁻¹12/5) = √[(1-5/13)/2] = √(8/26) = 2/√13 u = (1/2) ± (1/2)√13 [3/√13 - i 2/√13] = (1/2) ± (1/2) (3 - 2 i ) = (1/2) [ 1 ± (3 - 2 i ) ] Two solutions for u = √z : u = 2 - i , u = -1 + i Solutions for z = u²: z = 4-4 i -1 = 3 - 4 i, z = 1-2i -1 = -2i answer z ∈ { -2 i, 3 - 4 i }

[@mcwulf25]

First method with the Quadratic Formula

[@key_board_x]

If z is a complex number, then √z is a complex number too. √z = a + ib ← this is a complex number z = (a + ib)² z = a² + 2abi + i²b² z = a² - b² + 2abi z - √z = 1 - 3i a² - b² + 2abi - (a + ib) = 1 - 3i a² - b² + 2abi - a - ib = 1 - 3i (a² - b² - a) + i.(2ab - b) = 1 - 3i → then you can deduce that; (1): a² - b² - a = 1 (2): 2ab - b = - 3 (2): b.(2a - 1) = - 3 (2): b = - 3/(2a - 1) Restart from (1) a² - b² - a = 1 b² = a² - a - 1 → where: b = - 3/(2a - 1) [- 3/(2a - 1)]² = a² - a - 1 9/(2a - 1)² = a² - a - 1 9 = (2a - 1)².(a² - a - 1) 9 = (4a² - 4a + 1).(a² - a - 1) 9 = 4a⁴ - 4a³ - 4a² - 4a³ + 4a² + 4a + a² - a - 1 4a⁴ - 8a³ + a² + 3a - 10 = 0 → the aim, if we are to continue effectively, is to eliminate terms to the power 3 Let: a = x - (β/4α) → where: β is the coefficient for a³, in our case: - 8 α is the coefficient for a⁴, in our case: 4 4a⁴ - 8a³ + a² + 3a - 10 = 0 → let: a = x - (- 8/16) → a = x + (1/2) ← do not confuse with z in the original equation 4.[x + (1/2)]⁴ - 8.[x + (1/2)]³ + [x + (1/2)]² + 3.[x + (1/2)] - 10 = 0 4.[x + (1/2)]².[x + (1/2)]² - 8.[x + (1/2)]².[x + (1/2)] + [x² + x + (1/4)] + 3x + (3/2) - 10 = 0 4.[x² + x + (1/4)].[x² + x + (1/4)] - 8.[x² + x + (1/4)].[x + (1/2)] + x² + x + (1/4) + 3x + (3/2) - 10 = 0 4.[x⁴ + x³ + (1/4).x² + x³ + x² + (1/4).x + (1/4).x² + (1/4).x + (1/16)] - 8.[x³ + (1/2).x² + x² + (1/2).x + (1/4).x + (1/8)] + x² + x + (1/4) + 3x + (3/2) - 10 = 0 4.[x⁴ + 2x³ + (3/2).x² + (1/2).x + (1/16)] - 8.[x³ + (3/2).x² + (3/4).x + (1/8)] + x² + x + (1/4) + 3x + (3/2) - 10 = 0 4x⁴ + 8x³ + 6x² + 2x + (1/4) - 8x³ - 12x² - 6x - 1 + x² + x + (1/4) + 3x + (3/2) - 10 = 0 4x⁴ - 5x² - 9 = 0 4x⁴ - 5x² = 9 (2x²)² - 2.[2x² * (5/4)] = 9 (2x²)² - 2.[2x² * (5/4)] + (5/4)² = 9 + (5/4)² [2x² - (5/4)]² = 169/16 [2x² - (5/4)]² = (13/4)² 2x² - (5/4) = ± 13/4 2x² = (5/4) ± (13/4) 2x² = (5 ± 13)/4 x² = (5 ± 13)/8 First case: x² = (5 - 13)/8 = - 1 = i² x = ± i → recall: a = x + (1/2) a = (1/2) ± i Second case: x² = (5 + 13)/8 = 18/8 = 9/4 x = ± 3/2 → recall: a = x + (1/2) a = (1/2) ± (3/2) a = (1 ± 3)/2 First solution: a = (1/2) + i 2a = 1 + 2i 2a - 1 = 2i → recall: b = - 3/(2a - 1) b = - 3/2i b = - 3i/2i² b = 3i/2 → recall: z = a² - b² + 2abi z = [(1/2) + i]² - (3i/2)² + 2.[(1/2) + i].(3i/2).i z = (1/4) + i + i² - (9i²/4) + [(1/2) + i].3i² z = (1/4) + i - 1 + (9/4) - 3.[(1/2) + i] z = (1/4) + i - 1 + (9/4) - (3/2) - 3i → z = - 2i Second solution: a = (1/2) - i 2a = 1 - 2i 2a - 1 = - 2i → recall: b = - 3/(2a - 1) b = - 3/- 2i b = 3i/2i² b = - 3i/2 → recall: z = a² - b² + 2abi z = [(1/2) - i]² - (- 3i/2)² + 2.[(1/2) - i].(- 3i/2).i z = (1/4) - i + i² - (9i²/4) + [(1/2) - i].(- 3i²) z = (1/4) - i - 1 + (9/4) + 3.[(1/2) - i] z = (1/4) - i - 1 + (9/4) + (3/2) - 3i z = 3 - 4i Third solution: a = (1 + 3)/2 = 2 Recall: b = - 3/(2a - 1) b = - 3/3 b = - 1 → recall: z = a² - b² + 2abi z = 2² - (- 1)² + 2.[2 * - 1].i z = 4 - 1 - 4i z = 3 - 4i → already seen Fourt solution: a = (1 - 3)/2 = - 1 Recall: b = - 3/(2a - 1) b = - 3/- 3 b = 1 → recall: z = a² - b² + 2abi z = (- 1)² - (1)² + 2.[- 1 * 1].i z = 1 - 1 - 2i z = - 2i → already seen