Can We Solve A Nice Cubic Equation | Problem 413

[@seanfraser3125]

I used the secret 4th method: Multiply both sides by -i = i^3. We have (iz)^3 = 1 So iz is a third root of unity. Multiply each such root by -i to get the three solutions.

[@mcwulf25]

Nice

[@aplusbi]

Very nice!

[@aplusbi]

Thanks

[@bobbyheffley4955]

I used the third method.

[@JefiKnight]

It is always nice to double check Euler's work and make sure it still works! :)

[@key_board_x]

z³ = i ← this is a complex number The modulus of z³ is: m = √[(0)² + (1)²] = √1 = 1 → the modulus of z is: m^(1/3) → (1)^(1/3) = 1 The argument of z³ is: β = π/2 → the argument of z is: β/3 → (π/2)/3 = π/6 So you can see that the first root of z³ is: z1 = 1.[cos(π/6) + i.sin(π/6)] → to get the second root, you add 2π/3 z2 = 1.[cos{(π/6) + (2π/3)} + i.sin{(π/6) + (2π/3)}] → and to get the third one, you add 2π/3 once again z3 = 1.[cos{(π/6) + (2π/3) + (2π/3)} + i.sin{(π/6) + (2π/3) + (2π/3)}] It gives: z1 = cos(π/6) + i.sin(π/6) z2 = cos(5π/6) + i.sin(5π/6) z3 = cos(9π/6) + i.sin(9π/6) z3 = cos[(6π + 3π)/6] + i.sin[(6π + 3π)/6] z3 = cos[π + (3π/6)] + i.sin[π + (3π/6)] z3 = cos[π + (π/2)] + i.sin[π + (π/2)] z3 = - cos(π/2) - i.sin(π/2) After simplification: z1 = [(√3)/2] + i.(1/2) z2 = - [(√3)/2] + i.(1/2) z3 = - i To go further z1 = [(√3)/2].(1 + i) z2 = [(√3)/2].(- 1 + i) z3 = - i

[@mcwulf25]

Straight to method 3 for me except I put z = e^it Then 3it = i.(1 + 4n).pi/2 t = (1 + 4n).pi/6 Then un-Euler it to get the three solutions for z.

[@scottleung9587]

I used the third method.

[@trojanleo123]

Me too.