A tricky Mathematics Question from Oxford University Admission Exam

[@key_board_x]

x³ = i ← this is a complex number
The modulus of x³ is: m = √(0² + 1²) = 1, so the modulus of x is: m^(1/3) = 1^(1/3) = 1
The argument of x³ is: β = π/2, so the argument of x is (β/3) = (π/2)/3 = π/6
…therefore the first root of x³ is:
x1 = 1.[cos(π/6) + i.sin(π/6)] → then we add (2π/3) to get the second root
x2 = 1.[cos{(π/6) + (2π/3)} + i.sin{(π/6) + (2π/3)}] → and we add (2π/3) more to get the third root
x3 = 1.[cos{(π/6) + (2π/3) + (2π/3)} + i.sin{(π/6) + (2π/3) + (2π/3)}]
It gives
x1 = cos(π/6) + i.sin(π/6)
x1 = (√3)/2 + i.(1/2)
x1 = [(√3)/2].(1 + i)
x2 = cos(5π/6) + i.sin(5π/6) → recall: 5π/6 = (6π - π)/6 = (6π/6) - (π/6) = π - (π/6)
x2 = cos[π - (π/6)] + i.sin[π - (π/6)]
x2 = - cos(π/6) + i.sin(π/6)
x2 = - (√3)/2 + i.(1/2)
x2 = [(√3)/2].(- 1 + i)
x3 = cos(9π/6) + i.sin(9π/6)
x3 = cos(3π/2) + i.sin(3π/2)
x3 = 0 + i.(- 1)
x3 = - i
Summarize
x1 = [(√3)/2].(1 + i)
x2 = [(√3)/2].(- 1 + i)
x3 = - i

[@NichaelCramer]

Very nice. Thank you.

[@superacademy247]

Welcome 😊

[@ssalmero]

A better way is to use the polar form: x^3= e^î(pi/2+2npi), so x=e^i(pi/6+2npi/3), for n=0, 1, 2 are the three solutions of the equation.

[@RyanLewis-Johnson-wq6xs]

(-i)^3=i X=(i±Sqrt[3])/2= 0.5Sqrt[3]±0.5i X=-i

[@radupopescu9977]

(i)^(1/3) has 3 values. This values are those of x^3.

[@viggosimonsen]

Use polar coordinates - presto!

[@payoo_2674]

x^3=i
x^3=e^((1/2+2n)πi)=e^((1+4n)πi/2) n∈Z
x=(e^((1+4n)πi/2))^(1/3)
x=e^((1+4n)πi/6)
for n=0: x₁ = e^(πi/6) = cos(π/6)+i∙sin(π/6)) = √3/2+(1/2)i = (√3+i)/2
for n=1: x₂ = e^(5πi/6) = cos(5π/6)+i∙sin(5π/6)) = -√3/2+(1/2)i = (-√3+i)/2
for n=2: x₃ = e^(9πi/6) = cos(3π/2)+i∙sin(3π/2)) = 0+(-1)i = -i
For the next n solutions are repeated.

[@payoo_2674]

x³=i
Let x=a+bi {a,b∈R}
(a+bi)³=i
a³+3a²bi+3ab²i²+b³i³=i
a³-3ab²+(3a²b-b³)i=0+i → a³-3ab²=0 ① and 3a²b-b³=1 ②
① → a(a²-3b²)=0 → a(a-b√3)(a+b√3)=0 → a=0 or a=b√3 or a=-b√3
if a=0, than from ② -b³=1 → b=-1 → x₁=0+(-1)i=-i
if a=b√3, than from ② 9b³-b³=1 → 8b³=1 → b³=1/8 → b=1/2 → a=√3/2 → x₂=√3/2+(1/2)i=(√3+i)/2
if a=-b√3, than from ② 9b³-b³=1 → 8b³=1 → b³=1/8 → b=1/2 → a=-√3/2 → x₃=-√3/2+(1/2)i=(-√3+i)/2

[@nikolayplatnov5148]

Kidding ? i= exp (i(pi/2+2 pi N)).
X= exp (i(pi/6+2/3 pi N))