find the cube roots of the complex number.
27((cos(11pi/6)) + i(sin(11pi/6))
3x^2+5x^4+6
To find the cube roots of a complex number, we can use the formula for the cube roots of a complex number in polar form. Let's break down the given complex number into its polar form:
We have the complex number 27(cos(11π/6) + i(sin(11π/6)).
To convert this to polar form, we can identify the magnitude (r) and the argument (θ). The magnitude (r) is obtained by taking the square root of the sum of the squares of the real and imaginary parts:
r = sqrt[(cos(11π/6))^2 + (sin(11π/6))^2] = 27
The argument (θ) can be calculated using the inverse tangent function:
θ = atan2(sin(11π/6), cos(11π/6)) = -π/6
So, the complex number 27(cos(11π/6) + i(sin(11π/6)) can be represented in polar form as 27cis(-π/6).
Now, to find the cube roots, we need to take the cube root of the magnitude (r) and divide the argument (θ) by 3.
Taking the cube root of the magnitude: cbrt(27) = 3
Dividing the argument by 3: (-π/6) ÷ 3 = -π/18
Therefore, the cube roots of the complex number 27(cos(11π/6) + i(sin(11π/6)) are:
1) 3cis(-π/18)
2) 3cis(8π/9)
3) 3cis(-29π/18)
Note: "cis" is short for "cos + i*sin" and represents complex numbers in polar form.