nth roots of a complex number
Answer
` root(n)(z) = -1/6*sqrt(21)*cos(1/3*pi - 1/3*arctan(10/27*sqrt(3))) + 1/2*i*sqrt(7)*cos(1/3*pi - 1/3*arctan(10/27*sqrt(3))) + 1/6*i*sqrt(21)*sin(1/3*pi - 1/3*arctan(10/27*sqrt(3))) + 1/2*sqrt(7)*sin(1/3*pi - 1/3*arctan(10/27*sqrt(3))) \approx 0.5 + 1.44337567297406*i`
` root(n)(z) = -1/6*sqrt(21)*cos(1/3*pi - 1/3*arctan(10/27*sqrt(3))) - 1/2*i*sqrt(7)*cos(1/3*pi - 1/3*arctan(10/27*sqrt(3))) + 1/6*i*sqrt(21)*sin(1/3*pi - 1/3*arctan(10/27*sqrt(3))) - 1/2*sqrt(7)*sin(1/3*pi - 1/3*arctan(10/27*sqrt(3))) \approx -1.5 - 0.288675134594813*i`
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nth Roots of a complex number
Let z be a complex number which has the following polar form,
`z = r(cos\theta + i * sin\theta)`,
r = |z| is the modulus of z.
`\theta` is the argument of z.
To calculate the polar form, use this calculator:
calculate the polar form of a complex number
z has exactly n nth roots here denoted by `t_k` with `0 <=k<=n-1`,
`t_k = \text{}^nsqrt(r)(cos((\theta+2 \pi k)/n) + i * sin((\theta+2 \pi k)/n))`
To proove that, we use de Moivre's formula which states, for every integer n,`(cos\alpha+i*sin\alpha)^n = cos(n*\alpha) + i*sin(n*\alpha)`
We apply this formula to `t_k` :
`t_k^n = (\text{}^nsqrt(r))^n(cos(n*(\theta+2 \pi k)/n) + i * sin(n*(\theta+2 \pi k)/n))`
`t_k^n = r(cos(theta) + i * sin(\theta)) = z`
`t_k` is a n-th root of z.
See also
Polar form of a complex number
Modulus of a complex number
Argument of a complex number