# nth roots of a complex number

This is a complex number nth roots Calculator. Example, to calculate cubic roots of z, enter n = 3.

root(n)(z) = 1/3*sqrt(21)*cos(1/3*pi - 1/3*arctan(10/27*sqrt(3))) + 1/3*i*sqrt(21)*sin(1/3*pi - 1/3*arctan(10/27*sqrt(3))) \approx 1 + 1.15470053837925*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 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
Allowed: constants, operators and i. To multiply use a*b not ab

## 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.