12.6. Part 2. FINDING THE N-TH ROOTS OF A COMPLEX NUMBER
We want to find the solutions of the equation wn = z , where z is a complex number of the
form z = r (cos θ + j sin θ ) and n is fixed having one of the values = 2, 3, 4, 5, 6….
Assume solution of the form w = ρ (cos α + j sin α ) ⇒ wn = ρ (cos nα + j sin nα )
The equation wn = z becomes:
ρ n (cos nα + j sin nα ) = r (cos θ + j sin θ ) .Remember that two complex number are equal
if their real parts are equal on one hand and their imaginary parts are equal on the other
hand. That means: ρ n cos nα = w cos θ and ρ n j sin nα = j sin θ i.e.
cos nα = cos θ
ρ n = r and 
sin nα = sin θ
⇒ ρ = n r and nα = θ + k ⋅ 360 or
α=
θ + k ⋅ 360
n
with k = 0, ± 1, ± 2, ± 3, ± 4, ± 5,
For k=0 ⇒ α 0 =
For k= 1 ⇒ α1 =
θ
n
⇒ w1 = n r (cos α 0 + j sin α 0 ) that’s the first solution.
θ + 1 ⋅ 360
⇒ w2 = n r (cos α1 + j sin α1 ) that’s the second solution.
n
θ + 2 ⋅ 360
For k=2 ⇒ α 2 =
⇒ w3 = n r (cos α 2 + j sin α 2 ) that’s the third solution.
n
θ + 3 ⋅ 360
For k=3 ⇒ α 3 =
⇒ w4 = n r (cos α 3 + j sin α 3 ) that’s the fourth solution.
n
θ + (n − 1) ⋅ 360
For k= n − 1 ⇒ α n −1 =
⇒ wn = n r (cos α n −1 + j sin α n −1 ) that’s n-th
n
solution.
Note
θ + n ⋅ 360 θ n ⋅ 360 θ
For k= n ⇒ α n =
= +
= + 360 ⇒ wn +1 = n r (cos α n + j sin α n )
n
n
n
n
that’s (n+1)-th solution? No! It is the same as the one for k = 0 because
θ
θ
θ
θ
cos( + 360 ) = cos
, sin( + 360 ) = sin and therefore taking k > n does not yield
n
n
n
n
new solutions.
SUMMARY
The n solutions of the equation wn = r (cos θ + j sin θ ) are:
wn = n r (cos α + j sin α ) whith α =
θ + k ⋅ 360
n
and k = 0, 1, 2, 3, ..., (n − 1)