Roots of Complex Numbers

Key Questions

  • How do I find the nth root of a complex number?

    To evaluate the nth root of a complex number I would first convert it into trigonometric form:
    z=r[cos(theta)+isin(theta)]
    and then use the fact that:
    z^n=r^n[cos(n*theta)+isin(n*theta)]
    and:
    nsqrt(z)=z^(1/n)=r^(1/n)*[cos((theta+2kpi)/n)+isin((theta+2kpi)/n)]
    Where k=0..n-1

    For example: consider z=2+3.46i and let us try sqrt(z);
    z can be written as:
    z=4[cos(pi/3)+isin(pi/3)]
    So:
    k=0
    sqrt(z)=z^(1/2)=4^(1/2)[cos((pi/3+0)/2)+isin((pi/3+0)/2)]=
    =2[cos(pi/6)+isin(pi/6))]
    And:
    k=n-1=2-1=1
    sqrt(z)=z^(1/2)=4^(1/2)[cos((pi/3+2pi)/2)+isin((pi/3+2pi)/2)]=
    =2[cos(7pi/6)+isin(7pi/6))]
    Which gives, in total, two solutions.

    Gió · 1 · Jan 16 2015
  • How do I find the fourth root of a complex number?

    Answer:

    If you express your complex number in polar form as r(cos theta + i sin theta), then it has fourth roots:

    alpha = root(4)(r)(cos (theta/4) + i sin (theta/4)), i alpha, -alpha and - i alpha

    Explanation:

    Given a+ib, let r = sqrt(a^2+b^2), theta = "atan2"(b, a)

    Then a + ib = r (cos theta + i sin theta)

    This has one 4th root alpha = root(4)(r)(cos (theta/4) + i sin (theta/4))

    There are three other 4th roots: i alpha, -alpha and -i alpha

    George C. · 6 · Sep 2 2015
  • What are roots of unity?

    A root of unity is a complex number that when raised to some positive integer will return 1.

    It is any complex number z which satisfies the following equation:

    z^n = 1

    where n in NN, which is to say that n is a natural number. A natural number is any positive integer: (n = 1, 2, 3, ...). This is sometimes referred to as a counting number and the notation for it is NN.

    For any n, there may be multiple z values that satisfy that equation, and those values comprise the roots of unity for that n.

    When n = 1
    Roots of unity: 1

    When n = 2
    Roots of unity: -1, 1

    When n = 3
    Roots of unity = 1, (1 + sqrt(3)i)/2, (1 - sqrt(3)i)/2

    When n = 4
    Roots of unity = -1, i, 1, -i

    Ahmed E. · 3 · Nov 20 2014
  • How do I find the square root of a complex number?

    To evaluate the square root (and in general any root) of a complex number I would first convert it into trigonometric form:
    z=r[cos(theta)+isin(theta)]
    and then use the fact that:
    z^n=r^n[cos(n*theta)+isin(n*theta)]

    Where, in our case, n=1/2 (remembering that sqrt(x)=x^(1/2)).
    To evaluate the nth root of a complex number I would write:

    nsqrt(z)=z^(1/n)=r^(1/n)*[cos((theta+2kpi)/n)+isin((theta+2kpi)/n)]
    Where k=0..n-1

    For example: consider z=2+3.46i and let us try sqrt(z);
    z can be written as:
    z=4[cos(pi/3)+isin(pi/3)]
    So:
    k=0
    sqrt(z)=z^(1/2)=4^(1/2)[cos((pi/3+0)/2)+isin((pi/3+0)/2)]=
    =2[cos(pi/6)+isin(pi/6))]
    And:
    k=n-1=2-1=1
    sqrt(z)=z^(1/2)=4^(1/2)[cos((pi/3+2pi)/2)+isin((pi/3+2pi)/2)]=
    =2[cos(7pi/6)+isin(7pi/6))]
    Which gives, in total, two solutions.

    Gió · 1 · Jan 16 2015

Questions

Complex Numbers in Trigonometric Form