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