Powers of Complex Numbers

Key Questions

  • Given a complex number of form #a + bi#,it can be proved that any power of it will be of the form #c + di#.
    For example, #(a+bi)^2 = (a^2-b^2) + 2abi#

    Knowing that, its less scary to try and find bigger powers, such as a cubic or fourth.
    Whatsoever, any negative power of a complex number will look like this:
    #(a+bi)^-n = 1/(a+bi)^n = 1/(c+di)#
    This final form is not acceptable, as it has a division by #i#, but we can use a factoring method to make it better.
    #(m+n)(m-n) = m^2-n^2# #-># #(m+ni)*(m-ni) = m^2+n^2#

    #1/(c+di)*((c-di))/((c-di)) = (c-di)/(c^2+d^2)#//

  • You could use the complex number in rectangular form (#z=a+bi#) and multiply it #n^(th)# times by itself but this is not very practical in particular if #n>2#.
    What you can do, instead, is to convert your complex number in POLAR form: #z=r angle theta# where #r# is the modulus and #theta# is the argument.
    Graphically:
    enter image source here
    so that now the #n^(th)# power becomes:

    #z^n=r^n angle n*theta#

    Let's look at an example:
    Suppose you want to evaluate #z^4# where #z=4+3i#
    Using this notation you should evaluate: #(4+3i)^4# which is difficult and...well...boring!
    But if you change it in polar form you get:
    enter image source here

    Your number in polar form becomes: #z=5 angle 37°# and:
    #z^4=5^4 angle (4*37°)=625 angle 148°#

    You can now wonder what is the rectangular form of your result.
    We get there using the trigonometric form and do some math.
    Looking at your #1^(st)# graph you can see that:
    #a=r*cos(theta)#
    #b=r*sin(theta)#

    your complex number becomes now:
    #z=a+bi=r*cos(theta)+r*sin(theta)*i#
    That gives you:
    #z=-530+331i#

    (I rounded a little bit to make it clearer)

  • We can find the value of a power of i by using #i^2=-1#.

    Let us look at some examples.

    #i^5=i^2cdoti^2cdot i=(-1)cdot(-1)cdoti=i#

    #i^6=i^2cdoti^2cdoti^2=(-1)cdot(-1)cdot(-1)=-1#

    I hope that this was helpful.

Questions