How do I find the #n#th power of a complex number?

1 Answer
Dec 23, 2014

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)