Powers of Complex Numbers
Key Questions
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Given a complex number of form
a + bi ,it can be proved that any power of it will be of the formc + 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 byi , 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) //Bernardo Russo G. Ferreira · · Oct 2 2014
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You could use the complex number in rectangular form (
z=a+bi ) and multiply itn^(th) times by itself but this is not very practical in particular ifn>2 .
What you can do, instead, is to convert your complex number in POLAR form:z=r angle theta wherer is the modulus andtheta is the argument.
Graphically:
so that now then^(th) power becomes:z^n=r^n angle n*theta Let's look at an example:
Suppose you want to evaluatez^4 wherez=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:
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 your1^(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)
Gió
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Dec 23 2014
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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.
Wataru
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Sep 26 2014