Powers And Roots
In this section we’re going to take a look at a really nice way of quickly computin
and roots of complex numbers.
We’ll start with integer powers of z = reiθ since they are easy enough. If n is a
z = (re ) = rn ei
n
n iθ nθ
There really isn’t too much to do with powers other than working a quick examp
Example 1 Compute (3 + 3i)5 .
Hide Solution #
Of course, we could just do this by multiplying the number out, but this would
consuming and prone to mistakes. Instead we can convert to exponential form
(1) to quickly get the answer.
Here is the exponential form of 3 + 3i .
3
r = √9 + 9 = 3√2 tan θ = ⇒ Arg z =
3
π
3 + 3i = 3√2e i4
Note that we used the principal value of the argument for the exponential form
didn’t have to.
Now, use (1) to quickly do the computation.
5 5π
(3 + 3i)5 = (3√2) ei 4
5π 5π
= 972√2 (cos( ) + i sin( ))
4 4
= 972√2 (−
√2 √2
− i)
2 2
= −972 − 972i
In this section we’re going to take a look at a really nice way of quickly computin
and roots of complex numbers.
We’ll start with integer powers of z = reiθ since they are easy enough. If n is a
z = (re ) = rn ei
n
n iθ nθ
There really isn’t too much to do with powers other than working a quick examp
Example 1 Compute (3 + 3i)5 .
Hide Solution #
Of course, we could just do this by multiplying the number out, but this would
consuming and prone to mistakes. Instead we can convert to exponential form
(1) to quickly get the answer.
Here is the exponential form of 3 + 3i .
3
r = √9 + 9 = 3√2 tan θ = ⇒ Arg z =
3
π
3 + 3i = 3√2e i4
Note that we used the principal value of the argument for the exponential form
didn’t have to.
Now, use (1) to quickly do the computation.
5 5π
(3 + 3i)5 = (3√2) ei 4
5π 5π
= 972√2 (cos( ) + i sin( ))
4 4
= 972√2 (−
√2 √2
− i)
2 2
= −972 − 972i
