The Powers of $i$

In what follows, it is useful to keep in mind the powers of the imaginary unit $i$:

$\begin {aligned}i^1&=i \\i^2&=-1 \\i^3&=-i \\i^4&=1 \end {aligned}$ 

After that, they repeat, since $i^5=i^4\cdot i = i$. 

Computing Powers of Complex Numbers

Powers of complex numbers can be computed with the the help of the binomial theorem. Recall the binomial theorem, which tells how to compute powers of a binomial like $x+y$. It says:

 ${ (x+y) }^{ n }=\sum _{ k=0 }^{ n }{ \begin{pmatrix} n \\ k \end{pmatrix} } { x }^{ n-k }{ y }^{ k }=\sum _{ k=0 }^{ n }{ \begin{pmatrix} n \\ k \end{pmatrix} } { x }^{ k }{ y }^{ n-k }$

For example, consider the case $n=4.$ We have:

 $(x+y)^4 = x^4 +4x^3y+6x^2y^2 + 4xy^3+y^4$

We can use this to compute the fourth power of a complex number $a+bi$ by letting $x=a$ and $y=bi$. Then we have:

 $(a+bi)^4=a^4+4a^3bi+6a^2b^2i^2+4ab^3i^3+b^4i^4$

Now recalling the powers of $i$, we have: 

$(a+bi)^4 = a^4+4a^3bi-6a^2b^2+-4ab^3i+b^4$ 

If we gather the real terms and the imaginary terms, we have the complex number:

 $(a^4-6a^2b^2+b^4)+(4a^3b-4ab^3)i$

Example 1

Suppose you wanted to compute $(2+3i)^4$. Using the previous example as a guide, we have:

$(2+3i)^4=2^4+4\cdot 2^3\cdot 3i -6\cdot 2^2\cdot 3^2 -4\cdot 2 \cdot 3^3 i +3^4 $ 

which can be written as: 

$16 + 96i-216-216i+81 = -119-120i$ 

Example 2

Suppose you wanted to compute $(1+i)^3$. Using the binomial theorem directly, this can be written as:

$1^3+3\cdot 1^2 \cdot i + 3\cdot 1 \cdot i^2+i^3$ 

which can be simplified to: 

$1+3i-3-i=-2+2i$

Example 3

Suppose you wanted to compute $(2+i)^5$. Recall that the binomial coefficients (from the 5th row of Pascal's triangle) are $1, 5, 10, 10, 5, \text{and}\, 1.$ Using the binomial theorem directly, we have:

 $(2+i)^5 =2^5 + 5\cdot 2^4 i + 10\cdot 2^3 i^2 + 10\cdot 2^2 i^3 + 5\cdot 2 \cdot i^4 + i^5$

 This can in turn be written as:

 $32+80i-80-40i+10+i = -38+41i$