# How to Expand a Binomial that Contains Complex Numbers

The most complicated type of binomial expansion involves the complex number *i,* because you're not only dealing with the binomial theorem but dealing with imaginary numbers as well. When raising complex numbers to a power, note that *i*^{1}* = i,* *i*^{2} = –1, *i*^{3} = *–i,* and *i*^{4} = 1. If you run into higher powers, this pattern repeats: *i*^{5} = *i*, *i*^{6} = –1, *i*^{7} = –*i,* and so on. Because powers of the imaginary number *i* can be simplified, your final answer to the expansion should not include powers of *i.* Instead, use the information given here to simplify the powers of *i* and then combine your like terms.

For example, to expand (1 + 2*i*)^{8}, follow these steps:

Write out the binomial expansion by using the binomial theorem, substituting in for the variables where necessary.

In case you forgot, here is the binomial theorem:

Using the theorem, (1 + 2

*i*)^{8}expands toFind the binomial coefficients.

To do this, you use the formula for binomial expansion, which is written in the following form:

You may recall the term

*factorial*from your earlier math classes. If not, here is a reminder:*n*!, which reads as "n factorial," is defined asUsing the combination formula gives you the following:

Replace all

with the coefficients from Step 2.

1(1)

^{8}(2*i*)^{0}+ 8(1)^{7}(2*i*)^{1}+ 28(1)^{6}(2*i*)^{2}+ 56(1)^{5}(2*i*)^{3}+ 70(1)^{4}(2*i*)^{4}+ 56(1)^{3}(2*i*)^{5}+ 28(1)^{2}(2*i*)^{6}+ 8(1)^{1}(2*i*)^{7}+ 1(1)^{0}(2*i*)^{8}Raise the monomials to the powers specified for each term.

1(1)(1) + 8(1)(2i) + 28(1)(4i2) + 56(1)(8i3) + 70(1)(16i4) + 56(1)(32i5) + 28(1)(64i6) + 8(1)(128i7) + 1(1)(256i8)

Simplify any

*i*'s that you can.1(1)(1) + 8(1)(2i) + 28(1)(4)(–1) + 56(1)(8)(–i) + 70(1)(16)(1) + 56(1)(32)(i) + 28(1)(64)(–1) + 8(1)(128)(–i) + 1(1)(256)(1)

Combine like terms and simplify.

1 + 16i – 112 – 448i + 1,120 + 1,792i – 1,792 – 1,024i + 256

= –527 + 336i