Algebra
Textbooks
Boundless Algebra
Complex Numbers and Polar Coordinates
Complex Numbers
Algebra Textbooks Boundless Algebra Complex Numbers and Polar Coordinates Complex Numbers
Algebra Textbooks Boundless Algebra Complex Numbers and Polar Coordinates
Algebra Textbooks Boundless Algebra
Algebra Textbooks
Algebra
Concept Version 4
Created by Boundless

Complex Numbers and the Binomial Theorem

Powers of complex numbers can be computed with the the help of the binomial theorem.

Learning Objective

  • Connect complex numbers raised to a power to the binomial theorem


Key Points

    • The binomial theorem can be used to compute powers of complex numbers. To compute $(a+bi)^n$ we consider the expression $(x+y)^n $ where $x=a$ and $y=bi$. 
    • The powers of $i$ are $i^2=-1$, $i^3=-i$, $i^4=1$, and $i^5=i$, etc. 

Full Text

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:

 $32+80i-80-40i+10+i = -38+41i$

[ edit ]
Edit this content
Prev Concept
Multiplication of Complex Numbers
Complex Conjugates
Next Concept
Subjects
  • Accounting
  • Algebra
  • Art History
  • Biology
  • Business
  • Calculus
  • Chemistry
  • Communications
  • Economics
  • Finance
  • Management
  • Marketing
  • Microbiology
  • Physics
  • Physiology
  • Political Science
  • Psychology
  • Sociology
  • Statistics
  • U.S. History
  • World History
  • Writing

Except where noted, content and user contributions on this site are licensed under CC BY-SA 4.0 with attribution required.