Multiplying Complex Numbers
- Let's multiply complex numbers.
13.1: $i$ Squared
Write each expression in the form \(a+bi\), where \(a\) and \(b\) are real numbers.
- \(4i \boldcdot 3i\)
- \(4i \boldcdot \text-3i\)
- \(\text- 2i \boldcdot \text- 5i\)
- \(\text- 5i \boldcdot 5i\)
13.2: Multiplying Imaginary Numbers
Take turns with your partner to match an expression in column A with an equivalent expression in column B.
- For each match that you find, explain to your partner how you know it’s a match.
- For each match that your partner finds, listen carefully to their explanation. If you disagree, discuss your thinking and work to reach an agreement.
|\(5 \boldcdot 7i\)||-9|
|\(5i \boldcdot 7i\)||\(35i\)|
|\((\text- i)^2\)||\(\text- 8i\)|
13.3: Multiplying Complex Numbers
Write each product in the form \(a + bi\), where \(a\) and \(b\) are real numbers.
- \((\text- 3 + 9i)(5i)\)
- \((8 + i)(\text- 5 + 3i)\)
- \((3 + 2i)^2\)
- \((3 + 2i)(3 - 2i)\)
On October 16, 1843, while walking across the Broom Bridge in Dublin, Ireland, Sir William Rowan Hamilton came up with an idea for numbers that would work sort of like complex numbers. Instead of just the number \(i\) (and its opposite \(\text- i\)) squaring to give -1, he imagined three numbers \(i\), \(j\), and \(k\) (each with an opposite) that squared to give -1.
The way these numbers multiplied with each other was very interesting. \(i\) times \(j\) would give \(k\), \(j\) times \(k\) would give \(i\), and \(k\) times \(i\) would give \(j\). But the multiplication he imagined did not have a commutative property. When those numbers were multiplied in the opposite order, they’d give the opposite number. So \(j\) times \(i\) would give \(\text- k\), \(k\) times \(j\) would give \(\text- i\), and \(i\) times \(k\) would give \(\text- j\). A quaternion is a number that can be written in the form \(a+bi+cj+dk\) where \(a\), \(b\), \(c\), and \(d\) are real numbers.
Let \(w=2+3i-j\) and \(z=2i+3k\). Write each given expression in the form \(a+bi+cj+dk\).
To multiply two complex numbers, we use the distributive property:
\(\displaystyle (2 + 3i )(4 + 5i) = 8 + 10i + 12i + 15i^2\)
Remember that \(i^2 = \text- 1\), so:
\(\displaystyle (2 + 3i )(4 + 5i) = 8 + 10i + 12i - 15\)
When we add the real parts together and the imaginary parts together, we get:
\(\displaystyle (2 + 3i )(4 + 5i) = \text- 7 + 22i\)
- complex number
A number in the complex plane. It can be written as \(a + bi\), where \(a\) and \(b\) are real numbers and \(i^2 = \text-1\).
- imaginary number
A number on the imaginary number line. It can be written as \(bi\), where \(b\) is a real number and \(i^2 = \text-1\).
- real number
A number on the number line.