# Computations with Complex Numbers

## Task

Rewrite each of the following expressions involving complex numbers in the form $a+bi$ where $a$ and $b$ are real numbers.

- $$ (3 + 2i)(2 - 5i) $$
- $$ (5 + 4i)(17 - 13i) - (5 + 3i)(17 - 13i) $$
- $$ \left(\frac52 + \frac{7i}{2} \right)^2 - \left(\frac52 + \frac{i}{2} \right)^2 $$
- $$ (1 + i)(13 - 4i)(1 - i) $$
- $$ 1 + i + i^2 + i^3 $$

## IM Commentary

This task asks students to perform computations involving complex numbers using the properties of operations and the fact that $i^2 = -1$. Students can complete the task provided that they know this fact and can apply it to find that $i^3 = -i$. However, students who pay attention to the structure of each expression and use properties of operations (MP.7) will be able to avoid or shorten some tedious calculations.

A teacher who uses this problem as a classroom task should keep track of students' solution approaches and ask students to present different solutions to the same problem.

## Solution

- We have \begin{eqnarray*} (3 + 2i)(2 - 5i) & = & 3(2 - 5i) + 2i(2 - 5i) \\ & = & 6 - 15i + 4i - 10i^2 \\ & = & 6 - 15i + 4i + 10 \\ & = & 16 - 11i. \\ \end{eqnarray*} Note that we have used the fact that $i^2 = -1$ to write $-10i^2 = 10$.
- We can evaluate this expression by computing each product of binomials separately and then subtracting. However, we can make our job easier by noticing that each product contains a factor (17 - 13i), and factoring this out to obtain \begin{eqnarray*} (5 + 4i)(17 - 13i) - (5 + 3i)(17 - 13i) & = & ((5 + 4i) - (5 + 3i))(17 - 13i) \\ & = & i(17 - 13i) \\ & = & 17i - 13i^2 \\ & = & 13 + 17i. \\ \end{eqnarray*}
- Again, it is possible to square each binomial separately and then subtract, but since we have an expression of the form $x^2 - y^2$, let's try factoring the given expression as a difference of squares: \begin{eqnarray*} \left(\frac52 + \frac{7i}{2} \right)^2 - \left(\frac52 + \frac{i}{2} \right)^2 & = & \left(\left( \frac52 + \frac{7i}{2} \right) - \left( \frac52 + \frac{i}{2} \right) \right) \left( \left( \frac52 + \frac{7i}{2} \right) + \left( \frac52 + \frac{i}{2} \right) \right) \\ & = & 3i(5 + 4i) \\ & = & 15i + 12i^2 \\ & = & -12 + 15i \\ \end{eqnarray*}
- By the commutative and associative properties, we can pair and multiply the factors of this expression in whatever order we want. Since we see two complex conjugates - namely, (1 + i) and (1 - i) - we will try pairing and multiplying those first: \begin{eqnarray*} (1 + i)(13 - 4i)(1 - i) & = & (1 + i)(1 - i)(13 - 4i) \\ & = & (1 - i^2)(13 - 4i) \\ & = & 2(13 - 4i) \\ & = & 26 - 8i. \\ \end{eqnarray*}
- We know that $i^2 = -1$, and $i^3 = i^2 \cdot i = -i$. So $$ 1 + i + i^2 + i^3 = 1 + i + (-1) + (-i) = 0. $$

## Computations with Complex Numbers

Rewrite each of the following expressions involving complex numbers in the form $a+bi$ where $a$ and $b$ are real numbers.

- $$ (3 + 2i)(2 - 5i) $$
- $$ (5 + 4i)(17 - 13i) - (5 + 3i)(17 - 13i) $$
- $$ \left(\frac52 + \frac{7i}{2} \right)^2 - \left(\frac52 + \frac{i}{2} \right)^2 $$
- $$ (1 + i)(13 - 4i)(1 - i) $$
- $$ 1 + i + i^2 + i^3 $$