Complex Numbers - Addition and Subtraction

Complex Numbers
Addition and Subtraction

Yesterday we began discussing complex numbers, and I gave you a problem to try involving raising i to a power. You can see that post here: Complex Numbers – Examples and Powers of i

Today I will go over how to add and subtract complex numbers. But first I will provide a solution to yesterday’s problem. Here is the problem one more time, followed by a solution:

Example: Compute i⁵³.

Solution: When we divide 53 be 4 we get a remainder of 1. So i⁵³= i¹= i.

Notes: (1) For a review of how to compute remainders, see the following post: Remainders

(2) This computation can also be done quickly in your calculator, but be careful. Your calculator may sometimes “disguise” the number 0 with a tiny number in scientific notation. For example, when we type i^ 53 ENTER into out TI-84, we get an output of –3E–13 + i. The expression –3E–13 represents a tiny number in scientific notation which is essentially 0. So this should be read as 0 + i = i.

Addition and Subtraction

We add two complex numbers simply by adding their real parts, and then adding their imaginary parts.

(a + bi) + (c + di) = (a + c) + (b + d)i

Example: The sum of the complex numbers 2 – 3i and –5 + 6i is:

(2 – 3i) + (–5 + 6i) = (2 – 5) + (–3 + 6)i = –3 + 3i

Subtraction is similar.

(a + bi) – (c + di) = (a – c) + (b – d)i

Example 5: When we subtract 2 – 3i from –5 + 6i we get what complex number?

I will provide a solution to this question tomorrow. Meanwhile, please post your attempted solutions in the comments below.

For those of you that prefer videos…