Hey, everyone. Now that we know what complex numbers are, we can perform operations on them like addition and subtraction. Now I know you may be thinking that this is just yet another new skill you're going to have to learn, but adding and subtracting complex numbers is actually exactly the same as adding and subtracting algebraic expressions. So we're just going to take a skill that we already know and apply it to our complex numbers. Let's go ahead and jump in.
So just like with algebraic expressions, when you add and subtract complex numbers, you're just going to combine like terms. So when I'm given algebraic expressions to add, let's say 4 + 8x plus 2 + 3x, then I just need to remove my parentheses and then combine any like terms I have. So here I have 4 and 2, both constants that combine to 6, and then I have 8x and 3x, both x terms that combine to 11x. So when I'm working with algebraic expressions, I have some constants and some terms with variables that get combined to give me my final solution.
When I'm working with complex numbers, I still have constants, but instead of terms with a variable, I have terms with an i, my imaginary unit. So I'm really just going to treat that i as though it is a variable. So let's look at adding these complex numbers. So here I have \(4+8i+2+3i\). So let's go ahead and get rid of all of those parentheses. So I have \(4 + 8i + 2 + 3i\). And now I can just combine like terms. So I have constants \(4\) and \(2\), which are going to combine to give me \(6\). And then I have my imaginary terms, \(8i\) and \(+3i\), which are going to combine to give me \(+11i\).
Now one thing that we need to consider whenever we're adding or subtracting complex numbers is that we always want to express our final answer in standard form. Now remember, standard form for complex numbers is just \(a+bi\). So I got \(6 + 11i\) here, which is already in that standard form. So my final answer is \(6 + 11i\).
Let's take a look at subtracting some complex numbers. So, again, I have \(4 + 8i\), but now minus \(2 + 3i\). So let's go ahead and remove those parentheses. So I have \(4 + 8i\). But now since I'm subtracting, I need to make sure that I distribute that negative into my parentheses here. So I'm left with a minus \(2\) now and a minus \(3i\). Now I can simply combine like terms as I did before. So, again, my constants, \(4\) and then negative \(2\), these will combine to give me a positive \(2\). And then I have \(8i\) and negative \(3i\). So \(8i\) and negative \(3i\) are gonna combine to give me a positive \(5i\).
Now, again, we want to make sure that our answer is in standard form here, \(a + bi\), and it is. So my final answer is \(2 + 5i\). And that's all there is to adding and subtracting complex numbers. I'll see you in the next video.