# Video: Finding the Product of Two Complex Numbers in Algebraic Form

Multiply (−3 + 𝑖) by (2 + 5𝑖).

02:01

### Video Transcript

Multiply negative three plus 𝑖 by two plus five 𝑖.

There are a couple of different things we need to recall here. Firstly, we see that we have two complex numbers of the form 𝑎 plus 𝑏𝑖. Remember, 𝑎 is the real component of the complex number and 𝑏 is its imaginary component. We also recall that 𝑖 is the solution to the equation 𝑥 squared equals negative one such that 𝑖 is often written as the square root of negative one.

We’re looking to find the product of negative three plus 𝑖 and two plus five 𝑖. And when we write it like this, we spot that this looks a lot like the product of two binomials. In fact, we deal with this in much the same way. We distribute our parentheses using either the FOIL or grid method. Let’s use the FOIL method.

Remember, “F” stands for “first.” We multiply the first term in the first bracket by the first term in the second. Negative three times two is negative six. “O” stands for “outer.” We multiply the outer terms in our expressions. Negative three times five is negative 15. So negative three times five 𝑖 is negative 15𝑖. “I” then stands for “inner.” We multiply the inner terms. That gives us two 𝑖. And then, finally, “L” stands for “last.” We multiply the last term in the first bracket by the last term in the second. And that gives us five 𝑖 squared.

Now we’re not quite done here. We can simplify this further. Firstly, we spot that we can collect some like terms, negative 15𝑖 and two 𝑖. That gives us negative six minus 13𝑖 plus five 𝑖 squared. But then we recall that 𝑖 is equal to the square root of negative one. This must mean that we can say that 𝑖 squared is equal to negative one. We therefore replace 𝑖 squared with negative one. And we see that the product of these two complex numbers is negative six minus 13𝑖 minus five, which is negative 11 minus 13𝑖. The product of the two complex numbers negative three plus 𝑖 and two plus five 𝑖 is negative 11 minus 13𝑖.