Operations with Complex Numbers

Learning Objectives

• Add complex numbers.
• Subtract complex numbers.
• Multiply complex numbers.
• Find conjugates of complex numbers.
• Divide complex numbers.

Any time new kinds of numbers are introduced, one of the first questions that needs to be addressed is, “How do you add them?” In this topic, you’ll learn how to add complex numbers and also how to subtract, multiply, and divide them.

First, consider the following expression.

`(6x + 8) + (4x + 2)`

To simplify this expression, you combine the like terms, `6x` and `4x`. These are like terms because they have the same variable with the same exponents. Similarly, `8` and `2` are like terms because they are both constants, with no variables.

`(6x + 8) + (4x + 2) = 10x + 10`

In the same way, you can simplify expressions with radicals.

`(6sqrt3+8)+(4sqrt3+2)=10sqrt3+10`

You can add `6sqrt3` to `4sqrt3` because the two terms have the same radical, `sqrt3`, just as `6x` and `4x` have the same variable and exponent.

The number `i` looks like a variable, but remember that it is equal to `sqrt(-1)`. The great thing is you have no new rules to worry about. Whether you treat it as a variable or a radical, the exact same rules apply to adding and subtracting complex numbers A number in the form `a + bi`, where `a` and `b` are real numbers and `i` is the square root of `-1`. . You combine the imaginary parts (the terms with `i`), and you combine the real parts.

Again, consider the following expression. Before reading further, consider how you would simplify it.

`(5x)(-3x)`

You can simplify by multiplying the coefficients together, then the variables.

Multiplying two imaginary (but not complex!) numbers together works in a similar way, but there is an additional step. Start with the same method to multiply `5i` and `-3i`.

This seems fine so far, but the `i^2` can be simplified further.

When you multiply a square root by itself, you get the number under the radical. This is what square root means.

`(sqrt3)(sqrt3)=3`

`(sqrt15)(sqrt15)=15`

Well, `i` is also a square root. It’s equal to `sqrt(-1)`.

So, the final step to simplifying `(5i)(-3i) = -15i^2` is to replace `i^2` with `-1`.

Notice that the product of two imaginary numbers is a real number! We will see this again when we multiply two complex numbers.

The following expression is a little more complicated because two binomials are being multiplied. This means you have to use the Distributive Property of MultiplicationThe product of a sum (or a difference) and a number is the same as the sum (or difference) of the product of each addend (or each number being subtracted) and the number. For example: `3(4 + 2) = 3(4) + 3(2)` and `3(4 - 2) = 3(4) - 3(2)`.. (Remember that multiplying using the FOIL method (First, Outside, Inside, Last) is an application of the Distributive Property of Multiplication.) Once the binomials have been multiplied, simplify the expression by combining like terms.

Again, in the same way, you can multiply complex numbers. At the end, you will need to simplify `i^2`.

In this case, the product of two complex numbers is complex. But in the following example, the product is real, not complex. Can you figure out why?

Just as `6+8sqrt3` and `6-8sqrt3` are conjugates, `6 + 8i` and `6 - 8i` are conjugates. (Again, `i` is a square root, so this isn’t really a new idea.) When the numbers are complex, they are called complex conjugates Two complex numbers for which the real parts are equal and the imaginary parts are additive inverses. `a + bi` and `a - bi` are complex conjugates. . Because conjugates have terms that are the same except for the operation between them (one is addition and one is subtraction), the `i` terms in the product will add to `0`. In the example above, `-48i` was added to `48i`, and that sum is `0`, so there was no `i` term in the final product. That means the product of complex conjugates will always be a real (not complex) number.

So far, each operation with complex numbers has worked just like the same operation with radical expressions. This should no longer be a surprise. The number `i` is a radical, after all, so complex numbers are radical expressions!

Let’s look at division in two parts, like we did multiplication. First, let’s look at a situation in which the divisor is a monomial.

When the divisor (that is, the denominator in the fraction) is a complex number with non-zero real and imaginary parts, you must rationalize the denominator using the complex conjugate. Remember that the product of a complex number with its complex conjugate is always a real number, so the denominator will be a real number. That means the result will be equivalent, but rationalized.

Complex numbers can be added, subtracted, multiplied, and divided using the same ideas you used for radicals and variables. With multiplication and division, you may need to replace `i^2` with `-1` and simplify further.