Multiply and divide complex numbers

Multiplying Complex Numbers

Multiplying complex numbers is much like multiplying binomials. The major difference is that we work with the real and imaginary parts separately.

Example 4: Multiplying a Complex Number by a Real Number

Figure 5

Let’s begin by multiplying a complex number by a real number. We distribute the real number just as we would with a binomial. So, for example,

How To: Given a complex number and a real number, multiply to find the product.

1. Use the distributive property.
2. Simplify.

Example 5: Multiplying a Complex Number by a Real Number

Find the product [latex]4\left(2+5i\right)[/latex].

Solution

Distribute the 4.

[latex]\begin{cases}4\left(2+5i\right)=\left(4\cdot 2\right)+\left(4\cdot 5i\right)\hfill \\ =8+20i\hfill \end{cases}[/latex]

Try It 4

Find the product [latex]-4\left(2+6i\right)[/latex].

Solution

Multiplying Complex Numbers Together

Now, let’s multiply two complex numbers. We can use either the distributive property or the FOIL method. Recall that FOIL is an acronym for multiplying First, Outer, Inner, and Last terms together. Using either the distributive property or the FOIL method, we get

[latex]\left(a+bi\right)\left(c+di\right)=ac+adi+bci+bd{i}^{2}[/latex]

Because [latex]{i}^{2}=-1[/latex], we have

[latex]\left(a+bi\right)\left(c+di\right)=ac+adi+bci-bd[/latex]

To simplify, we combine the real parts, and we combine the imaginary parts.

[latex]\left(a+bi\right)\left(c+di\right)=\left(ac-bd\right)+\left(ad+bc\right)i[/latex]

How To: Given two complex numbers, multiply to find the product.

1. Use the distributive property or the FOIL method.
2. Simplify.

Example 6: Multiplying a Complex Number by a Complex Number

Multiply [latex]\left(4+3i\right)\left(2 - 5i\right)[/latex].

Solution

Use [latex]\left(a+bi\right)\left(c+di\right)=\left(ac-bd\right)+\left(ad+bc\right)i[/latex]

[latex]\begin{cases}\left(4+3i\right)\left(2 - 5i\right)=\left(4\cdot 2 - 3\cdot \left(-5\right)\right)+\left(4\cdot \left(-5\right)+3\cdot 2\right)i\hfill \\ \text{ }=\left(8+15\right)+\left(-20+6\right)i\hfill \\ \text{ }=23 - 14i\hfill \end{cases}[/latex]

Try It 5

Multiply [latex]\left(3 - 4i\right)\left(2+3i\right)[/latex].

Solution

Dividing Complex Numbers

Division of two complex numbers is more complicated than addition, subtraction, and multiplication because we cannot divide by an imaginary number, meaning that any fraction must have a real-number denominator. We need to find a term by which we can multiply the numerator and the denominator that will eliminate the imaginary portion of the denominator so that we end up with a real number as the denominator. This term is called the complex conjugate of the denominator, which is found by changing the sign of the imaginary part of the complex number. In other words, the complex conjugate of [latex]a+bi[/latex] is [latex]a-bi[/latex].

Note that complex conjugates have a reciprocal relationship: The complex conjugate of [latex]a+bi[/latex] is [latex]a-bi[/latex], and the complex conjugate of [latex]a-bi[/latex] is [latex]a+bi[/latex]. Further, when a quadratic equation with real coefficients has complex solutions, the solutions are always complex conjugates of one another.

Suppose we want to divide [latex]c+di[/latex] by [latex]a+bi[/latex], where neither a nor b equals zero. We first write the division as a fraction, then find the complex conjugate of the denominator, and multiply.

Multiply the numerator and denominator by the complex conjugate of the denominator.

Apply the distributive property.

Simplify, remembering that [latex]{i}^{2}=-1[/latex].

Example 9: Substituting a Complex Number into a Polynomial Function

Let [latex]f\left(x\right)={x}^{2}-5x+2[/latex]. Evaluate [latex]f\left(3+i\right)[/latex].

Solution

Figure 6

Substitute [latex]x=3+i[/latex] into the function [latex]f\left(x\right)={x}^{2}-5x+2[/latex] and simplify.

Analysis of the Solution

We write [latex]f\left(3+i\right)=-5+i[/latex]. Notice that the input is [latex]3+i[/latex] and the output is [latex]-5+i[/latex].

Simplifying Powers of i

The powers of i are cyclic. Let’s look at what happens when we raise i to increasing powers.

We can see that when we get to the fifth power of i, it is equal to the first power. As we continue to multiply i by itself for increasing powers, we will see a cycle of 4. Let’s examine the next 4 powers of i.

Licenses & Attributions