# 2.4 Complex numbers  (Page 3/8)

 Page 3 / 8

## Multiplying a complex number by a complex number

Multiply: $\text{\hspace{0.17em}}\left(4+3i\right)\left(2-5i\right).$

$\begin{array}{ccc}\hfill \left(4+3i\right)\left(2-5i\right)& =& 4\left(2\right)-4\left(5i\right)+3i\left(2\right)-\left(3i\right)\left(5i\right)\hfill \\ & =& 8-20i+6i-15\left({i}^{2}\right)\hfill \\ & =& \left(8+15\right)+\left(-20+6\right)i\hfill \\ & =& 23-14i\hfill \end{array}$

Multiply: $\text{\hspace{0.17em}}\left(3-4i\right)\left(2+3i\right).$

$18+i$

## Dividing complex numbers

Dividing two complex numbers is more complicated than adding, subtracting, or multiplying because we cannot divide by an imaginary number, meaning that any fraction must have a real-number denominator to write the answer in standard form $\text{\hspace{0.17em}}a+bi.\text{\hspace{0.17em}}$ 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 $\text{\hspace{0.17em}}a+bi\text{\hspace{0.17em}}$ is $\text{\hspace{0.17em}}a-bi.\text{\hspace{0.17em}}$ For example, the product of $\text{\hspace{0.17em}}a+bi\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}a-bi\text{\hspace{0.17em}}$ is

$\begin{array}{ccc}\hfill \left(a+bi\right)\left(a-bi\right)& =& {a}^{2}-abi+abi-{b}^{2}{i}^{2}\\ & =& {a}^{2}+{b}^{2}\hfill \end{array}$

The result is a real number.

Note that complex conjugates have an opposite relationship: The complex conjugate of $\text{\hspace{0.17em}}a+bi\text{\hspace{0.17em}}$ is $\text{\hspace{0.17em}}a-bi,$ and the complex conjugate of $\text{\hspace{0.17em}}a-bi\text{\hspace{0.17em}}$ is $\text{\hspace{0.17em}}a+bi.\text{\hspace{0.17em}}$ 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 $\text{\hspace{0.17em}}c+di\text{\hspace{0.17em}}$ by $\text{\hspace{0.17em}}a+bi,$ where neither $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ nor $\text{\hspace{0.17em}}b\text{\hspace{0.17em}}$ 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.

$\frac{\left(c+di\right)}{\left(a+bi\right)}\cdot \frac{\left(a-bi\right)}{\left(a-bi\right)}=\frac{\left(c+di\right)\left(a-bi\right)}{\left(a+bi\right)\left(a-bi\right)}$

Apply the distributive property.

$=\frac{ca-cbi+adi-bd{i}^{2}}{{a}^{2}-abi+abi-{b}^{2}{i}^{2}}$

Simplify, remembering that $\text{\hspace{0.17em}}{i}^{2}=-1.$

$\begin{array}{l}=\frac{ca-cbi+adi-bd\left(-1\right)}{{a}^{2}-abi+abi-{b}^{2}\left(-1\right)}\hfill \\ =\frac{\left(ca+bd\right)+\left(ad-cb\right)i}{{a}^{2}+{b}^{2}}\hfill \end{array}$

## The complex conjugate

The complex conjugate    of a complex number $\text{\hspace{0.17em}}a+bi\text{\hspace{0.17em}}$ is $\text{\hspace{0.17em}}a-bi.\text{\hspace{0.17em}}$ It is found by changing the sign of the imaginary part of the complex number. The real part of the number is left unchanged.

• When a complex number is multiplied by its complex conjugate, the result is a real number.
• When a complex number is added to its complex conjugate, the result is a real number.

## Finding complex conjugates

Find the complex conjugate of each number.

1. $2+i\sqrt{5}$
2. $-\frac{1}{2}i$
1. The number is already in the form $\text{\hspace{0.17em}}a+bi.\text{\hspace{0.17em}}$ The complex conjugate is $\text{\hspace{0.17em}}a-bi,$ or $\text{\hspace{0.17em}}2-i\sqrt{5}.$
2. We can rewrite this number in the form $\text{\hspace{0.17em}}a+bi\text{\hspace{0.17em}}$ as $\text{\hspace{0.17em}}0-\frac{1}{2}i.\text{\hspace{0.17em}}$ The complex conjugate is $\text{\hspace{0.17em}}a-bi,$ or $\text{\hspace{0.17em}}0+\frac{1}{2}i.$ This can be written simply as $\text{\hspace{0.17em}}\frac{1}{2}i.$

Find the complex conjugate of $\text{\hspace{0.17em}}-3+4i.$

$-3-4i$

Given two complex numbers, divide one by the other.

1. Write the division problem as a fraction.
2. Determine the complex conjugate of the denominator.
3. Multiply the numerator and denominator of the fraction by the complex conjugate of the denominator.
4. Simplify.

## Dividing complex numbers

Divide: $\text{\hspace{0.17em}}\left(2+5i\right)\text{\hspace{0.17em}}$ by $\text{\hspace{0.17em}}\left(4-i\right).$

We begin by writing the problem as a fraction.

$\frac{\left(2+5i\right)}{\left(4-i\right)}$

Then we multiply the numerator and denominator by the complex conjugate of the denominator.

$\frac{\left(2+5i\right)}{\left(4-i\right)}\text{\hspace{0.17em}}\cdot \frac{\left(4+i\right)}{\left(4+i\right)}$

To multiply two complex numbers, we expand the product as we would with polynomials (using FOIL).

Note that this expresses the quotient in standard form.

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