16.4.2: Operations with Complex Numbers

Learning Objectives

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

Introduction

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.

Adding and Subtracting Complex Numbers

First, consider the following expression.

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

To simplify this expression, you combine the like terms, \(\ 6 x\) 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.

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

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

\(\ (6 \sqrt{3}+8)+(4 \sqrt{3}+2)=10 \sqrt{3}+10\)

You can add \(\ 6 \sqrt{3}\) to \(\ 4 \sqrt{3}\) because the two terms have the same radical, \(\ \sqrt{3}\), 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. You combine the imaginary parts (the terms with \(\ i\)), and you combine the real parts.

Example

Add. \((-3+3 i)+(7-2 i)\)

Solution

\(\ -3+3 i+7-2 i=-3+7+3 i-2 i\)	Rearrange the sums to put like terms together.
\(\ -3+7=4\) and \(\ 3 i-2 i=(3-2) i=i\)	Combine like terms.

\(\ (-3+3 i)+(7-2 i)=4+i\)

Example

Subtract. \((-3+3 i)-(7-2 i)\)

Solution

\(\ (-3+3 i)-(7-2 i)=-3+3 i-7+2 i\)	Be sure to distribute the subtraction sign to all terms in the subtrahend.
\(\ -3-7+3 i+2 i\)	Rearrange the terms to put like terms together.
\(\ -3-7=-10\) and \(\ 3 i+2 i=(3+2) i=5 i\)	Combine like terms.

\(\ (-3+3 i)-(7-2 i)=10+5 i\)

Exercise

Subtract. \(\ (5+3 i)-(3-i)\)

1. \(\ 2+4 i\)
2. \(\ 6\)
3. \(\ 2+2 i\)
4. \(\ 8+2 i\)

Answer

1. \(\ 2+4 i\)

   Correct. Distributing the subtraction to the second complex number gives \(\ 5+3 i-3+i\). Rearranging to put like terms together gives \(\ 5-3+3 i+i\), and combining like terms gives \(\ 2+4 i\).

2. \(\ 6\)

   Incorrect. You may have combined \(\ 5+3\) from the first number (ignoring the \(\ i\)) and \(\ 3-1\) from the second number (ignoring the \(\ i\)), giving that result of \(\ 8-2=6\). Instead, you should distribute the subtraction across the second complex number to get \(\ 5+3 i-3+i\). Rearranging to put like terms together gives \(\ 5-3+3 i+i\), and combining like terms gives the correct answer \(\ 2+4 i\).

3. \(\ 2+2 i\)

   Incorrect. You probably forgot to distribute the subtraction to the imaginary part of the second complex number, leaving it as \(\ -i\) instead of \(\ +i\). Distributing the subtraction to the second complex number gives \(\ 5+3 i-3+i\). Rearranging to put like terms together gives \(\ 5-3+3 i+i\), and combining like terms gives the correct answer \(\ 2+4 i\).

4. \(\ 8+2 i\)

   Incorrect. You may have added instead of subtracted. Distributing the subtraction to the second complex number gives \(\ 5+3 i-3+i\). Rearranging to put like terms together gives \(\ 5-3+3 i+i\), and combining like terms gives the correct answer \(\ 2+4 i\).

Multiplying Complex Numbers

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

\[\ (5 x)(-3 x) \nonumber \]

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

\[\ \begin{aligned}
(5 x)(-3 x) &=(5)(-3)(x)(x) \\
&=-15 x^{2}
\end{aligned} \nonumber \]

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 \(\ \text { 5i }\) and \(\ -3 i\).

\[\ \begin{aligned}
(5 i)(-3 i) &=(5)(-3)(i)(i) \\
&=-15 i^{2}
\end{aligned} \nonumber \]

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.

\[\ \begin{array}{l}
(\sqrt{3})(\sqrt{3})=3 \\
(\sqrt{15})(\sqrt{15})=15
\end{array} \nonumber \]

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

\[\ \begin{aligned}
i^{2} &=(i)(i) \\
&=(\sqrt{-1})(\sqrt{-1}) \\
&=-1
\end{aligned} \nonumber \]

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

\[\ \begin{aligned}
(5 i)(-3 i) &=(5)(-3)(i)(i) \\
&=-15 i^{2} \\
&=-15(-1) \\
&=15
\end{aligned} \nonumber \]

Example

Multiply. \(\ (3 i)(2 i)\)

Solution

\(\ (3 i)(2 i)=(3)(2)(i)(i)=6 i^{2}\)	Multiply the coefficients of \(\ i\) together, and then multiply \(\ i\) times \(\ i\).
\(\ 6 i^{2}=6(-1)\)	\(\ {i}^{2}\) with -1.
\(\ 6(-1)=-6\)	Multiply.

\(\ (3 i)(2 i)=-6\)

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

Exercise

Multiply and simplify. \(\ (3 i)(-i)\)

1. \(\ 3\)
2. \(\ -3\)
3. \(\ 3i\)
4. \(\ -3 i^{2}\)

Answer

1. \(\ 3\)

   Correct. \(\ (3 i)(-i)=3(-1)(i)(i)=-3 i^{2}=-3(-1)=3\)

2. \(\ -3\)

   Incorrect. You probably either missed the sign on the second factor, or you incorrectly replaced \(\ i^{2}\). \(\ (3 i)(-i)=3(-1)(i)(i)=-3 i^{2}=-3(-1)=3\)

3. \(\ 3i\)

   Incorrect. You probably incorrectly multiplied \(\ (i)(-i)\), or incorrectly replaced \(\ i^{2}\). \(\ (3 i)(-i)=3(-1)(i)(i)=-3 i^{2}=-3(-1)=3\)

4. \(\ -3 i^{2}\)

   Incorrect. This is a correct multiplication, but \(\ i^{2}\) can be simplified further. \(\ (3 i)(-i)=3(-1)(i)(i)=-3 i^{2}=-3(-1)=3\)

Using the Distributive Property

The following expression is a little more complicated because two binomials are being multiplied. This means you have to use the Distributive Property of Multiplication. (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.

\[\ \begin{aligned}
(6 x+8)(4 x+2) &=6 x(4 x+2)+8(4 x+2) \\
&=6 x \cdot 4 x+6 x \cdot 2+8 \cdot 4 x+8 \cdot 2 \\
&=24 x^{2}+12 x+32 x+16 \\
&=24 x^{2}+44 x+16
\end{aligned} \nonumber \]

Again, in the same way, you can multiply complex numbers. At the end, you will need to simplify \(\ i^{2}\).

Example

Multiply and simplify. \(\ (6+8 i)(4+2 i)\)

Solution

\(\ \begin{array}{l} (6+8 i)(4+2 i) \\ 6(4+2 i)+8 i(4+2 i) \\ 6 \cdot 4+6 \cdot 2 i+8 i \cdot 4+8 i \cdot 2 i \\ 24+12 i+32 i+16 i^{2} \end{array}\)	Two binomials are being multiplied, so you need to use the Distributive Property of Multiplication. We could have used FOIL and gone directly to the line \(\ 6 \cdot 4+6 \cdot 2 i+8 i \cdot 4+8 i \cdot 2 i\).
\(\ \begin{array}{l} 24+44 i+16 i^{2} \\ 24+44 i+16(-1) \\ 24+44 i-16 \\ 8+44 i \end{array}\)	Combine like terms. Replace \(\ i^{2}\) with -1 and simplify.

\(\ (6+8 i)(4+2 i)=8+44 i\)

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?

Example

Multiply and simplify. \(\ (6+8 i)(6-8 i)\)

Solution

\(\ \begin{array}{l} (6+8 i)(6-8 i) \\ 6 \cdot 6+6 \cdot-8 i+8 i \cdot 6+8 i \cdot-8 i \\ 36-48 i+48 i-64 i^{2} \\ 36-64 i^{2} \\ 36-64(-1) \\ 36+64 \\ 100 \end{array}\)

Use FOIL to expand the product. Combine like terms. Replace \(\ i^{2}\) with -1 and simplify.

\(\ (6+8 i)(6-8 i)=100\)

Just as \(\ 6+8 \sqrt{3}\) and \(\ 6-8 \sqrt{3}\) are conjugates, \(\ 6+8 i\) and \(\ 6-8 i\) 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. 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, \(\ -48 i\) was added to \(\ 48 i\), 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.

Exercise

Multiply. \(\ (9+i)(9-i)\)

1. \(\ 82+18 i\)
2. \(\ 80-18 i\)
3. \(\ 80\)
4. \(\ 82\)

Answer

1. \(\ 82+18 i\)

   Incorrect. Don’t forget to use the FOIL method so that every term in the first complex number is multiplied by every term in the other complex number. Since the factors are complex conjugates, their product should be a real number. \(\ (9+i)(9-i)=81-9 i+9 i-i^{2}=81-i^{2}=81-(-1)=81+1=82\).

2. \(\ 80-18 i\)

   Incorrect. Don’t forget to use the FOIL method so that every term in the first complex number is multiplied by every term in the other complex number. Since the factors are complex conjugates, their product should be a real number. \(\ (9+i)(9-i)=81-9 i+9 i-i^{2}=81-i^{2}=81-(-1)=81+1=82\).

3. \(\ 80\)

   Incorrect. You probably multiplied correctly but forgot to subtract \(\ i^{2}\), or you forgot the negative when you multiplied \(\ i\) and \(\ -i\). \(\ (9+i)(9-i)=81-9 i+9 i-i^{2}=81-i^{2}=81-(-1)=81+1=82\)

4. \(\ 82\)

   Correct. \(\ (9+i)(9-i)=81-9 i+9 i-i^{2}=81-i^{2}=81-(-1)=81+1=82\)

Division of Complex Numbers

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.

Example

Simplify. \(\ -24 i \div 6\)

Solution

\(\ -24 i \div 6=\frac{-24 i}{6}=\frac{-4 i}{1} \cdot \frac{6}{6}\)	Treat the division as a fraction. Simplify the fraction using a factor that the numerator and denominator have in common.
\(\ -24 i \div 6=-4 i\)	Since the result has no denominator, no more simplification is needed.

Example

Simplify. \(\ 32 i \div 6 i\)

Solution

\(\ 32 i \div 6 i=\frac{32 i}{6 i}=\frac{16}{3} \cdot \frac{2 i}{2 i}\)	Treat the division as a fraction. Simplify the fraction using a factor that the numerator and denominator have in common. Note in this case, \(\ i\) is part of the common factor.
\(\ 32 i \div 6 i=\frac{16}{3}\)	The fraction is in simplest form.

Example

Simplify. \(\ 56 \div-7 i\)

Solution

\(\ 56 \div-7 i=\frac{56}{-7 i}=\frac{8}{-i} \cdot \frac{7}{7}\)	Treat the division as a fraction. Simplify the fraction using a factor that the numerator and denominator have in common.
\(\ \frac{8}{-i}\)	In this case, the denominator still has \(\ i\) in it. Since \(\ i\) is a radical, you should simplify further by rationalizing the denominator.
\(\ \begin{aligned} \frac{8}{-i} \cdot \frac{i}{i} &=\frac{8 i}{-i^{2}} \\ &=\frac{8 i}{-(-1)} \\ &=8 i \end{aligned}\)	Since the denominator is just one term, you don’t need to think about complex conjugates. Just multiply by 1 in the form \(\ \frac{i}{i}\) and simplify. (Remember, the product of two imaginary numbers is real, so the denominator is real.)

\(\ 56 \div-7 i=8 i\)

Exercise

Simplify. \(\ 12 \div 10 i\)

1. \(\ \frac{6}{5}\)
2. \(\ -\frac{6}{5} i\)
3. \(\ \frac{6}{5} i\)
4. \(\ -\frac{6}{5}\)

Answer

1. \(\ \frac{6}{5}\)

   Incorrect. When you simplify the numerical parts of \(\ \frac{12}{10 i}\), you still have \(\ i\) in the denominator. Rationalize the denominator of \(\ \frac{6}{5 i}\) by multiplying by \(\ \frac{i}{i}\): \(\ \frac{6}{5 i}=\frac{6}{5 i} \cdot \frac{i}{i}=\frac{6 i}{5 i^{2}}=\frac{6 i}{5(-1)}=-\frac{6}{5} i\)

2. \(\ -\frac{6}{5} i\)

   Correct. After simplifying the numerical parts of \(\ \frac{12}{10} i\), you still have to rationalize the denominator because \(\ i\) is left in the denominator: \(\ \frac{12}{10 i}=\frac{6}{5 i} \cdot \frac{2}{2}=\frac{6}{5 i}=\frac{6}{5 i} \cdot \frac{i}{i}=\frac{6 i}{5 i^{2}}=\frac{6 i}{5(-1)}=-\frac{6}{5} i\)

3. \(\ \frac{6}{5} i\)

   Incorrect. When you rationalize the denominator, you have \(\ i^{2}\) in the denominator. Since \(\ i^{2}=-1\), the quotient is negative. \(\ \frac{12}{10 i}=\frac{6}{5 i} \cdot \frac{2}{2}=\frac{6}{5 i}=\frac{6}{5 i} \cdot \frac{i}{i}=\frac{6 i}{5 i^{2}}=\frac{6 i}{5(-1)}=-\frac{6}{5} i\)

4. \(\ -\frac{6}{5}\)

   Incorrect. You may have forgotten the radical \(\ i\) in the denominator, or you may have made a mistake when you rationalized the denominator. When you rationalize the denominator, you must also multiply the numerator by \(\ i\). \(\ \frac{12}{10 i}=\frac{6}{5 i} \cdot \frac{2}{2}=\frac{6}{5 i}=\frac{6}{5 i} \cdot \frac{i}{i}=\frac{6 i}{5 i^{2}}=\frac{6 i}{5(-1)}=-\frac{6}{5} i\)

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.

Example

Simplify. \(\ (56-8 i) \div(14+10 i)\)

Solution

\(\ \begin{aligned} (56-8 i) \div(14+10 i) &=\frac{56-8 i}{14+10 i} \\ &=\frac{(28-4 i)(2)}{(7+5 i)(2)} \\ &=\frac{28-4 i}{7+5 i} \cdot \frac{2}{2} \\ &=\frac{28-4 i}{7+5 i} \end{aligned}\)	Treat the division as a fraction. Simplify the fraction using a factor that the numerator and denominator have in common, if any. Be careful to use the distributive property: the numbers must be a factor of all terms.
\(\ \begin{aligned} \frac{28-4 i}{7+5 i} &=\frac{28-4 i}{7+5 i} \cdot \frac{7-5 i}{7-5 i} \\ &=\frac{(28-4 i)(7-5 i)}{(7+5 i)(7-5 i)} \end{aligned}\)	In this case, the denominator still has \(\ i\) in it. To rationalize the denominator, multiply by the complex conjugate of the denominator. In this case, the complex conjugate is \(\ (7-5 i)\).
\(\ \begin{aligned} \frac{(28-4 i)(7-5 i)}{(7+5 i)(7-5 i)} &=\frac{28(7)+28(-5 i)+(-4 i)(7)+(-4 i)(-5 i)}{7(7)+7(-5 i)+5 i(7)+5 i(-5 i)} \\ &=\frac{196-140 i-28 i+20 i^{2}}{49-35 i+35 i-25 i^{2}} \\ &=\frac{196-168 i+20 i^{2}}{49-25 i^{2}} \end{aligned}\)	(For complex conjugates, the real parts are equal and the imaginary parts are additive inverses.) Expand the numerator and the denominator. Remember, the denominator should be a real number (no \(\ i\) term) if you chose the correct complex conjugate and performed the multiplication correctly.
\(\ \begin{aligned} \frac{196-168 i+20 i^{2}}{49-25 i^{2}} &=\frac{196-168 i+20(-1)}{49-25(-1)} \\ &=\frac{196-168 i-20}{49+25} \\ &=\frac{176-168 i}{74} \end{aligned}\)	Replace \(\ i^{2}\) with -1 and simplify. Be sure to replace \(\ i^{2}\) in both the numerator and the denominator!
\(\ \frac{176-168 i}{74}=\frac{176}{74}-\frac{168}{74} i\)	The quotient can be written in the form \(\ a+b i\) using fractions for both \(\ a\) and \(\ b\).
\(\ \begin{aligned} \frac{176}{74}-\frac{168 i}{74} &=\frac{88}{37} \cdot \frac{2}{2}-\left(\frac{84}{37} \cdot \frac{2}{2}\right) i \\ &=\frac{88}{37}-\frac{84}{37} i \end{aligned}\)	Always check the final product to see if you can simplify further. In this case, both fractions can be simplified.

\(\ (56-8 i) \div(14+10 i)=\frac{88}{37}-\frac{84}{37} i\)

Exercise

Simplify. \(\ (10+6 i) \div(5-3 i)\)

1. \(\ \frac{16}{17}+\frac{30}{17} i\)
2. \(\ 2-2 i\)
3. \(\ 2+\frac{15}{4} i\)
4. \(\ \frac{16}{17}+60 i\)

Answer

1. \(\ \frac{16}{17}+\frac{30}{17} i\)

   Correct. Write as a rational expression and multiply both numerator and denominator by the complex conjugate of the divisor: \(\ \frac{10+6 i}{5-3 i} \cdot \frac{5+3 i}{5+3 i}=\frac{50+30 i+30 i+18 i^{2}}{25-9 i^{2}}=\frac{50+60 i-18}{25+9}=\frac{32+60 i}{34}\). This simplifies to \(\ \frac{16}{17}+\frac{30}{17} i\).

2. \(\ 2-2 i\)

   Incorrect. You probably just divided the real parts, then divided the coefficients of the imaginary parts. This works with addition and subtraction, but not division. Try writing the division as a rational expression, and multiply both numerator and denominator by the complex conjugate of the divisor. The correct answer is \(\ \frac{16}{17}+\frac{30}{17} i\).

3. \(\ 2+\frac{15}{4} i\)

   Incorrect. You probably correctly set up the rational expression and rationalized: \(\ \frac{10+6 i}{5-3 i} \cdot \frac{5+3 i}{5+3 i}=\frac{50+30 i+30 i+18 i^{2}}{25-9 i^{2}}\). Remember at this point that \(\ i^{2}=-1\), so subtracting \(\ 9 i^{2}\) is the same as adding 9. The correct answer is \(\ \frac{16}{17}+\frac{30}{17} i\).

4. \(\ \frac{16}{17}+60 i\)

   Incorrect. You probably correctly set up the rational expression and rationalized: \(\ \frac{10+6 i}{5-3 i} \cdot \frac{5+3 i}{5+3 i}=\frac{50+30 i+30 i+18 i^{2}}{25-9 i^{2}}=\frac{50+60 i-18}{25+9}=\frac{32+60 i}{34}\). Remember at this point the denominator (34) divides both terms in the numerator, so this simplifies to \(\ \frac{32}{34}+\frac{60}{34} i=\frac{16}{17}+\frac{30}{17} i\).

Operations with Complex Numbers

To add or subtract, combine like terms.

To multiply monomials, multiply the coefficients and then multiply the imaginary numbers \(\ i\). If \(\ i^{2}\) appears, replace it with -1.

To multiply complex numbers that are binomials, use the Distributive Property of Multiplication, or the FOIL method. Multiply the resulting terms as monomials.

To divide, treat the quotient as a fraction.

• Simplify the numerical parts, and then rationalize the denominator, if needed.
• Replace \(\ {i}^{2}\) in both the numerator and the denominator with -1, as needed.
• Write the answer in \(\ a+b i\) form, which may require further simplification of \(\ a\) and \(\ b\) when they are fractions.

Summary

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.