# How to Multiply Imaginary Numbers

### How to Multiply Powers of I

##### Example 1

Simplify the following product:

$$i^6 \cdot i^3$$

Step 1)

Use the rules of exponents (in other words add 6 +3)

$$i^{\red{6 + 3}} = i ^9$$

Step 2)

$$i^9 \\ i ^1 \\ \boxed{i}$$

##### Example 2

Simplify the following product:

$$3i^5 \cdot 2i^6$$

Step 1)

Group the real coefficients and the imaginary terms

$$\blue3 \red i^5 \cdot \blue2 \red i^6 \\ ( \blue 3 \cdot \blue 2) ( \red i^5 \cdot \red i^6)$$

Step 2)

Multiply the real numbers and use the rules of exponents on the imaginary terms

$$( \blue 6 ) ( \red i^{ 5 + 6}) \\ ( \blue 6 ) ( \red i^{ 11 })$$

Step 3)

$$( \blue 6 ) ( \red i^{ 11 }) \\ ( \blue 6 ) ( \red i^{ 3 }) \\ ( \blue 6 ) ( \red {-i})$$

Step 4)

Ungroup terms

$$\boxed{-6i}$$

### How to Multiply Imaginary Numbers

##### Example 3

Simplify the following product

$$3\sqrt{-6} \cdot 5 \sqrt{-2}$$

Step 1)

Group the real coefficients (3 and 5) and the imaginary terms

$$( \blue{ 3 \cdot 5} ) ( \red{ \sqrt{-6}} \cdot \red{ \sqrt{-2} } )$$

Step 2)

Multiply the real numbers and separate out $$\sqrt{-1}$$ also known as $$i$$ from the imaginary numbers

$$(\blue {15}) (\red{ \sqrt{-1}} \sqrt{6} \cdot \red{\sqrt{-1}}\sqrt{2} ) \\ (\blue {15}) (\red i \sqrt{6} \cdot \red i \sqrt{2} )$$

Step 3)

Multiply real radicals and imaginary numbers

$$(\blue {15}) (\red i \color{green}{\sqrt{6}} \cdot \red i \color{green}{ \sqrt{2} } ) \\ (15) ( \red i \cdot \red i \cdot \color{green}{\sqrt{ 12} }) \\ (15) ( \red i^2 \cdot \color{green}{\sqrt{ 12} })$$

Step 4)

Simplify

$$15 ( \red i^2 \cdot \color{green}{\sqrt{4 } \sqrt{3} }) \\ 15 ( -1 \cdot \color{green}{2 \sqrt{3} }) \\ \boxed{ -30\sqrt{3}}$$

Look carefully at the two sample problems below.

$\sqrt{2} \cdot \sqrt{6} \\ \sqrt{2 \cdot 6} \\ \sqrt{12} \\ \sqrt{4} \cdot \sqrt{3} \\ \boxed{2 \sqrt{3}}$

Multiplying two imaginary numbers

$\sqrt{-2} \cdot \sqrt{-6} \\ \sqrt{-2 \cdot -6} \\ \sqrt{12} \\ \sqrt{4} \cdot \sqrt{3} \\ \boxed{2 \sqrt{3}}$

# How did we get the same result?

We got the same answer because we did something wrong in Sample Problem B

$\sqrt{-2} \cdot \sqrt{-6} \\ \red{ \sqrt{-2 \cdot -6}} \\ \sqrt{12} \\ \sqrt{4} \cdot \sqrt{3} \\ \boxed{2 \sqrt{3}}$

You learned that you can rewrite the multiplication of radicals/square roots like $$\sqrt{2} \cdot \sqrt{6}$$ as $$\sqrt{2\cdot 6}$$
However, you can not do this with imaginary numbers( ie negative radicands).
$\sqrt{a} \cdot \sqrt{b} = \sqrt{a \cdot b} \\ \text{ if only if }\red{a>0 \text{ and } b >0 }$ So, if the radicand is negative you cannot apply that rule.

In , the radicands are negative and it is therefore incorrect to write :

$\cancelred{ \sqrt{-2} \cdot \sqrt{-6} = \sqrt{-2 \cdot -6} }$

### Practice Problem Multiply Powers of I

Step 1

Use the rules of exponents (in other words add 4 +11)

$$i^4 \cdot i^{11} = i^{ \red{4 + 11} } \\ i^{15}$$

Step 2

$$i^{ \red{15} } \\ i^{ \red{3} } \\ \red{-} i$$

More questions like Problem 1 ...
Step 1

$$i^1 \cdot i^{19} = i^{ \red{1 + 19} } \\ i^{20}$$

Step 2

$$i^{ \red{20} } \\ i^{ \red{4} } \\ \boxed{1}$$

Step 1

$$i^{15} \cdot i^{17} = i^{ \red{15 + 17} } \\ i^{32}$$

Step 2

$$i^{ \red{32} } \\ i^{ \red{2} } \\ \boxed{-1}$$

Step 1

$$(i^{16})^2 = i^{\red{16 \cdot 2}} \\ i^{32}$$

Step 2

$$i^{ \red{32} } \\ i^{ \red{2} } \\ \boxed{-1}$$

Step 1

This problem is like example 2

Group the real coefficients and the imaginary terms

$$\blue3 \red i^6 \cdot \blue 7 \red i^8 \\ ( \blue 3 \cdot \blue 7) ( \red i^6 \cdot \red i^8)$$
Step 2

Multiply the real numbers and use the rules of exponents to simplify the imaginary ones

$$( \blue {21}) ( \red i^{ 6 + 8}) \\ ( \blue {21})( \red i^{ 14 })$$
Step 3

Simplify the Imaginary Number

$$( \blue {21})(i^{ \red{ 14 } }) \\ ( \blue {21})(i^{ \red{ 2 }}) \\ ( \blue {21})( \red{-1 })$$
Step 4

Ungroup terms

$$( \blue {21} )( \red{-1 }) \\ -21$$
More questions like Problem 2 ...
Step 1

This problem is like example 2

Group the real coefficients and the imaginary terms

$$\blue {2} \red i^{11} \cdot \blue{10} \red i^6 \\ ( \blue 2 \cdot \blue {10}) ( \red i^{11} \cdot \red i^6)$$
Step 2

Multiply the real numbers and use the rules of exponents to simplify the imaginary ones

$$( \blue {20}) ( \red i^{ 11 + 6}) \\ ( \blue {20})( \red i^{ 17 })$$
Step 3

Simplify the Imaginary Number

$$( \blue {20})( \red i^{ 17 }) \\ ( \blue {20})(i^{ \red{ 3 }}) \\ ( \blue {20})( \red{-i })$$
Step 4

Ungroup terms

$$( \blue {20} )( \red{-i }) \\ \boxed{ -20i}$$
Step 1

This problem is like example 2.

Group the real coefficients and the imaginary terms

$$(-3 i^{2})^3 \\ (\blue{-3})^3 \red{(i^2)}^3 \\$$
Step 2

Apply the the rules of exponents to imaginary and real numbers

$$(\blue{-3})^3 (\red{i^2})^3 \\ (\blue{-27}) (\red{i^8})$$
Step 3

Simplify the Imaginary Number

$$(\blue{-27}) (\red{i^8}) \\ (\blue{-27}) (\red{i^0}) \\ (\blue{-27}) (1)$$
Step 4

Ungroup terms

$$\boxed{ -27 }$$
Step 1
Group the real coefficients and the imaginary terms

$$( \blue{ 4 \cdot 2} ) ( \red{ \sqrt{-15}} \cdot \red{ \sqrt{-3} } )$$

Step 2

Multiply the real numbers and separate out $$\sqrt{-1}$$ also known as $$i$$ from the imaginary numbers

$$(\blue {8}) (\red{ \sqrt{-1}} \sqrt{15} \cdot \red{\sqrt{-1}}\sqrt{3} ) \\ (\blue {8}) (\red{ i} \sqrt{15} \cdot \red{i }\sqrt{3} )$$

Multiply real radicals and imaginary numbers

$$(\blue {8}) (\red i \color{green}{\sqrt{15}} \cdot \red i \color{green}{ \sqrt{3} } ) \\ (8) ( \red i \cdot \red i \cdot \color{green}{\sqrt{ 45 } }) \\ (8) ( \red i^2 \cdot \color{green}{\sqrt{ 45 } })$$

Step 3

Simplify

$$(8) ( \red i^2 \cdot \color{green}{\sqrt{ 45 } }) \\ 8 ( -1 \cdot \color{green}{\sqrt{9} \sqrt{5} }) \\ 8 ( -1 \cdot \color{green}{3 \sqrt{5} }) \\ \boxed{ -24\sqrt{5}}$$

Step 1
Group the real coefficients and the imaginary terms

$$( \blue{ 5 \cdot 7} ) ( \red{ \sqrt{-12}} \cdot \red{ \sqrt{-15} } )$$

Step 2

Multiply the real numbers and separate out $$\sqrt{-1}$$ also known as $$i$$ from the imaginary numbers

$$(\blue {35}) (\red{ \sqrt{-1}} \sqrt{12} \cdot \red{\sqrt{-1}}\sqrt{15} ) \\ (\blue {35}) (\red{ i} \sqrt{12} \cdot \red{{i}}\sqrt{15} )$$

Multiply real radicals and imaginary numbers. (Note: It is often easier to simplify radicals before multiplying them. So, in this case we are doing a bit of the work that we often save for step 4)

$$(\blue {35}) (\red{ i} \sqrt{12} \cdot \red{{i}}\sqrt{15} ) \\ 35 ( \red{ i^2} \cdot \color{green}{ 2 \sqrt{3}} \cdot \color{green}{ \sqrt{3} \sqrt{5}} ) \\ 35 ( \red{ i^2} \cdot \color{purple}{ 2 \sqrt{3}} \cdot \color{purple}{ \sqrt{3} \sqrt{5}} ) \\ 35 ( \red{ i^2} \cdot \color{green}{ 2 } \cdot \color{purple}{3} \color{green}{ \sqrt{5}} ) \\ 35 ( \red{ i^2} \cdot 6 \color{green}{ \sqrt{5}} )$$

Step 3

Simplify

$$35 ( \red{ i^2} \cdot 6 \color{green}{ \sqrt{5}} ) \\ (35) ( -1 \cdot 6 \color{green}{ \sqrt{5}} ) \\ (35) ( - 6 \color{green}{ \sqrt{5}} ) \\ (35) ( - 6 \color{green}{ \sqrt{5}} ) \\ \boxed{ -210\sqrt{5}}$$

Step 1
Group the real coefficients and the imaginary terms

$$( \blue{ -2 \cdot 7 \cdot 5} ) ( \red{ \sqrt{-15}} \cdot \red{ \sqrt{-3} } \cdot \red{ \sqrt{-10}} )$$

Step 2

Multiply the real numbers and separate out $$\sqrt{-1}$$ also known as $$i$$ from the imaginary numbers

$$(\blue {-70}) (\red{ \sqrt{-1}} \sqrt{15}\cdot \red{\sqrt{-1}}\sqrt{3} \cdot \red{\sqrt{-1}}\sqrt{10} ) \\ (\blue {-70}) (\red{i} \sqrt{15}\cdot \red{i } \sqrt{3} \cdot \red{i}\sqrt{10} )$$

Multiply real radicals and imaginary numbers. (Note: It is often easier to simplify radicals before multiplying them. So, in this case we are doing a bit of the work that we often save for step 4)

$$(\blue {-70}) (\red{i^3} \color{green}{ \sqrt{15}} \cdot \color{green}{ \sqrt{3} } \cdot \color{green}{ \sqrt{10}} ) \\ (\blue {-70}) (\red{i^3} \cdot \color{green}{ \sqrt{45} } \cdot \color{green}{ \sqrt{10}} ) \\ (\blue {-70}) (\red{i^3} \cdot \color{purple}{ 3\sqrt{5} } \cdot \color{green}{ \sqrt{10}} ) \\ \\ (\blue {-70}) (\red{i^3} \cdot \color{green}{ 3\sqrt{50}} )$$

Step 3

Simplify

$$-70 ( \red{ i^3} \cdot 3 \color{green}{ \sqrt{50}} ) \\ -70 ( -i \cdot 3 \color{green}{ \sqrt{50}} ) \\ -70 ( -i \cdot 3 \color{green}{ \sqrt{25}\sqrt{2}} ) \\ -70 ( -i \cdot 3 \cdot\color{green}{5 \sqrt{2}} ) \\ -70 ( -15i \cdot\color{green}{ \sqrt{2}} ) \\ \\ \boxed{ 1050i\sqrt{2}}$$

$$\text{ Jen's Solution} \\ (3 \cdot 4) (\sqrt{-2} \cdot \sqrt{-8} ) \\ ( 12 ) (\sqrt{-2 \cdot -8} ) \\ \red{ ( 12 ) (\sqrt{16 } ) } \\ ( 12 ) ( 4 ) \\ 48$$
$\sqrt{-2} \cdot \sqrt{-8} \red{ \ne } \sqrt{-2 \cdot -8}$