## How to Multiply and Divide Complex Numbers

### Multiplying complex numbers

The process of multiplying complex numbers is quite similar to multiplying binomials. The most significant difference is that we work separately on the real and imaginary parts.

We have two methods for multiplying complex numbers: the distributive property and the FOIL technique. Remember that FOIL stands for multiplying together the First, Outer, Inner, and Last terms. We get the same answer whether we use the distributive property or the FOIL method:

$$(a \ + \ bi) \ \times \ (c \ + \ di) \ = \ ac \ + \ adi \ + \ bci \ + \ bdi^2$$

We know that $$i^2 \ = \ -1$$, so we have: $$ac \ + \ adi \ + \ bci \ - \ bd$$

Now simplify to get the final answer: $$ac \ + \ adi \ + \ bci \ - \ bd \ = \ (ac \ - \ bd) \ + \ i(ad \ + \ bc)$$

### Dividing complex numbers

Adding, subtracting, and multiplying is easier than dividing two complex numbers. This is because we can't divide by an imaginary number, so any fraction must have a real-number denominator. We need to find a term by which we can multiply the numerator and the denominator to eliminate the imaginary part of the denominator and get a real number as the denominator. This term is called the complex conjugate of the denominator. It is found by changing the sign of the imaginary part of the complex number. In other words, the conjugated form of $$a \ + \ bi$$ is $$a \ - \ bi$$, and the conjugated form of $$a \ - \ bi$$ is $$a \ + \ bi$$

Also, complex solutions to a quadratic equation with real coefficients are always complex conjugates of each other.

Let's say we want to divide $$a \ + \ bi$$ by $$c \ + \ di$$ when $$c, \ d \ ≠ \ 0$$. First, we write the division as a fraction. Then, we find the complex conjugate of the denominator and multiply:

$$\frac{a \ + \ bi}{c \ + \ di}$$

Multiply both the numerator and the denominator by the denominator's complex conjugate.

$$\frac{a \ + \ bi}{c \ + \ di} \times \frac{c \ - \ di}{c \ - \ di} \ = \ \frac{(a \ + \ bi) \ (c \ - \ di)}{(c \ + \ di) \ (c \ - \ di)}$$

Use the distributive property: $$\frac{ac \ - \ adi \ + \ bci \ - \ bdi^2}{c^2 \ - \ d^2i^2}$$

Now Simplify, ($$i^2 \ = \ -1$$)

$$\frac{ac \ - \ adi \ + \ bci \ + \ bd}{c^2 \ + \ d^2} \ = \ \frac{(ac \ + \ bd) + \ i(-ad \ + \ bc)}{c^2 \ + \ d^2}$$

### Exercises for Multiplying and Dividing Complex Numbers

1) Find the answer: $$(2i) \ \times \ (5i)$$

2) Find the answer: $$(2 \ + \ 3i) \ \times \ (4 \ + \ 2i)$$

3) Find the answer: $$(5 \ - \ i) \ \times \ (2 \ + \ 3i)$$

4) Find the answer: $$(-7 \ + \ 4i) \ \times \ (8 \ - \ i)$$

5) Find the answer: $$3i \ \times \ (7 \ + \ 12i)$$

6) Find the answer: $$\frac{-3 \ - \ 2i}{4 \ + \ 3i}$$

7) Find the answer: $$\frac{3 \ - \ i}{-2 \ + \ 4i}$$

8) Find the answer: $$\frac{5 \ - \ 9i}{-3 \ + \ 2i}$$

9) Find the answer: $$\frac{-9i}{5 \ + \ 6i}$$

10) Find the answer: $$\frac{5 \ + \ i}{-3i}$$

1) Find the answer: $$(2i) \ \times \ (5i)$$

$$\color{red}{(2i) \ \times \ (5i) \ = \ -10}$$

2) Find the answer: $$(2 \ + \ 3i) \ \times \ (4 \ + \ 2i)$$

$$\color{red}{2(4) \ + \ 2(2i) \ + \ (3i)(4) \ + \ (3i)(2i) \ = \ 8 \ + \ 4i \ + \ 12i \ + \ 6(-1) \ = \ 2 \ + \ 16i}$$

3) Find the answer: $$(5 \ - \ i) \ \times \ (2 \ + \ 3i)$$

$$\color{red}{5(2) \ + \ 5(3i) \ - \ (i)(2) \ - \ (i)(3i) \ = \ 10 \ + \ 15i \ - \ 2i \ - \ 3(-1) \ = \ 13 \ + \ 13i}$$

4) Find the answer: $$(-7 \ + \ 4i) \ \times \ (8 \ - \ i)$$

$$\color{red}{(-7)(8) \ + \ (-7)(-i) \ + \ (4i)(8) \ + \ (4i)(-i) \ = \ -56 \ + \ 7i \ + \ 32i \ - \ 4(-1)}$$ $$\color{red}{= \ -52 \ + \ 39i}$$

5) Find the answer: $$3i \ \times \ (7 \ + \ 12i)$$

$$\color{red}{(3i)(7) \ + \ (3i)(12i) \ = \ 21i \ + \ 36(i^2) \ = \ 21i \ + \ 36(-1) \ = \ 21i \ - \ 36}$$

6) Find the answer: $$\frac{-3 \ - \ 2i}{4 \ + \ 3i}$$

$$\color{red}{\frac{-3 \ - \ 2i}{4 \ + \ 3i} \ = \ \frac{-3 \ - \ 2i}{4 \ + \ 3i} \times \frac{4 \ - \ 3i}{4 \ - \ 3i} \ = \ \frac{-12 \ + \ 9i \ - \ 8i \ + \ 6(-1)}{16 \ - \ 9(-1)} \ = \ \frac{-18 \ + \ i}{25} \ = \ -\frac{18}{25} \ + \ \frac{i}{25}}$$

7) Find the answer: $$\frac{3 \ - \ i}{-2 \ + \ 4i}$$

$$\color{red}{\frac{3 \ - \ i}{-2 \ + \ 4i} \ = \ \frac{3 \ - \ i}{-2 \ + \ 4i} \times \frac{-2 \ - \ 4i}{-2 \ - \ 4i} \ = \ \frac{-6 \ - \ 12i \ + \ 2i \ - \ 4}{4 \ - \ (-16)} \ = \ \frac{-10 \ - \ 10i}{20} \ = \ -\frac{10}{20} \ - \ \frac{10i}{20} \ = \ -\frac{1}{2} \ - \ \frac{i}{2}}$$

8) Find the answer: $$\frac{5 \ - \ 9i}{-3 \ + \ 2i}$$

$$\color{red}{\frac{5 \ - \ 9i}{-3 \ + \ 2i} \ = \ \frac{5 \ - \ 9i}{-3 \ + \ 2i} \times \frac{-3 \ - \ 2i}{-3 \ - \ 2i} \ = \ \frac{-15 \ - \ 10i \ + \ 27i \ + \ 18(-1)}{9 \ - \ 4(-1)} \ = \ \frac{-33 \ + \ 17i}{13} \ = \ -\frac{33}{13} \ + \ \frac{17i}{13}}$$

9) Find the answer: $$\frac{-9i}{5 \ + \ 6i}$$

$$\color{red}{\frac{-9i}{5 \ + \ 6i} \ = \ \frac{-9i}{5 \ + \ 6i} \times \frac{5 \ - \ 6i}{5 \ - \ 6i} \ = \ \frac{-45i \ + \ 54(-1)}{25 \ - \ 36(-1)} \ = \ \frac{-45i \ - \ 54}{61} \ = \ -\frac{45i}{61} \ - \ \frac{54}{61}}$$

10) Find the answer: $$\frac{5 \ + \ i}{-3i}$$

$$\color{red}{\frac{5 \ + \ i}{-3i} \ = \ \frac{5 \ + \ i}{-3i} \times \frac{i}{i} \ = \ \frac{5i \ + \ (-1)}{-3(-1)} \ = \ \frac{5i}{3} \ - \ \frac{1}{3}}$$

## Multiplying and Dividing Complex Numbers Practice Quiz

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