# Complex Numbers

## Introduction

Just as we can perform operations with real numbers, we can also add, subtract, multiply, and divide complex numbers. Recall the order of operations for real numbers here: Operations with Numbers

When performing operations with complex numbers, you may encounter three types of complex numbers:

• Real number as complex number - imaginary part is equal to 0

Examples: $$2 , -6, 7.462437, \frac{3}{5}$$

Tip: Since all real numbers are also complex numbers, you can add $$0i$$ to these numbers and they'll still be considered complex numbers!

• Imaginary number as complex number - real part is equal to 0

Examples: $$i, 3i, \sqrt{2}i, \frac{1}{2}i$$

• Complex number composed of real and imaginary part

Examples: $$3+4i, \frac{5}{2}-0.1i, -7.25-11i$$

To add or subtract complex numbers, just combine like terms as if the number $$i$$ is a variable, then evaluate.

• For any two complex numbers, say $$x=a+bi$$ and $$y=c+di$$, we can add them as follows:

$$x+y=(a+bi)+(c+di)=(a+c)+(b+d)i$$

• For any two complex numbers, say $$x=a+bi$$ and $$y=c+di$$, we can subtract them as follows:

$$x-y=(a+bi)-(c+di)=(a-c)+(b-d)i$$

Example: Evaluate $$(-4+5i)-(6-6i)+(2i+7)$$.

Solution:

1. Combine like terms:

$$(-4+5i)-(6-6i)+(2i+7)$$

$$=(-4-6+7)+(5i-(-6i)+2i)$$

2. Simplify and evaluate:

$$=-3+(5-(-6)+2)i$$

$$=-3+13i$$

## Multiplication

To multiply complex numbers, you can first use one the FOIL process to expand the two binomials. The FOIL process is a quick method of apply the distributive law to expand two binomials:

$$(a+b)(c+d)=ac+ad+bc+bd$$

where FOIL is an acronym for:

• First
• Outer
• Inner
• Last

since we multiply the "firsts" in the two binomials - $$a$$ and $$c$$ in the above case, then the "outer" terms - $$a$$ and $$d$$, then "inner" terms - $$b$$ and $$c$$, then "last" terms - $$b$$ and $$d$$, and add all these products up.

Once you've finished expanding, the final step is to combine like terms. To summarize, you can multiply any two complex numbers, say, $$(a+bi)$$ and $$(c+di)$$, as follows:

$$(a+bi)(c+di)=(a)(c)+(a)(di)+(bi)(c)+(bi)(di)=ac+adi+bci+bd(-1)=(ac-bd)+(ad+bc)i$$

*Note: in the second last step, we were able to replace $$i\times i=i^2$$ with -1 since, by definition, $$i^2=(\sqrt{-1})^2=-1$$

Example: Evaluate $$(3+7i)(2-i)$$.

Solution:

1. First, expand using the FOIL method:

$$(3+7i)(2-i)$$

$$=(3)(2)+(3)(-i)+(7i)(2)+(7i)(-i)$$

2. Next, multiply out all the terms:

$$=6-3i+14i-7i^2$$

$$=6-3i+14i-7(-1)$$

$$=6-3i+14i+7$$

3. Evaluate by collecting like terms and simplifying:

$$=13+11i$$

## Division

Complex division is often represented using fractions, in the general form of $$\frac{a+bi}{c+di}$$ for any two complex numbers. Notice that the denominator is the sum of two terms, so we cannot find the quotient directly.

Recall that the conjugate of a complex number $$z=a+bi$$ is: $$\bar{z}=a-bi$$. This will be helpful to us when dividing complex numbers since multiplying a complex number by its conjugate gives us a real number product.

For any two complex numbers, say $$x=a+bi$$ and $$y=c+di$$, we can divide $$x$$ by $$y$$  (i.e. evaluate $$\frac{a+bi}{c+di}$$) by following these steps:

1. Determine the conjugate of the denominator (which is $$c-di$$ here). Then multiply the numerator and denominator by this conjugate:

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

*Note: this step does not change the original expression since we're actually just multiplying it by $$\frac{c-di}{c-di}=1$$!

2. Expand the expression by multiplying:

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

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

3. Simplify by collecting like terms and reducing fractions (if applicable):

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

Example: Evaluate $$\frac{3+5i}{2+4i}$$.

See the below video for the solution:

## Practice

Time to do some practice!