## 17Calculus Precalculus - Complex Numbers

On this site we give you an overview of complex numbers and a discussion of Euler's Formula in order to get you up and running, ready for calculus. If you want to learn more about complex numbers, we recommend Dr Chris Tisdell's free ebook Introduction to Complex Numbers and his YouTube playlist related to the book.
For a gentler introduction to complex numbers, check out this youtube playlist we have put together for you.

Complex numbers are sometimes called imaginary numbers but they are neither complex nor imaginary. The good news is that you already know how to work with complex numbers if you know how to work with radicals.

You are familiar with the expression $$1+3\sqrt{2}$$ from previous algebra classes. For complex numbers we do something that you have been told in the past was not allowed. We replace the two under the square root with a negative number, and not just any negative number, we replace it with $$-1$$ to get $$1+3\sqrt{-1}$$. And that's about it. We now have a complex number.

Now, to make it easier to write this complex number, we replace the square root of $$-1$$ with the letter $$i$$. So we define $$i$$ as $$i=\sqrt{-1}$$. We can then write $$1+3\sqrt{-1}$$ as $$1+3i$$. This is the usual way we write complex numbers and, from now on, you will almost never see $$\sqrt{-1}$$. Now let's define some terms.

For the complex number $$1+3i$$

$$1$$

real part

$$3i$$

imaginary part

Before we go on, let's watch a quick video about how to simplify complex numbers. This video has several examples as well.

### MIP4U - Simplify Square Roots to Imaginary Numbers [4min-18secs]

video by MIP4U

Adding, subtracting, multiplying and dividing using complex numbers work exactly like radicals. For example, to simplify $$(1+3\sqrt{2}) + (6+8\sqrt{2})$$, you would combine the parts without the radicals and then combine the parts with the radicals. So the result is $$(1+3\sqrt{2}) + (6+8\sqrt{2}) = (1+6) + (3+8)\sqrt{2} = 7+11\sqrt{2}$$.

With complex numbers, $$(1+3i) + (6+8i) = (1+6) + (3+8)i = 7+11i$$. [ Remember: $$i=\sqrt{-1}$$ ]
Subtraction works just as you would expect.

Multiplying Complex Numbers

Multiplying works the same as with radicals. For example, let's multiply the same two numbers we used above, first with square root of 2 and then with complex numbers.
Example: $$(1+3\sqrt{2}) (6+8\sqrt{2}) = 1(6+8\sqrt{2}) + 3\sqrt{2}(6+8\sqrt{2}) =$$ $$6+8\sqrt{2} +18\sqrt{2} +24(2) =$$ $$54+26\sqrt{2}$$
Notice we had $$(\sqrt{2})^2 = 2$$.

Now lets multiply complex numbers. Example: $$(1+3i) (6+8i) = 1(6+8i) + 3i(6+8i) =$$ $$6+8i +18i +24i^2 =$$ $$6+26i - 24 = -18+26i$$
Notice that when we got $$i^2$$ and since $$i=\sqrt{-1}$$, this gives us $$i^2 = (\sqrt{-1})^2 = -1$$. So $$24i^2 = 24(-1)=-24$$.

Before we go on, let's watch a video to give us more information.

### Dr Chris Tisdell - Complex numbers are AWESOME [16min-46secs]

video by Dr Chris Tisdell

Why Do We Need Complex Numbers?

A natural question is, what are complex numbers used for? Why even have them? Here is an explanation, in two quick videos, that will explain why we need them without a lot of technical jargon.
The first video sets up the rationale behind why we might even consider complex numbers.
The second video explains why we need them.

### Derek Owens - Complex Numbers - Why We Need Them [4min-35secs]

video by Derek Owens

### Derek Owens - Complex Numbers - Why We Need Them (continued) [10min-15secs]

video by Derek Owens

Surprisingly, there are many areas where complex numbers are useful. As you know from the previous video, there are some equations that could not be solved without complex numbers. Here is a video showing an example of solving for complex multiple roots of 1.

### Khan Academy - Exponential form to find complex roots [11min-52secs]

Rationalizing

Remember with square roots, you would perform an operation called rationalizing. This usually occurs when you want to remove a square root from the denominator of a fraction. The idea is to multiply by something called the conjugate. This is what it looks like. If you have $$1+3\sqrt{2}$$, the conjugate is $$1-3\sqrt{2}$$, i.e. you change the sign in the middle. The beauty of this that when you you multiply these together, the square root disappears.
Example: $$(1+3\sqrt{2})(1-3\sqrt{2}) =$$ $$1(1-3\sqrt{2}) +3\sqrt{2}(1-3\sqrt{2}) =$$ $$1-3\sqrt{2} + 3\sqrt{2} -9(2) = -17$$

Notice in the last example, that the square root of two is gone. That is the point of rationalizing. We can do the same thing with complex numbers.
Example: $$(1+3i)(1-3i) =$$ $$1(1-3i) +3i(1-3i) =$$ $$1-3i + 3i -9i^2 = 1-9(-1) = 10$$

Want More?

Do you want more detail on complex numbers? We recommend Dr Chris Tisdell's free ebook Introduction to Complex Numbers. This first YouTube playlist is related to the book.

### Dr Chris Tisdell - introduction to complex numbers playlist

video by Dr Chris Tisdell

For a gentler introduction to complex numbers, here is another playlist we have put together for you.

### Dr Chris Tisdell - A Gentle Introduction To Complex Numbers

video by Dr Chris Tisdell

### complex numbers 17calculus youtube playlist

Really UNDERSTAND Precalculus

### Trig Formulas

The Unit Circle

The Unit Circle [wikipedia] Basic Trig Identities

Set 1 - basic identities

$$\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }$$

$$\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }$$

$$\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }$$

$$\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }$$

Set 2 - squared identities

$$\sin^2t + \cos^2t = 1$$

$$1 + \tan^2t = \sec^2t$$

$$1 + \cot^2t = \csc^2t$$

Set 3 - double-angle formulas

$$\sin(2t) = 2\sin(t)\cos(t)$$

$$\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }$$

Set 4 - half-angle formulas

$$\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }$$

$$\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }$$

Trig Derivatives

 $$\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }$$ $$\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }$$ $$\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }$$ $$\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }$$ $$\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }$$ $$\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }$$

Inverse Trig Derivatives

 $$\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }$$ $$\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }$$

Trig Integrals

 $$\int{\sin(x)~dx} = -\cos(x)+C$$ $$\int{\cos(x)~dx} = \sin(x)+C$$ $$\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C$$ $$\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C$$ $$\int{\sec(x)~dx} =$$ $$\ln\abs{\sec(x)+\tan(x)}+C$$ $$\int{\csc(x)~dx} =$$ $$-\ln\abs{\csc(x)+\cot(x)}+C$$

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