The post What Are Complex Numbers? appeared first on Magoosh Math.

]]>Complex numbers include everyday real numbers like 3, -8, and 7/13, but in addition, we have to include all of the *imaginary* numbers, like *i*, 3*i*, and -π*i*, as well as combinations of real and imaginary. You see, complex numbers are what you get when you mix real and imaginary numbers together — a very complicated relationship indeed!

By definition a **complex number** is any number of the form *a* + *bi*, where both *a* and *b* are real numbers, and *i* is the **imaginary unit**, defined by its main property: *i*^{2} = -1.

We’ll have a lot more to say about *i* in this article. And we’ll explore the definition and properties of complex numbers. There’s a lot to say about these amazing and mysterious gadgets, so buckle up and hold on tight!

Mathematicians have a tendency to invent new tools as the need arises. Need to count losses as well as profits? Invent the negative numbers. Need to keep track of parts of a whole? Introduce fractions.

Need to take a square root of a negative number? Dream up imaginary numbers!

For example, the equation *x*^{2} = -1 cannot be solved by any real number. Why? Because if you square either a positive or a negative real number, the result is always positive. In particular, *x* = -1 is *not* a solution to the equation because (-1)^{2} = 1.

But if we could *imagine* a number like √-1, then we can solve the equation! Let’s give it a name: *i* for imaginary.

*i* = √-1

The trouble is: *i* is still *not* a real number! Oh well, just because something doesn’t exist, that’s no reason to discriminate against it. In fact, if we’re willing to think like a mathematician, then we can use *i* just like any other number if we’re careful and follow the rules.

For example, if you follow the basic rule: *i*^{2} = -1, then you get two solutions to the equation *x*^{2} = -1. Both *x* = *i* and *x* = –*i* work!

(±*i*)^{2} = -1

_{By SFerdon}

Up until the mid-sixteenth century, certain kinds of equations were impossible to solve. For example, some kinds of **cubic equations** (degree 3 polynomial equations) resisted all known methods at the time.

Then along comes NiccolĂ˛ Fontana (Tartaglia), who managed to crack the code of the cubic equation! He, along with his rival Gerolamo Cardano and Cardano’s student Lodovico Ferrari, were using strange “imaginary” numbers to find real solutions to the cubic.

The rest, as they say, is history!

So remember, a complex number has the form *a* + *bi*.

For example, -7 + 3*i* and 8.23 + (43/11)*i* are complex numbers.

There are a few basic rules and properties that we need to establish in order to work with these kinds of numbers.

**The Most Important Rule:***i*^{2}= -1**The Second Most Important Rule:**Purely real terms and purely imaginary terms are not like terms.

We’ve seen that first rule already, so let me say something about the second rule.

Essentially, you have to keep the real and imaginary parts separate at all times. The complex number -7 + 3*i* is NOT the same as -4*i*. Just think of *i* almost as a variable (like *x* or *y* in a polynomial expression).

If there is a negative under a radical, then chances are that you’ll need to use complex numbers to simplify it.

Just remember that the square root of -1 is equal to *i*.

And don’t forget to pull out any perfect squares while you’re at it!

**Example:** √-200 = √(-1)(100)(2) = *i*10√2 (or 10√2 *i* in standard form)

Complex numbers have to follow most of the same rules as real numbers, such as the commutative, associative, and distributive laws.

The rules for adding, subtracting, and multiplying follow pretty easily.

**Adding:**(*a*+*bi*) + (*c*+*di*) = (*a*+*c*) + (*b*+*d*)*i*.This formula just means we have to group like terms.

**Example:**(4 – 3*i*) + (-1 + 8*i*) = (4+(-1)) + (-3+8)*i*= 3 + 5*i***Subtracting:**(*a*+*bi*) – (*c*+*di*) = (*a*–*c*) + (*b*–*d*)*i*Again, just group like terms. Don’t forget to distribute the minus!

**Example:**(4 – 3*i*) – (-1 + 8*i*) = 4 – 3*i*+ 1 – 8*i*= 5 – 11*i***Multiplying:**(*a*+*bi*) × (*c*+*di*) = (*ac – bd*) + (*ad*+*bc*)*i*Instead of memorizing that formula, just think “FOIL” multiplication, and remember that

*i*^{2}= -1**Example:**(4 – 3*i*)(-1 + 8*i*) = (4)(-1) + (4)(8*i*) + (-3*i*)(-1) + (-3*i*)(8*i*)= -4 + 32

*i*+ 3*i*+ (-24)*i*^{2}= -4 + 35

*i*+ (-24)(-1)= -4 + 35

*i*+ 24= 20 + 35

*i***Dividing:**On the other hand, the rule for division is a bit complicated. You have to use the**complex conjugate**and then simplify your result completely.

The **complex conjugate** of a complex number *a* + *bi* is: *a* – *bi*.

When you divide (*a* + *bi*) / (*c* + *di*), you have to multiply both the numerator and denominator by the complex conjugate of the denominator and then simplify.

It helps to know the **difference of squares** formula by heart, as it will play a role in cleaning up the denominator.

(*x* + *y*)(*x* – *y*) = *x*^{2} – *y*^{2}

Let’s see how this works by example. Notice in the first line where the complex conjugate of the denominator has to be multiplied to the top and bottom.

As you might realize, there’s a lot more to be said about complex numbers! This article represents just the tip of a very large iceberg.

Complex numbers are used to describe the electromagnetic fields and waves that allow your cell phone to operate. They help to define the fundamental particles of our universe, such as the electron and proton. And they can even generate beautiful fractal images.

I hope that you have gained a better understanding of imaginary and complex numbers! Keep the basic rules and definitions in mind, and perhaps you will learn to love these fascinating and complicated mathematical tools as much as I do!

The post What Are Complex Numbers? appeared first on Magoosh Math.

]]>The post Math Practice: Working with Radical Expressions appeared first on Magoosh Math.

]]>So what exactly are radical expressions? Basically any mathematical expression that involves a **radical**. Radical comes from the Latin *radix*, which means *root*. So we think of square roots, cube roots, and even 100th roots of numbers.

An *n*th root of a number *x* is a number *y* such that the *n*th power of *y* gives you back *x*. *Square root* is just another name for the 2nd root. Does it sound confusing? Don’t worry, we’ll see how to use this definition in the examples below.

So for example, √9 = 3, because 3^{2} = 9. And = 2, because 2^{5} = 32.

See how radicals are like the opposite of powers? (Technically, we say that finding the *n*th root is the *inverse operation* to finding the *n*th power.)

**Radical expressions** may include variables or only numbers. The expression within the radical is called the **radicand**. The tiny number in the upper left of the radical symbol is the **index**. Square roots have index 2, implied but not written in the radical notation.

Radical expressions are equivalent to expressions with fractional powers. The rule is pretty simple:

In other words, any time you see an exponent like 1/*n*, then you can regard that as an *n*th root, and vice versa.

This is especially important when variables are involved, because it shows how roots essentially turn into division of exponents by the index.

For more about this point, as well as a list of practice problems, go check out Math Practice: Negative and Fractional Exponents.

Sometimes radical expressions can be simplified. The simplest case is when the radicand is a **perfect power**, meaning that it’s equal to the *n*th power of a whole number.

For example the **perfect squares** are: 1, 4, 9, 16, 25, 36, etc., because 1 = 1^{2}, 4 = 2^{2}, 9 = 3^{2}, 16 = 4^{2}, 25 = 5^{2}, 36 = 6^{2}, and so on. Therefore, we have √1 = 1, √4 = 2, √9 = 3, etc.

**Perfect cubes** include: 1, 8, 27, 64, etc. So, , and so on.

Variables with exponents also count as perfect powers if the exponent is a multiple of the index. For instance, *x*^{2} is a perfect square. But so are *y*^{4} and *w*^{18}.

The key concept is that an *n*th root of a perfect *n*th power will completely simplify to a non-radical expression.

For example, .

The square root of 81 is 9, while the square root of *x*^{6} is (*x*^{6})^{1/2} = *x*^{3}.

If the radicand is *not* a perfect power, then you still might be able to factor it into part that *is* a perfect power. Then you can simplify that part of it and leave the rest within the radical.

Let me show you what I mean.

Suppose you want to take the cube root of 24*x*^{4}. Well 24 is not a perfect cube… *but* one of its factors, 8, definitely is! Similarly, *x*^{4} is not a perfect cube, but we can factor it as *x*^{3}·*x*, and *x*^{3} is a perfect cube.

Notice how the cube root of 8*x*^{3} simplifies to 2*x*, which we then write *outside* of the radical expression (in green). The leftover 3*x* cannot simplify and must remain within the radical.

Finally, we have to discuss another method of simplifying radicals called **rationalizing the denominator**. This is a technique for rewriting a radical expression in which the radical shows up on the bottom of a fraction (denominator).

There are two basic cases to worry about. Also, for pretty much every problem in your Algebra class, you only have to do this for square roots. (Rationalizing roots with higher index can get a *lot* more complicated.)

- The denominator is a single term involving a radical. In this case, multiply both top and bottom of the fraction by the radical expression.
- The demoninator has two terms, like
*a*+ √b. Here, you must multiply both the top and bottom of the fraction by the**conjugate**,*a*– √b. And if you had*a*– √b to begin with, its conjugate would be*a*+ √b. (Just flip the sign of the radical.)

Here’s an example showing the technique in practice.

You may have noticed that we sometimes were able to simplify part of the radical expression by pulling out a perfect power.

** Be Careful!** This technique only works because we are dealing with

So you *may* do this: √36 · *x* = 6√*x*.

But you **MAY NOT** do this: √36 + *x* = 6 + √*x* (*Big No-no*).

In other words, radical expressions do not break apart over plus (+) or minus (–) signs.

Even the best students make this mistake from time to time. So just be aware and try to catch it early before it becomes a habit.

Ok, let’s see what you know about simplifying radicals!

Simplify each radical expression.

- 4√24 – 7√150

- Pull out the perfect squares.
4√4 · 6 – 7√25 · 6 = 4(2√6) – 7(5√6) = 8√6 – 35√6 = -27√6

- Multiply both the top and bottom by the radical.

- Don’t forget to multiply both the top and bottom of the fraction by the conjugate!

- Pull out perfect square factors.

The post Math Practice: Working with Radical Expressions appeared first on Magoosh Math.

]]>The post What Is the Binomial Theorem? appeared first on Magoosh Math.

]]>The Binomial Theorem is an important topic within the High School Algebra curriculum (*Arithmetic with Polynomials and Rational Expressions HSA-APR.C.5*). It also plays a significant role in college mathematics courses, such as Calculus, Discrete Mathematics, Statistics, as well as certain applications in Computer Science.

So, without further ado, let’s take a look at the Binomial Theorem!

Suppose

nis any natural number. Then

While the formula may look intimidating at first, there’s nothing to be afraid of. Let’s break it down!

The left side, (*x* + *y*)* ^{n}*, is an arbitrary

The right side is the formula for expanding (*x* + *y*)* ^{n}*.

- It’s a
**sum**(that’s what the “sigma” Σ symbol means) of certain kinds of terms. The second line of the formula shows how the sum expands explicitly. - Each term begins with a number called a
**binomial coefficient**. I’ve used one kind of notation in this formula, but you may have seen others.

Don’t worry too much if you don’t understand the formula… there are other ways to find these numbers! (For more about the binomial coefficients, you might want to start here: Binomial Coefficients.) - The next factors in each term are powers of
*x*and*y*. Notice that the two powers always add up to*n*. That is, (*n*—*k*) +*k*=*n*. This will be important as you work out problems.

Remember the rule, (*a* + *b*)^{2} = *a*^{2} + 2*ab* + *b*^{2}? Well, we can work this out now using the Binomial Theorem together with Pascal’s Triangle.

With *n* = 2, *x* = *a*, and *y* = *b* in the formula, we find:

The values of the binomial coefficients can be found using a Calculator (using the “_{n}C_{r}” function), or by *Pascal’s Triangle*.

You may be wondering why we used the Theorem in this case, since we already know a quick way to handle this situation, *FOIL multiplication*:

(*a* + *b*)^{2} = (*a* + *b*)(*a* + *b*) = *a*^{2} + *ab* + *ba* + *b*^{2} = *a*^{2} + 2*ab* + *b*^{2}

That’s all well and good for squaring (*n* = 2) and cubing (*n* = 3) binomials. But what about the 5^{th} power, or the 19^{th}, and so on? Multiplying out by hand would be way too time-consuming and error-prone for large exponents without the Binomial Theorem.

There is an amazing mathematical tool called **Pascal’s Triangle**, which easily generates all of the binomial coefficients!

Go get a blank page of notebook paper. Go on… I’ll wait. Now, on the top line of your paper, in the center, write a single “1.” Then write “1”s in two diagonals extending below your first “1” to the left and to the right. You can go down the page as far as you want to, but I stopped after seven more lines. Label each row, starting at the top with “*n* = 0,” *n* = 1,” etc.

Next, starting at row *n* = 2 (which is actually the 3^{rd} row!), the number that should fill in the blank space is found by adding the two numbers immediately above-left and above-right. That is, 1 + 1 = 2. Now your row *n* = 2 should look like:

1 2 1

Continue in this way down the triangle, always adding the two numbers above-left and above-right to generate the next number.

I’ve included arrows to show how each number is built from those above it:

The reason we care about Pascal’s Triangle is that its entries are exactly the values of the binomial coefficients, as long as you start counting at *n* = 0 down the rows and *k* = 0 across the diagonals.

So for example, look at row *n* = 4. The numbers are: 1 4 6 4 1. This means that:

Let’s do a few examples.

Expand: (3*x* + 2)^{3}

You *could* do this one by multiplying everything out…

However, let’s see how the Binomial Theorem works in this case.

First identify the power, *n* = 3. Then find the appropriate row in Pascal’s Triangle.

1 3 3 1

These numbers are the binomial coefficients for this example. Make sure you can spot where we use them in the computation below.

Also note that the original terms of the binomial are 3*x* and 2. Those items will replace *x* and *y*, respectively, in the formula.

Expand: (2*m* – 5*p*^{2})^{5}

Again, look for the correct row of Pascal’s Triangle. The power is *n* = 5.

1 5 10 10 5 1

You must also rewrite the subtraction as addition in order to properly use the formula.

(2*m* — 5*p*^{2})^{5} = (2*m* + (–5*p*^{2}))^{5}

Now with *x* = 2*m*, and *y* = –5*p*^{2}, we can proceed. (Note, this problem is *not* a good candidate for just multiplying out, as there would be five factors to deal with!)

Expand using the Binomial Theorem (binomial coefficients displayed in red in the second line). Be careful as you work out the exponents! (For a quick refresher on the rules of exponents, you might try: Quick Tips on Using the Exponent Rules.)

As you can see, the solutions tend to get very large as the power *n* gets larger. Sometimes, however, a problem may ask only for a single term of a complicated expression. The Binomial Theorem can be used to find just that one term without having to work out the expression completely!

Find the degree 9 term of (4x^{3} + 1)^{6}.

We can avoid working out the entire expression, by identifying which value of *k* corresponds to what’s being asked.

Here, we need the exponent on *x* to be 9. According to the formula, the exponent of term *k* would be:

3(6 — *k*) = 18 — 3*k*

Setting that equal to 9 and solving, we find that *k* = 3.

Thus, the appropriate term is:

The post What Is the Binomial Theorem? appeared first on Magoosh Math.

]]>