Trigonometry, Part 1

This begins many posts on trigonometry. Trigonometry is often called circular functions in some textbooks. I have found that this topic is troublesome for many of my students. I hope you find these posts useful.

Let’s begin with the unit circle. The below picture is from Wolfram MathWorld:

It is very important that you understand this circle as drawing it and analysing the specifics of the problem at hand will greatly add to your understanding of the problem.

The unit circle, as its name implies, is a circle of radius 1, centered at the origin of a cartesian coordinate system. A right triangle can be formed from a point on the unit circle by dropping a vertical line from the point down to the x-axis. The line from the origin to the point completes the triangle with an angle πœƒ from the positive x-axis. The x coordinate of the point is the length of the base of this triangle. From the basic definition of the trig function cos πœƒ:

\[\text{cos}\ \theta \ =\frac{\text{adjacent}}{\text{hypotenuse}} =\frac{x\ \text{coordinate}}{1} =x\]

Similarly,

\[\text{sin}\ \theta \ =\frac{\text{opposite}}{\text{hypotenuse}} =\frac{y\ \text{coordinate}}{1} =y\]

Now I will mainly use radians as the angle measure as this is most frequently used in science and engineering. Conventionally, angles are measured from the positive x-axis. They are positive if you go counter-clockwise and negative if you go clockwise. As 2πœ‹ radians are a full circle, multiples of 2πœ‹ can be added or subtracted from an angle and you get the same angle. For example πœ‹/4 + 2πœ‹ = 9πœ‹/4, πœ‹/4 – 2πœ‹ = -7πœ‹/4, πœ‹/4 + 4πœ‹ = 17πœ‹/4, πœ‹/4 – 4πœ‹ = -15πœ‹/4. (In degrees, this is the same as adding multiples of 360Β°). These are all the same angle and they intersect the unit circle at the same place, so their sines are all equal as well as their cosines.

At this point, we can make some observations about the signs of the cosine, sine and tangent of an angle based on what quadrant the angle falls in. Remember that the tangent of an angle, in terms of the sine and cosine, is sin πœƒ/cos πœƒ:

It is very important to be aware of the quadrant you are working in. When solving trigonometric equations, your calculator may give you an answer that is algebraically correct but is in the wrong quadrant based on the information from the problem at hand. I will illustrate this in future posts.

Looking the right triangle in the first figure above, remember the Pythagorean theorem that the sum of the squares of the two right angle sides equals the square of the hypotenuse. Applying this theorem to the right triangle in the figure, you get the most used trig identity, the Pythagorean Identity:

sinΒ²(πœƒ) + cosΒ²(πœƒ) = 1

You will probably use this enough so that you will automatically remember it.

In my next post, I will develop more trig identities which will be obvious by looking at the unit circle.

Complex Numbers, Part 5

Blast from the past: remember the quadratic formula

\[
x = \frac{-b \pm \sqrt{{b}^{2}-4ac}}{2a}
\]

and you would look at the discriminant (the bΒ² – 4ac part) to determine if there were 1, 2, or no solutions? You were told that if bΒ² – 4ac < 0, then there were no solutions. Well that was a lie (please don’t ask me about Santa Claus). It turns out that there are always solutions to a quadratic equation

axΒ² + bx + c = 0, where a, b, and c are real and a β‰  0.

Let’s look at zΒ² + 4z + 29 = 0 and solve for z. Using the quadratic formula, we get

\[z = \frac{-4 \pm \sqrt{{4}^{2} – 4(1)(29)}}{2(1)} = \frac{-4 \pm \sqrt{-100 }}{2}\]

The square root of -100 used to be a problem but no longer:

\[\frac{-4 \pm \sqrt{-100 }}{2} = \frac{-4 \pm \sqrt{100 }\sqrt{-1}}{2} = = \frac{-4 \pm 10i}{2}= -2 \pm 5i \]

Notice that the answer is a complex conjugate pair. This always happens for any polynomial equation with real coefficients. If a complex number is a root, so is its conjugate. The answer above also means that

zΒ² + 4z + 29 = (z + 2 – 5i)(z + 2 + 5i)

So now if bΒ² – 4ac < 0, the quadratic equation has a conjugate pair as the solution.

If solutions to polynomial equations are complex conjugates, that means that a cubic equation has to have at least one real root because a cubic equation can have at most, three solutions. In fact, any odd powered polynomial equation has to have at least one real root. This makes sense graphically because the highest power of an odd-powered polynomial will eventually dominate all the other terms as x goes very negative or very positive. The highest powered term will have opposite signs for negative and positive values of x, so the polynomial will eventually cross the x-axis.

Let’s do one more problem. If a polynomial has single roots 2 Β± i and 3, what is that polynomial? To solve this, just form the factors with these roots:

(z -2 –i)(z -2 +i)(z -3) = (zΒ² – 4z + 5)(z -3) = zΒ³ – 7zΒ² + 17z – 15

Complex Numbers, Part 4

Now let’s do some maths with complex numbers. First, notice what happens with successive powers of i:

i0 = 1 (anything to the 0 power is 1)
i1 = i
i2 = -1 (by definition)
i3 = i2 Γ— i = –i
i4 = i2 Γ— i2 = 1

If you keep increasing the power, you will keep getting the pattern:
1, i, -1, –i. For higher powers, you can find where you are in the pattern by dividing the power by 4 and looking at the remainder. If the remainder is 0, then the answer is 1. If the remainder is 1, the answer is i. A remainder of 2 gives -1, and a remainder of 3 gives –i. For example, i69 , 69Γ·4 = 17 with a remainder of 1. So i69 = i.

Now lets’ convert between rectangular and polar forms.

Convert the following to polar form:

\[\begin{array}{l}
{z}_{1} = 1 \,- \sqrt{3}i\\
{z}_{2} = -1+ \sqrt{3}i\end{array}\]

First let’s find the modulus r. Both of these complex numbers have the same modulus because of the squaring of each term:

\[
{r}_{1} = \sqrt{{1}^{2}+{(-\sqrt{3})}^{2}} = {r}_{2} = \sqrt{{(-1)}^{2} + {(\sqrt{3})}^{2}} = \sqrt{4} = 2\]

Now let’s compute Arg(z) for each. Using my calculator, I get the same answer for both:

\[\begin{array}{l}
{\theta }_{1} = \text{tan}^{-1}{\left(\frac{-\sqrt{3}}{1}\right)} = -\frac{\pi }{3}\\
{\theta }_{2} = \text{tan}^{-1}{\left(\frac{\sqrt{3}}{-1}\right)} = -\frac{\pi }{3}\end{array}\]

The second one is wrong. In the first one, the real part is positive and the imaginary part is negative. This puts the complex number in the fourth quadrant in an Argand plot, so -πœ‹/3 is the correct angle. But in the second one, the real part is negative and the imaginary part is positive. This puts the complex number in the second quadrant, so clearly -πœ‹/3 is wrong:

Your calculator does not know where the minus sign is so it’s just programmed to give an answer in the range -πœ‹/2 to πœ‹/2, (-90Β° to 90Β°). You however, being smarter than your calculator, know that the minus sign in z2, puts it in the second quadrant. You need to adjust the given angle to put it in the second quadrant as illustrated.

So the polar form of z1 is 2 cis (-πœ‹/3). The polar form of z2 is 2 cis 2πœ‹/3.

Now let’s go from polar to rectangular. Convert

\[4\sqrt{2}\text{cis} \left (-\frac{3πœ‹}{4}
\right )\]

into its rectangular form.

In my last post I gave the equations

\[x = r \text{ cos} πœƒ, y = r \text{ sin} πœƒ\]

So z =4√2cos(-3πœ‹/4) + 4√2 sin(-3πœ‹/4)i = -4 – 4i. You do not need to worry about quadrant issues converting from polar to rectangular.

In my next post, I will solve some quadratic equations that previously did not have any solutions when we were stuck in the real domain. Now life is more complex.

Complex Numbers, Part 3

Complex numbers cannot be plotted on the traditional Cartesian coordinate system because of the imaginary part. A cartesian system is used to plot a pair of real numbers. Since a pair of complex numbers consists of four real numbers, a pair of complex numbers cannot be plotted. However, we can plot a single complex number on a modified version of a Cartesian system where the x-axis is relabelled as the real axis and the y-axis is the imaginary axis. This modified coordinate system is called an Argand diagram after its inventor Jean-Robert Argand, a Swiss mathematician.

Here is a plot of a generic complex number z = x + yi:

An Argand plot can be used to see the results of adding and subtracting complex numbers. These are interesting and you can see these in any textbook but the real power of an Argand plot is to see the relationship between the rectangular form of a complex number (the form we have been using so far) and the polar form.

Remember that polar coordinates are a different way to identify the location of a point in a two-dimensional plane. Instead of giving x and y coordinates, you can give an angle from the positive x-axis to define a direction, then a distance along that direction to define where a point is located. The same is true for an Argand plot of a complex number, and this polar form is more convenient to use than the rectangular form in certain scenarios. Below is the same plot as above, but I’ve added the polar coordinates:

From your past experience with triangles and trigonometry, you can see that the following relationships are true:

\[\begin{array}{l}
r = |z| = \sqrt{{x}^{2}+{y}^{2}}\\
x = r \text{ cos} πœƒ, y = r \text{ sin} πœƒ\\
πœƒ =\text{tan}^{-1}\frac{y}{x}\end{array}\]

From the above, you can see that in terms of r and πœƒ,

z = r cosπœƒ + ir sinπœƒ

The polar form notation that is used in VCE Australia is

z = r cisπœƒ

where “cis” is shorthand for “cos + i sin”. There are other ways to express a polar complex number as well. In electronics, rβˆ πœƒ is frequently used.

Some definitions: modulus is the length of a complex number, that is, r. Arg(z) is the principal value of πœƒ. This is pronounced as “the argument of z“. As any particular πœƒ can have multiples of 2πœ‹ radians or 360Β° added to it, arg(z) restricts πœƒ to -πœ‹ < arg(z) ≀ πœ‹. Be careful when computing πœƒ with the inverse tangent formula. A calculator will only give values between -πœ‹/2 and πœ‹/2 (-90Β° and 90Β°), that is the first and fourth quadrants. You need to interpret this result to the correct angle by looking at the signs of x and y to know which quadrant your particular complex number falls in. I will do some sample problems in my next post to illustrate this.

Complex Numbers, Part 2

Addition, subtraction, and multiplication of complex numbers extend the rules for real numbers. Division is done a bit differently but still follow rules you already know for real numbers.

Addition: To add two complex numbers, just add the real parts and the imaginary parts separately. Example:

(2 – 5i) + (-3 +2i) = (2 – 3) + (-5 + 2)i = -1 – 3i

Subtraction: same as addition, you just separately subtract the real parts and the imaginary parts. Example:

(2 – 5i) – (-3 +2i) = (2 – (-3)) + (-5 – (+2))i = 5 – 7i

Multiplication: Just as for real algebraic expressions like (x + y)(2x + 3y), you just multiply two complex numbers the same way then add like terms. Just remember that iΒ² = -1 so a negative sign will appear when i is multiplied by i. Example:

(2 – 5i) Γ— (-3 +2i) = 2 Γ— (-3) + 2 Γ— 2i -5i Γ— (-3) – 5i Γ— 2i
= -6 + 4i + 15i -10iΒ² = -6 + 19i -10(-1) = 4 + 19i

The same method applies when multiplying a real number times a complex number:

2 Γ— (2 – 5i) = 4 – 10i

Before I talk about division, I need to introduce another definition. In my last post, I solved a quadratic equation with the solution -2 Β± 3i. So there are two solutions, one with the + and the other with the -. These two solutions are called a complex conjugate pair. -2 + 3i is the complex conjugate of -2 – 3i and vice versa. It turns out that if a polynomial equation has a complex solution, the conjugate of that solution is also a solution. That is, complex solutions to polynomial equations always come in complex conjugate pairs. If z is a complex number, zΜ„ is used to represent its conjugate.

So now let’s look at complex division. To divide by a complex number (which include real numbers) by a complex number with a non-zero imaginary part, multiply the numerator and denominator by the conjugate of the denominator. Example:

\[(2-5i)\div (-3+2i)=\frac{2-5i}{-3+2i}\times \frac{-3-2i}{-3-2i}=\frac{-6-4i+15i-10}{9+6i-6i+4}\] \[=\frac{-16+11i}{13}=-\frac{16}{13}+\frac{11}{13}i\]

This example illustrates a few more things about complex numbers. First, at the end, dividing a complex number by a pure real number (or multiplying for that matter), you just divide (or multiply) each part of the complex number by the real number.

The other thing you may have noticed is in the multiplication in the denominator. Multiplying a complex number by its conjugate results in a real number. This resulting number has graphical significance which you will see later. In general, if z = x + yi, then zΜ„ = xyi and z zΜ„ = xΒ² + yΒ².

In my last post, I found that -2 Β± 3i are solutions to zΒ² +4zΒ + 13 = 0. Let’s check one of these solutions:

(-2 + 3i)Β² + 4(-2 + 3i) +13 = 4 -12i -9 – 8 + 12i + 13 = 0 + 0i

I’ll leave it to you to show that the conjugate also solves the equation.

Note that two complex numbers are equal only if their real parts are equal and their imaginary parts are equal.

In my next post, I’ll look at how we can plot complex numbers.

Complex Numbers, Part 1

Complex numbers extend the real numbers (ℝ) to include a second element, yi:

z = x + yi

where x and y are real numbers and i is the imaginary unit that is defined in terms of its square, iΒ² = -1. Yes, this does mean that i is the square root of -1. All this time you’ve been told that taking the square root of a negative number is illegal. It is if your calculations are limited to being real numbers, but with this special definition of i, it is now legal. Welcome to the world of complex numbers! The symbol for the set of all complex numbers is β„‚.

With this new definition and this new domain of numbers, there are new definitions and rules for operations. Operations on complex numbers are mostly extending the rules you already know for real numbers. First, let’s look at some definitions.

Just as x is the default variable used for an unknown real number, z is usually used for complex numbers. If z = x + yi, x is called the real part of z and y is called the imaginary part. There is shorthand notation for this:

Re(z) = x, Im(z) = y

where x and y are real numbers: x ∈ ℝ and y ∈ ℝ.

Now if Im(z) = 0, then you just have a real number. So real numbers are a subset of complex numbers. That is ℝ βŠ‚ β„‚

So were complex numbers created just to give maths students something to do? Absolutely not! (Though it would be nice to have some quiet time by giving my students complex problems). Since there is an extra number in a complex number, there is additional information there that can describe a physical quantity. Most often in science and engineering, this extra information relates to a phase angle or a frequency. I will show examples of this in future posts. Though the imaginary part of a complex number is called “imaginary”, a complex number can describe a real physical quantity.

Expanding our working maths domain to complex numbers, allows for solutions to equations that previously had “no solution”. Consider

xΒ² +4x + 13 = 0

This equation would previously be said to have no solutions. But now, we can find complex solutions. So perhaps we should change the variable to

zΒ² +4z + 13 = 0

Using the quadratic formula to solve this gives

\[ z = \frac{-b \pm \sqrt{{b}^{2}-4ac}}{2a} =\frac{-4 \pm \sqrt{{4}^{2}-4(1)(13)}}{2(1)} = \frac{-4 \pm \sqrt{-36}}{2} \]

Now -36 = (-1)(36) and the rule that the square root of a product is the product of the square roots applies:

\[z =\frac{-4 \pm \sqrt{-36}}{2} = \frac{-4 \pm \sqrt{36}\sqrt{-1}}{2} = \frac{-4 \pm 6i}{2} = -2 \pm 3i\]

Before we can confirm that -2 Β± 3i is a solution to the equation, we need to know how to do basic operations on complex numbers. That will be the subject of my next post.

VCE Maths

I know it’s been a while since my last post. I have not forgotten – I am taking the blog site in a slightly different direction and I am finding out how to reconfigure the website to accomodate this.

Future topics will specifically cover VCE maths. For my US friends, this is the maths covered in the Junior and Senior years of high school. Here in Australia (in the state of Victoria), VCE maths contribute to a score that has a lot of bearing on which university you can go to — similar to the SAT in the US. For that reason, topics covering VCE maths will be more useful to my local readers. However, these topics will also be covered in the US, just organised differently. The categories and tags structure will identify the topic.

Functional Notation, Part 2

Last time we saw that we can replace y in an equation with f(x) when y is alone on the left side of an equation:

y = f(x) = 3xΒ² – 5x + 1

The above is an example of the function definition. Once defined, you replace all the x‘s on the right side with whatever is in the brackets on the left side, even if it is not a number. For example,

f(2) = 3(2)Β² – 5(2) + 1 = 3
f(a) = 3aΒ² – 5a + 1

Even if the thing in the brackets is another expression, for example, an expression that is used in calculus a lot is x + h:

f(x+h) = 3(x + h)Β² – 5(x + h) + 1

And you can even use another function of x inside the brackets of another function. Like x, the letter f is used in the first instance for a function, but if other functions need to be defined as well, other letters are used:

f(x) = 3xΒ² – 5x + 1
g(x) = xΒ² – 7
f[g(x)] = f(xΒ² – 7) = 3(xΒ² – 7)Β² – 5(xΒ² – 7) + 1
g[f(x)] = g(3xΒ² – 5x + 1) = (3xΒ² – 5x + 1)Β² – 7

The domain of a function is all the valid values of x that can be used. Many times, the domain of a function (like f(x) and g(x) above) is just any real number. But there are functions where you cannot use just any number. For example, consider

\[
{f}{(}{x}{)}\hspace{0.33em}{=}\hspace{0.33em}\frac{3}{{x}{-}{2}}
\]

There is one value of x you cannot use. That value is 2 because that will make the denominator 0, and as you know, this will bring the maths police to your door. So the domain of this function is all real numbers except for 0.

Now consider

\[
{f}{(}{x}{)}\hspace{0.33em}{=}\hspace{0.33em}\sqrt{{x}{-}{2}}
\]

Another illegal operation is taking the square root of a negative number. The requirement for this function is that x – 2 has to be 0 or greater. For this to be true, x must be greater than or equal to 2. The phrase ” greater than or equal to” can be replaced by the maths symbol β‰₯. So the domain of this function is x β‰₯ 2.

There are other reasons why the domain of a function is restricted, but the most common things to look for is dividing by 0 or taking the square root (or any even root) of a negative number.

Functional Notation, Part 1

Any maths beyond algebra relies on something called functional notation. I used this in one of my posts on Newton’s Laws but more needs to be said if you are to be comfortable with it.

We have looked at many equations to date. Most involved x and y. For example,

y = 3xΒ² – 5x + 1

This will plot as a parabola on an xy coordinate system. The plot is a picture of all (x, y) pairs of numbers that, when substituted in the above equation, will result in a true statement. For example, (0, 1) would be a point on that parabola because 1 = 3(0)Β² – 5(0) + 1 = 1.

Most of the equations we have looked at have y on the left side and all the other things with x and stand-alone numbers are on the right side. In this form, it is easy to choose a number to replace x with, then do the maths with that number on the right side to find the corresponding y to make the equation true. For example, let’s choose “1” in place of x. The corresponding y will be

y = 3xΒ² – 5x + 1 = 3(1)Β² – 5(1) + 1 = 3 – 5 + 1 = -1

So (1, -1) is also a point on this equation’s plot.

This is frequently done: choose a value for x, then replace x with that value and do the maths on the right side and find the corresponding y. Notice that we are free to choose a value for x, but once we do, the value for the corresponding y is fixed. For that reason, x is called the independent variable and y is called the dependent variable: y depends on the x we choose.

If y depends on the x we choose, then another way to say this is that “y is a function of x“. The new functional notation makes use of this by replacing y with f(x), read “function of x“. So the functional form of the equation above is

f(x) = 3xΒ² – 5x + 1

This will plot exactly the same but we would replace the y label on the vertical axis with f(x):

So now it is easy to ask the question “What is the value of the function if x = 0″ by just replacing the “x” in f(x) with “0”, that is, f(0):

f(0) = 3(0)Β² – 5(0) + 1 = 1

So, f(0) = 1. We saw above that f(1) = -1. So now you will see the general coordinate as (x, f(x)) instead of (x, y). This is just a difference in notation – the plot stays the same.

There are some properties of functions and a few more definitions that will be explored next time.

Financial Maths, Part 5

Last time we saw that our $1000 invested at 3% compounded annually will result in $1343.92 in 10 years. To get even better results, the interest rate can be compounded more frequently than annually. Let’s say the interest is applied mid-year. Then the interest earned during the first half of the year will be added to the principal amount and that total will be used to apply the second half interest to.

This will change the formula AnΒ =Β A0(1 +Β r)n a bit though. First of all, we can’t apply the full per annum interest rate to the initial investment as only half a year has passed. So only half the interest rate will be used. Also, the period length is now half a year so the number of periods refers to how many half-years have passed.

So in our example, the annual interest rate is 3%, so the 6 month interest rate, r, is 3/2 = 1.5% since there are 2 six-month periods in a year. If we want to know how much we will have after 10 years, the number of periods, n, is now 10Γ—2 = 20 since there are 20 half -year periods in 10 years. So we can use the same formula with r = 1.5% = 0.015 and n = 20:

A20Β =Β A0(1 +Β r)20Β = 1000(1 + 0.015)20Β = $1346.86

So the extra compounding has made us a bit more money.

You might ask (OK – I’ll ask for you), would we make more money by compounding more frequently? Yes we would!

Let’s compound every quarter-year. this means the interest rate we apply each quarter is 3/4 = 0.75% and the number of periods after 10 years is 10Γ—4 = 40:

A40Β =Β A0(1 +Β r)40Β = 1000(1 + 0.0075)40Β = $1348.35

Looks like we want more yet. What about monthly? Here r = 3/12 = 0.25% = 0.0025 and n = 10Γ—12 = 120:

A120Β =Β A0(1 +Β r)120Β = 1000(1 + 0.0025)120Β = $1349.35

Better, but notice that this is not much better that quarterly. It appears that we will reach a limit as to how much we can make. Let’s try compounding daily. Here r = 3/365 = 0.0082% = 0.000082 and n = 10Γ—365 = 3650:

A3650Β =Β A0(1 +Β r)3650Β = 1000(1 + 0.000082)3650Β = $1349.84

Well that’s disappointing. There is only a 0.49 difference between compound monthly and daily after 10 years.

There is a maths formula that computes the amount of interest if the investment is compounded continuously. This is the limit of what you can make by compounding. For our example, after 10 years, the most compounding will get us after 10 years is $1349.86, just 2 cents more than compounding daily.