Trigonometry, Part 6

Remember what the term argument means? The argument of sin (2x + 7) is 2x + 7, that is an argument of function is the thing the function is operating on. In this case, the sine function is operating on 2x + 7.

So when the argument of a trig function is other than x, the basic method of solving an equation with this function is basically the same as what we did in the last post but an extra bit of algebra is required.

For example, solve 2sin(2x +πœ‹) -1 = 0 for 0 ≀ x ≀ 2πœ‹. The argument of the sine is 2x + πœ‹. We want to first, solve the equation just as we did before, but in terms of the argument 2x + πœ‹:

2sin(2x +πœ‹) -1 = 0 ⟹ sin(2x +πœ‹) = 1/2

From our table of common values, we see that the angle πœ‹/6 has a sine of 1/2. So the basic starting point to find all solutions is 2x +πœ‹ = πœ‹/6. To correctly limit the number of solutions, let’s convert the given domain 0 ≀ x ≀ 2πœ‹ to be in terms of 2x +πœ‹. To do this, algebraically change the inequality so that the middle expression is 2x +πœ‹:

0 ≀ x ≀ 2πœ‹ ⟹ 0 ≀ 2x ≀ 4πœ‹ ⟹ πœ‹ ≀ 2x + πœ‹ ≀ 5πœ‹

So we need to find all the angles between πœ‹ and 5πœ‹ that have a sine of 1/2. Again, a unit circle diagram will help. I will start with πœ‹/6, but realise that πœ‹/6 is not in the required modified domain:

So starting with the basic angle of πœ‹/6, I add multiples of 2πœ‹ until we are outside the given domain. For πœ‹/6, adding 2πœ‹ gives 13πœ‹/6. Adding 2πœ‹ again gives 25πœ‹/6. Adding another 2πœ‹ gives 37πœ‹/6, but 37πœ‹/6 is greater than 5πœ‹ so it is not a valid solution.

So far, the intermediate solutions are 13πœ‹/6 and 25πœ‹/6. The other basic angle found from the unit circle is 5πœ‹/6. Again, this angle is not in our modified domain so we need to add multiples of 2πœ‹ to find the angles within πœ‹ and 5πœ‹. So just like we did for πœ‹/6, adding 2πœ‹ to 5πœ‹/6 gives 17πœ‹/6. Adding another 2πœ‹ gives 29πœ‹/6. Adding another 2πœ‹ puts us outside the domain.

So we have four intermediate solutions:

2x + πœ‹ = 13πœ‹/6
2x + πœ‹ = 25πœ‹/6
2x + πœ‹ = 17πœ‹/6
2x + πœ‹ = 29πœ‹/6

To find the final solutions, we need to solve each of these equations for x:

2x + πœ‹ = 13πœ‹/6 ⟹ x = 7πœ‹/12
2x + πœ‹ = 25πœ‹/6 ⟹ x = 19πœ‹/12
2x + πœ‹ = 17πœ‹/6 ⟹ x = 11πœ‹/12
2x + πœ‹ = 29πœ‹/6 ⟹ x = 23πœ‹/12

A generic method for solving many trig equations is:

  1. Find the basic angles that solve the equation in terms of the argument of the trig function in the equation.
  2. Modify the given domain to be in terms of the argument.
  3. Add or subtract multiples of 2πœ‹ to the basic angles to find all angles within the modified domain. Keep in mind that the basic angles themselves may or may not be in the domain.
  4. Set the argument equal to all solutions and solve for the variable used in the problem (usually πœƒ or x)

There is another way to analyse trig equations which I will show on my next post.

Trigonometry, Part 5

Now let’s turn to solving equations with trig functions. Before we get into it though, let me introduce the inverse trig functions.

Just like square and square root, logs and exponentials, add and subtract, etc are inverse operations, there are corresponding inverse functions for the trig functions. These functions are identified usually by a -1 exponent. For example, sin-1(0.25) is asking the question: what angle has a sine of 0.25? Since sin-1x is an inverse function, then sin-1(sin x) = sin(sin-1 x) = x. That is, they undo each other. The same notation is used for the inverse cosine and tangent: cos-1x, tan-1x. Some older calculators may use the notation arcsin(arcsine), arccos(arccosine), and arctan(arc tangent), or they may further abbreviate these as asin, acos, atan. However, these are the same things as the corresponding inverse functions.

Solving trig equations, for the most part, use the same skills you already know for other types of algebraic equations. However, you need to be aware of the cyclic nature of trig functions and the varying signs of these functions in different quadrants, in order to get the correct and complete answers. For example, solve

√2Μ…cos(x) + 1 = 0, 0 ≀ x ≀ 2πœ‹

First notice that frequently, a domain of the equation is specified. This is the 0 ≀ x ≀ 2πœ‹ part. In most trig equations, where the unknown is in the trig expression, a domain needs to be specified or there will be an infinite number of solutions. By the way, the expression that is being acted upon by the trig function is called the argument of that function. For example the argument of sin(xΒ² + 7) is xΒ² + 7. In our example, the argument of the cosine is simply x.

So to solve this, you use your algebra skills to get,

\text{cos}( x) =-\frac{1}{\sqrt{2}}

So we can take the inverse cosine of both side and use our calculator to find x, and you would find that x =3πœ‹/4. However, let’s look at the equation before we take the inverse cosine. You may be asked to solve this without a calculator. You should notice that the cos(x) is equal to an entry in the table of common values I presented in my last post (except for the minus sign). It appears that our answers are associated with the angle πœ‹/4.

Now we have two trig identities that will solve this for us, namely

cos(πœ‹ – πœƒ) = -cosπœƒ
cos(πœ‹ + πœƒ) = -cosπœƒ

So, as we want the negative of cos(πœ‹/4), then the two angles that will solve this equation within the domain 0 ≀ x ≀ 2πœ‹ are πœ‹ – πœ‹/4 = 3πœ‹/4 and πœ‹ + πœ‹/4 = 5πœ‹/4. Notice that your calculator only gave one of these answers.

Instead of using the trig identities directly, I prefer to use the unit circle to guide me to all correct answers. I highly recommend drawing a unit circle for a particular problem to help guide you to the correct solutions. For this problem:

For practice, draw the unit circle for the same problem, with the provided domain as -πœ‹ ≀ x ≀ πœ‹. You should be able to see that the two answers would be -3πœ‹/4 and 3πœ‹/4.

I will do more examples in my next post.

Trigonometry, Part 4

I will now do some examples of using the trig identities covered in the previous posts. But before I do, I want to show you a table that gives the values of the main trig functions for common angles:

Angle, πœƒsinπœƒcosπœƒtanπœƒ
πœ‹/6, 30Β°1/2√3Μ…/21/√3Μ…
πœ‹/4, 45Β°1/√2Μ…1/√2Μ…1
πœ‹/3, 60°√3Μ…/21/2√3Μ…
πœ‹/2, 90Β°10Undefined
πœ‹, 180Β°0-10
3πœ‹/2, 270Β°-10Undefined

Now let’s do some examples:

  1. If cos(πœƒ) = 0.8829 and πœƒ is in the first quadrant, find cos(3πœ‹/2 – πœƒ).

According to the identity developed before, cos(3πœ‹/2 – πœƒ) = -sinπœƒ. But what is sin(πœƒ)? To find this, we need the Pythagorean identity, sinΒ²(πœƒ) + cosΒ²(πœƒ) = 1:

\text{sin}^{2} \theta +\text{cos}^{2} \theta \ =1\ \ \ \Longrightarrow \ \ \ \text{sin}( \theta ) =\sqrt{1-\text{cos}^{2} \theta }
\end{equation*}\] \[\begin{equation*}
\text{sin}( \theta ) =\sqrt{1-0.8829^{2}} =0.4696

Therefore, cos(3πœ‹/2 – πœƒ) = -sinπœƒ = -0.4696

2. If sin(πœƒ) = 0.1736, and πœƒ is in the first quadrant, find tan(3πœ‹/2 + πœƒ).

According to the identity developed before, tan(3πœ‹/2 + πœƒ) = -1/tanπœƒ = -cosπœƒ/sinπœƒ. Again, we need the Pythagorean identity to find cosπœƒ:

\text{cos}( \theta ) =\sqrt{1-0.1736^{2}} =0.9848

Therefore, tan(3πœ‹/2 + πœƒ) = -0.9848/0.1736 = -5.6729.

3. Find the exact value of cos(5πœ‹/6) (without a calculator).

The clues here are that I have been developing trig identities and I just gave you a table of exact values of trig functions for common angles. Perhaps we can equate 5πœ‹/6 in terms of a common angle. Notice that 6πœ‹/6 – πœ‹/6 = 5πœ‹/6. That is 5πœ‹/6 = πœ‹ – πœ‹/6. We have a trig identity for this: cos(πœ‹ – πœƒ) = -cosπœƒ. In our case πœƒ = πœ‹/6 and cos(πœ‹/6) = √3Μ…/2 so cos(5πœ‹/6) = -√3Μ…/2.

By the way, you can find a decimal approximation to √3Μ…/2, but this will just be an approximation no matter how many decimal places you include since √3Μ…/2 is an irrational number and has non-repeating decimals. So the exact value is √3Μ…/2 since we have agreed that √3Μ… is the notation for the exact square root of 3.

4. If sinπœƒ = 4/5 and πœ‹/2 < πœƒ < πœ‹, find the exact value of cosπœƒ

The part πœ‹/2 < πœƒ < πœ‹ means that the angle is in the second quadrant which means the cosine is negative. As before, we can find the cosine via the Pythagorean identity:

\text{cos}( \theta ) =\sqrt{1-(4/5)^{2}} =3/5

But as the angle is in the second quadrant, cosπœƒ = – 3/5.

Trigonometry, Part 3

The identities shown in my last post showed how measuring πœƒ from the negative x-axis affected its trig functions. Today’s post shows how measuring πœƒ from the y-axis affects its trig functions.

Consider the diagram below:

The angle πœƒ is shown measured from both the x and the y axes. The angle πœƒ measured from the y-axis is πœ‹/2 – πœƒ when conventionally measured from the positive x-axis. Notice that the right triangles formed from both measurements are identical – just in different orientations. The x coordinate of the πœ‹/2 – πœƒ angle on the unit circle, cos(πœ‹/2 – πœƒ), is the same as the y coordinate of the πœƒ angle, sinπœƒ. Similarly the y coordinate of the πœ‹/2 – πœƒ angle on the unit circle, sin(πœ‹/2 – πœƒ), is the same as the x coordinate of the πœƒ angle, cosπœƒ. This illustrates the following identities:

cos(πœ‹/2 – πœƒ) = sinπœƒ
sin(πœ‹/2 – πœƒ) = cosπœƒ
tan(πœ‹/2 – πœƒ) = sin(πœ‹/2 – πœƒ)/cos(πœ‹/2 – πœƒ) = cosπœƒ/sinπœƒ = 1/tanπœƒ = cotπœƒ

I’ve introduced another trig function here, the cotangent, abbreviated cot. The cotangent is the reciprocal of the tangent.

When solving equations involving trig functions (which we will eventually do), these and the following identities can be used to convert sines to cosines and vice versa.

Now let’s measure πœƒ from the other side of the y-axis. This gives us the conventional angle πœ‹/2 + πœƒ:

The only change here is that the x coordinate on the unit circle is now negative. So the resulting identities are:

cos(πœ‹/2 + πœƒ) = -sinπœƒ
sin(πœ‹/2 + πœƒ) = cosπœƒ
tan(πœ‹/2 + πœƒ) = sin(πœ‹/2 + πœƒ)/cos(πœ‹/2 + πœƒ) = cosπœƒ/-sinπœƒ = -1/tanπœƒ = -cotπœƒ

Now the angle to the negative y-axis is 3πœ‹/2 (270Β°). An angle measured to the left from the negative y-axis is the conventional angle 3πœ‹/2 – πœƒ. The coordinates of this angle on the unit circle are swapped and negative of the angle πœƒ in the first quadrant. So the picture looks like this:

And the corresponding identities are:

cos(3πœ‹/2 – πœƒ) = -sinπœƒ
sin(3πœ‹/2 – πœƒ) = -cosπœƒ
tan(3πœ‹/2 – πœƒ) = sin(3πœ‹/2 – πœƒ)/cos(3πœ‹/2 – πœƒ) = -cosπœƒ/-sinπœƒ = 1/tanπœƒ = cotπœƒ

I will leave it as an exercise for you to show that for an angle measured to the right of the negative y-axis, the corresponding identities are:

cos(3πœ‹/2 + πœƒ) = sinπœƒ
sin(3πœ‹/2 + πœƒ) = -cosπœƒ
tan(3πœ‹/2 + πœƒ) = sin(3πœ‹/2 + πœƒ)/cos(3πœ‹/2 + πœƒ) = -cosπœƒ/sinπœƒ = -1/tanπœƒ = -cotπœƒ

Next time, I will use the identities shown so far in some example problems.

Trigonometry, Part 2

Now let’s use the unit circle to see some of the common trig identities. These identities (rules) will be used in future posts.

Let’s assume we have an acute angle πœƒ. An acute angle is one that is between 0 and πœ‹/2 (or 0 to 90Β°). The following identities are valid for any angle, not just acute ones – it is just easier to see the logic in the diagram if we assume this.

The following picture shows the relationship between an angle πœƒ in the first quadrant, and an angle in the second quadrant which is symmetric with πœƒ:

You can see that to measure this symmetric angle from the postive x-axis, you just subtract it from πœ‹. The coordinates of the intersected point on the unit circle are negative for the x coordinate but the same y coordinate as the original angle πœƒ. So the following identities are evident from this picture:

cos(πœ‹ – πœƒ) = -cosπœƒ
sin(πœ‹ – πœƒ) = sinπœƒ
tan(πœ‹ – πœƒ) = -cosπœƒ/sinπœƒ = -tanπœƒ

Again, these are true for any angle, not just acute ones.

As an example, let πœƒ = πœ‹/3, (60Β°). The following is true for πœ‹/3:

cos(πœ‹/3) = 1/2
sin(πœ‹/3) = √3Μ…/2
tan(πœ‹/3) = √3Μ…

Now πœ‹ – πœ‹/3 = 2πœ‹/3. So using these identities, we know that

cos(2πœ‹/3) = -1/2
sin(2πœ‹/3) = √3Μ…/2
tan(2πœ‹/3) = -√3Μ…

Now let’s look at a symmetric angle in the third quadrant. To measure this angle from the positive x-axis, you add it to πœ‹. The corresponding coordinates of the intersected point on the unit circle are both the negative of the coordinates for πœƒ. So the following identities are shown in this picture:

So these identities are

cos(πœ‹ + πœƒ) = -cosπœƒ
sin(πœ‹ + πœƒ) = -sinπœƒ
tan(πœ‹ + πœƒ) = -cosπœƒ/-sinπœƒ = tanπœƒ

Using our same example, πœ‹ + πœ‹/3 = 4πœ‹/3. Using these identities:

cos(4πœ‹/3) = -1/2
sin(4πœ‹/3) = -√3Μ…/2
tan(4πœ‹/3) = √3Μ…

One more quadrant to go:

As was mentioned before, angles measured clockwise from the positive x-axis are negative. So the following trig identities are shown in the figure above:

cos(-πœƒ) = cosπœƒ
sin(-πœƒ) = -sinπœƒ
tan(-πœƒ) = cosπœƒ/-sinπœƒ = -tanπœƒ


cos(-πœ‹/3) = 1/2
sin(-πœ‹/3) = -√3Μ…/2
tan(-πœ‹/3) = -√3Μ…

There are a couple more identities I would like to show but I’ll save that for next time.

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:

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


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

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:

{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:

{\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:

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.