Quantum Relativity

Calculus Chapter 5

By Mark Lawrence

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Well, we're all done with derivatives for a little while. We're going to get off onto a side topic now - different ways of doing arithmetic - then we'll return and see how to do calculus on these new types of "numbers."

We're going to start with a quick review of the square root. The square root of 1, √1, is 1. This means 1*1 = 1. √4 is 2, that is, 2*2 = 4.  √2 is what's called an irrational number. Irrational does not mean that √2 has a short temper and is subject to inexplicable fits of rage. Irrational means you cannot express √2 as the ratio of two numbers, like 3/2 or 17/12.  √2 is the number 1.4142135623730950488016887242097. . ., which means that 1.414 * 1.414 = 1.999 is very close to 2, and as we use more and more of the series we get closer and closer to 2. The sqrt( A*B ) = sqrt( A) * sqrt( B ). That is, sqrt( 9*25 ) = sqrt( 9 ) * sqrt( 25 ) = 3*5.

Now comes an interesting exercise: what is √-1? That is, if i*i = -1, what is i? Well, this question has no answer, so we just assume that this number exists and call  √-1 = i. i*i = -1. That's pretty much all there is to this chapter. From here on, we're going to look at this in more detail, but there will be no more revelations.

Now we can take the square root of any negative number, because any negative number, say -121, is -1 * 121, that is -1 * a positive number, so the square root of -A = i sqrt( A ).

Here's some stuff you can do with i:

i*i = i2 = -1.
i*i*i = i3 = i*i * i = -i.
i*i*i*i = i4 = i*i * i*i = -1*-1 = 1.
1 / i = i4 / i = i3 = -i.
-i * -i = i3 * i3 = i6 = i4 * i2 = 1 * -1 = -1.

Of course, just as -2 * -2 = 4, -i * -i = -1.

At this point it would be very natural to ask "What's the square root of i?" Well, i has a square root, but we're not quite ready to talk about that just yet. Soon.

Any number that has i as part of it is called an imaginary number. There's sortof only one imaginary number, we call it i, and i*i = -1. We can make other imaginary numbers, like 6i or 17i, and we can multiply them to get i*i*6*17 = -102. We can add them to get 6i + 17i = 23i. We can subtract them to get 6i - 17i = -11i. We can divide them to get 6i / 17i = 6/17. Ok, that's about it for imaginary numbers.

These numbers are called imaginary because in the dark ages new thoughts were viewed with much suspicion. At late as 1600, negative numbers were considered unreal, and all calculations which involved negative numbers were re-done using only positive numbers before making the calculation public. In fact, to this day accountants continue with their several hundred year old tradition of putting negative numbers in brackets rather than writing a negative number.

The idea of the square root of one of these unreal numbers was considered simply outrageous. Euler (pronounced like Oiler) invented the notation i in about 1777, almost 100 years after Newton and Leibnetz invented calculus. Gauss and DeCartes in about 1800 wrote enough about imaginary and complex numbers that everyone else decided it was time to accept them as numbers. The word "imaginary" is left over from a time before about 1800, and is generally considered an unfortunate choice of name for these numbers.

We could ask the question, "Do imaginary numbers actually exist?" Well, since we live in a quantum universe, it's reasonable to say that maybe only integers like 3, 5, 7 exist, maybe even fractions like ½ don't "exist," exist in the sense that there's something you can point to and say, "That's a ½." We're good with irrational numbers like √2, even though we can't actually write the number down and we're not certain exactly what value it has. We're good with using a symbol like X to represent an unknown number or quantity. We're good with a symbol like ex, which represents an infinite number of things to compute, even though we only live a finite amount of time in what is very likely a finite universe. We're good with using a symbol like d/dx to represent a mathematical operation, which operation likely has no physical meaning. All things considered, "i" is just one more symbol for a mathematical idea, which seems to me to be no more or less imaginary than pretty much anything else in this book. You can judge for yourself, I expect. Well, that's enough philosophy, back to, um, reality.

Imaginary numbers get a lot more interesting when we add a real number to them - as in 6 + 12i. A number of this form, which is a real number added to an imaginary number, is called a complex number. Of course, a complex number could be 0+3i, so the real part of this complex number is zero. We call this number pure imaginary. Or, a complex number could be 5+0i, so the imaginary part of this complex number is zero. We call this number real. Thus, we see that the complex numbers include both the real numbers and the imaginary numbers.

Why are we doing this complex number stuff? Well, all of electrical engineering is done with complex arithmetic. All optics calculations are done with complex arithmetic. All quantum mechanics is done with complex arithmetic. So, we study complex numbers.

We can add two complex numbers, and we get another complex number as the result. So, (6+7i) + (9+11i) = (6+9) + (7+11)i = 15 + 18i. That is, we add the real parts, and we separately add the imaginary parts. We can subtract two complex numbers, so (6 + 7i) - (9 + 11i) = (6-9) + (7-11)i = -3 - 4i. We can multiply two complex numbers. This is a little tricky, but makes perfect sense after you think about it for just a second. To multiply two complex numbers, we multiply in all possible combinations, so (6 + 7i) * (9 + 11i) = 6*9 + 6*11i + 7i*9 + 7i * 11i = 54 + 66i + 63i - 77 = -23 + 129i. This makes perfect sense, because if we were multiplying two sums of real numbers, we'd do the same thing: (6 + 7) * (9 + 11) = 6*9 + 6*11 + 7*9 + 7*11 = 54 + 66 + 63 + 77 = 120 + 140 = 260 = (6 + 7) * (9 + 11) = (13) * (20) = 260. The only trick when multiplying complex numbers is to remember that i*i = -1.

What do complex numbers look like? This was a key point in the evolution of complex numbers - DeCartes and Gauss started graphing them, and this turned everyone into believers. Here's how we graph them. We use the X axis to represent the real part of a complex number, and the Y axis to represent the imaginary part of a complex numbers. So, 3 + 2i is a point on the graph which is out 3 units in the X direction, and up 2 units in the Y direction. Below in figure 5.1 I've graphed a few complex numbers for you.

Figure 5.1

Now we need what is known as a metric - that is, we need to know how to calculate the distance between points. On a normal piece of graph paper, we use the Pythagorean Theorem to calculate distance. We will call distance ds. The Pythagorean Theorem tells us that on normal graph paper, that is in Euclidean space, ds2 = x2 + y2. Let's try it out for the point (0,i). 02 + i2 = -1. This is a real problem, as there's no such thing as negative distance2. We have to fix this problem somehow.

So, we invent a new idea - it's called complex conjugation. If we have a complex number, call it C, which is (a+bi), then its complex conjugate is called C*, and is (a-bi). "C*" is read aloud as "C-star." That is, to conjugate a complex number, we change the sign of the imaginary part. So, the complex conjugate of (2+3i), which we write as (2+3i)*, is (2-3i).

Now, we can have what's called a positive-definite metric, that is, distance is never negative. We'll take our point 0+i, and we'll define the distance squared to the point as ds2 = (0+i) (0+i)* = (0+i) (0-i) = +1. Notice that since we're using "*" to mean complex conjugation, I can no longer use it for multiplication everywhere I might like to. So, we're going to have to get used to the idea that we're just always multiplying things that are standing around next to each other. Now we can try a few more points. For example, the distance 2 from the origin to (3+2i) is (3+2i) (3+2i)* = (3+2i) (3-2i) = 3*3 - 3*2i + 3*2i - 2i*2i = 3*3 + 2*2 = 13. Here, we see that all the imaginary terms magically canceled out or had their signs fixed up properly. So, this complex conjugation stuff is just what we needed.

Just to be sure, let's try one more point, (-2-2i). ds2 = (-2-2i) (-2-2i)* = (-2-2i) (-2+2i) = -2*-2 - 2*2i + 2*2i - 2i*2i = 8.

This distance stuff is so important that we have a special symbol for it. The distance this time, not the distance squared, ds = | A |. This notation, | A |, is read out loud as "the absolute value of A" or "the modulus of A", and it means sqrt( A* A ), that is, the square root of the complex conjugate of A times A.

Now we're ready to learn how to divide complex numbers. Actually, we can't do this. But, we can cheat. If we want A/B, where both are complex numbers, we can multiply the top and the bottom by B*, the complex conjugate of B.  So, for example,

(3+2i) (3+2i) (-2+2i)  -3-4 -4i + 6i -7+2i

  = -7/8 + i/4
(-2-2i) (-2-2i) (-2+2i) 8 8

Now we're going to consider complex numbers in what are called polar coordinates. To convert a point on a graph to polar coordinates, we find how far the point is from the origin, and the angle made between the point and the origin. The distance from the origin is called r, for radius, and the angle with the x-axis is commonly called θ. See Figure 5.2 below.

Figure 5.2

For the point (3+2i), the distance from the origin, r, is | 3 + 2i | = sqrt( 3*3 + 2*2 ) = sqrt( 13 ) = 3.6. θ, the angle with the x-axis, is 33.7°. To find the angle, we use the arc-tangent function on a calculator. So, this complex number can be represented as (3+2i), or as 3.6 @ 33.7°. They mean the same thing, which is the point shown above.

Figure 5.3

Now we're going to do a quick review of trigonometry. In high school, when they teach this stuff, there's about a million formulas to memorize. Well, we get to do it now with complex arithmetic. This is going to make it really easy - there's only one thing to memorize, and it's pretty simple. Above, in Figure 5.3, there's a triangle. The three sides are labeled SA for Side Adjacent, SO for Side Opposite, and HYP for Hypotenuse. This is a right triangle, and one of the angles is labeled θ. Here's what you have to memorize. Get a piece of paper, and in one column towards the right half of the paper, double spaced, write "SO SA SO SA HYP HYP" down the page. Now, draw a line under each of the six words. Now, starting at the bottom going up, write the same thing, "SO SA SO SA HYP HYP." Now, starting at the top, in another column to the left, write "Sine Cosine Tangent Cotangent Secant Cosecant." Actually, I'm usually lazy and write "Sin Cos Tan Ctn Sec Csc." Anyway, you have to memorize how to draw this table. This table defines the six trigonometric functions. I've typed it in below in Table 5.1.

Table 5.1
The Trigonometric Functions
Sine SO

Cosine SA

Tangent SO


Secant HYP



For our angle θ above, the sine of θ is defined as the length of the size opposite to θ divided by the length of the hypotenuse. The length of the hypotenuse is | 3 + 2i | = 3.6, and the length of the side opposite is 2, the imaginary part of (3+2i). So, the sine of θ is 2 / 3.6, about .556. The cosine of θ is the side adjacent divided by the hypotenuse. The side adjacent is the real part, which is 3, so the cosine of θ is 3 / 3.6, about .833. Now, if we've done everything correctly, sine2 + cosine2 = 1, and sure enough, .5562 + .8332 = 1.00. It's obvious from our chart that sine2 + cosine2 = 1, because SO2/HYP2 + SA2/HYP2 = (SO2 + SA2) / HYP2 = HYP2 / HYP2 = 1.

There's also these other functions, Tangent, Cotangent, Secant, and Cosecant. Tangent is used only rarely. Cotangent is pretty much never used. I don't even know what Secant and Cosecant are for. So, as far as I'm concerned, you don't have to worry about them. They won't be on the final.

This is calculus, so we work most of the time in radians. There are 2π radians in a circle, just as there are 360 degrees in a circle. A right angle is 90 degrees or π/2 radians. A straight line is 180 degrees or π radians. Radians = π / 180 * degrees. Degrees = 180 / π * radians. 180 / π = 57.3, that is, there are about 57 degrees in a radian.

Sine and Cosine are functions which oscillate forever - they look like waves. Below is a picture of Sine (dashed line) and Cosine (solid line). Cosine( 0 ) is 1; Sine( 0 ) is 0.

Figure 5.4 - Sine (dashed) and Cosine (solid)

The Tangent function is SO/SA. You can easily convince yourself this is Sine / Cosine. Since Cosine is zero sometimes, the Tangent function is infinity sometimes. Below is a graph of the Tangent function. The Tangent function is zero for angles which are an even multiple of π, that is, an even multiple of 180 degrees. Tangent is infinity for angles which are an odd multiple of π/2, that is 90 degrees away from the zeros. Just before π/2, Tangent is nearly infinity. Just after π/2, Tangent is nearly negative infinity. Tangent is not defined for precisely π/2, but we can feel certain that whatever value it might have, it's not a small number.

Figure 5.5 - the Tangent function

The derivative of Sine is Cosine. The derivative of Cosine is -Sine. You don't have to memorize that - soon it will be obvious. However, you can also just look at their graphs and see that this seems about right.

Another important trigonometry function is ArcTangent, which means you know SO/SA, and you want to know θ. ArcTangent is the inverse of Tangent. There's also ArcSine and ArcCosine, but these are rarely used.

If you want to know θ, you can't just use the ArcTangent function. First of all, if the hypotenuse is verticle, that is the Side Adjacent has length 0, then you have to divide by zero before you can use the function. Because of this, most computer programming languages have two functions, one called arctan, and one called something like arctan2. Arctan2 take two arguments, the Side Adjacent and the Side Opposite, which allows the function to check and not divide by zero. Secondly, the arctan function returns the wrong values for most of the circle. What you really want is this:

θ = arctan2( SO, SA );
if( SA < 0 ) θ = θ + π;
else if( SO < 0 ) θ = θ + 2*π;
return θ;

If you look at the graph of Tangent above you can convince yourself pretty quickly that this is the right way to find θ.

That's it for our review of trigonometry - just memorize "SO SA SO SA HYP HYP," and it's good to remember that sine2 + cosine2 = 1.


5.1: Do all the following arithmetic:
    a) (4 + 11i) + (7 - 6i)
    b) (4 + 11i) - (7 - 6i)
    c) (4 + 11i) * (7 - 6i)
    d) (11 + 4i) * (6 - 7i)
    e) (6+5i)*
    f) (-8-3i)*
    g) | -3 + 4i |
    h) (4 + 11i) / (7 - 6i)
    i) (7 - 6i) / (4 + 11i)

answer: (11 + 5i), (-3 + 17i), (94 + 53i), (94 - 53i), (6-5i), (-8+3i), 5, (38/85 + i 101/85), (38/137 - i 101/137)

5.2: Graph the following numbers:
    a) -1
    b) 2i
    c) -i
    d) 2 + i
    e) 2 - i
    f) -1 -i

Graph Paper for Problem 5.2

5.3: You may use a calculator for this problem. If you don't actually own a calculator (Heavens!), there's a calculator in Windows, see Start / Programs / Accessories. When the Calculator starts, select View / Scientific. Make certain your calculator is in radians mode. Find all the following values:

    a) Sine( 0 )
    b) Cosine( 0 )
    c) Sine( π/6 )
    d) Cosine( π/6 )
    e) Sine( π/4 )
    f) Cosine( π/4 )
    g) Sine( π/2 )
    h) Cosine( π/2 )
    i) arctan( 1 )
    j) arctan( 1.73 )
    k) arctan( .578 )

answer: 0, 1, 1/2, √3/2 = .85, √2/2 =.707, √2/2, 1, 0, π/4, π/3, π/6.

5.4: Repeat problem 5.3, but in degrees. Do all the conversions.

answer: sine(0), cosine(0), sine(30), cosine(30), sine(45), cosine(45), sine(90), cosine(90), 45, 60, 30.

Chapter 4    Chapter 5   Chapter 6