Complex Numbers For High School Students For the Love of Mathematics - - PowerPoint PPT Presentation

Complex Numbers For High School Students

For the Love of Mathematics and Computing Saturday, October 14, 2017 Presented by: Rich Dlin

Presented by: Rich Dlin Complex Numbers For High School Students 1 / 29

About Rich

Graduated UWaterloo with BMath in 1993 Worked as a software developer for almost 10 years Began teaching in 2002 Taught in 4 different high schools over a span of 5 years Department head of mathematics for 10 years Earned MMT from UWaterloo in 2013 Visiting Lecturer at UWaterloo beginning September 2017 Part-time artist, musical theatre performer/director, bodybuilder, philosopher Fully believes that all of these are consistent with being a mathematician.

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Laying a Foundation For Complex Numbers

We’ll start with some seemingly trivial exploration, because as it turns out, the exploration of the trivial can be decidedly non-trivial! Most of us instinctively solve the following equation, either by inspection

• r more formally using the (cherished) algorithms we teach as part of an

algebra curriculum: x + 8 = 13

(I know ... shockingly difficult)

A basic understanding of the meaning of the statement allows us to see that the solution is x = 5. But as soon as we start teaching algorithms, we sneakily introduce significant yet abstract notions that are rarely if ever addressed. So let’s address them!

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Identities

Consider the following equations in x: x + y = x and xy = x, x = 0 Once again, in each case, we instinctively know the value of y that solves the equation, even though we do not know the value of x (nor can we).

Identities (Not the Triggy kind)

The number 0 is called the Identity Element under addition, because for any x, x + 0 = 0 + x = x. Similarly, the number 1 is called the Identity Element under multiplication, because for any x, 1 × x = x × 1 = x.

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Inverses

Inverses

if a is the identity element for an operation ⊲, then the inverse of x under ⊲ is x−1. where x ⊲ x−1 = x−1 ⊲ x = a. Examples: The inverse of 5 under multiplication is 1

• 5. We say that 5 and 1

5 are

multiplicative inverses of each other. The inverse of 5 under addition is −5. We say that 5 and −5 are additive inverses of each other. So then the operations of division and subtraction are really just multiplication and addition, respectively, but using the appropriate inverse. For example 5 − 2 = 5 + (−2), and 7 ÷ 9 = 7 × 1

9.

This is actually extremely significant.

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Solving Equations

Let’s solve some equations. But let’s really think while we do. 3a − 7 = 42 (1) 3a − 7 + 7 = 42 + 7 (2) 3a + 0 = 49 (3) 3a = 49 (4) 1 3 × 3a = 1 3 × 49 (5) 1 × a = 49 3 (6) a = 49 3 (7) The above is really a logical argument It begins with the assumption that “There exists a ∈ R, such that 7 less than triple the value of a results in 42. So we have proven this implication (given the Real numbers as a universe of discourse): If 3a − 7 = 42 then a = 49

3 .

(Interesting thought: What are we proving when we “check the answer”?)

We did it by using some fundamental numeric concepts, not the least

• f which is the power of identity elements and inverses.

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Solving Harder Equations

Here’s one that, in Ontario, is usually seen first in grade 10.

Baby’s First Quadratic Equation

3x2 − 7 = 41 3x2 = 48 x2 = 16

(Ok ... maybe not baby’s first ...)

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Solving Harder Equations

So we have x2 = 16 But what now? The truth is at this stage most people think we just use square root to determine the solution. Students instinctively deduce x = 4. Why? Because they implicitly use a universe of discourse of non-negative Real numbers. We then remind them that we may consider negatives also. Many students (and teachers?) then say something like “16 has two square roots”. But the number 16 does not have two square roots (entirely because square root is not defined that way)!

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Solving Harder Equations

In fact the solution is the logical statement

• x =

√ 16 = 4

• OR
• x = −

√ 16 = −4

• .

We use properties of multiplication and negative numbers to deduce the second root of the equation, namely −4. This is not intuitive. At first. Understanding this requires a background in understanding negative numbers, and how they behave under multiplication. We take these “understandings” for granted, which is interesting - why should we?

◮ Are negative numbers intuitive? ◮ We call them Real, but are they truly realistic? Presented by: Rich Dlin Complex Numbers For High School Students 9 / 29

Thinking More Deeply About Equations

Now let’s turn to a somewhat mystifying equation: x2 = −1 Is there a solution? In the Real numbers, there is no number capable of squaring to a negative. Therefore, no Real solution. But wait! We have already shown that we are capable of imagining unrealistic numbers ... ... so surely we can imagine a universe of discourse with a solution?

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Imagining Numbers

Consider this symbol: Is it the number three? If we were to poll a class of thirty students, it is extremely likely that unless they thought it was a trick question, they would all answer yes. However it is not so simple. The actual answer is NO. It is a symbol that we have all agreed would represent the number three, so that when we see it, or write it, it is tied to the notion of three in our

• minds. But what is three?

Can you define this number?

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Imagining Numbers

An Attempt to Define “Three”

A counting concept used to represent the total number of elements in a set which could be subdivided into non-empty sets, each containing fewer elements than the original, exactly one of which could be subdivided again in the same manner, none of which could be subdivided in the same manner again. This is cumbersome A clever eye will realize that this definition depended very much on the definitions of “one” and “zero” With these numbers already defined, the definition for three could be made more concise.

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Imagining Numbers

Definition of 3

Three is a counting concept used to represent a quantity of objects that is

• ne more than one more than one, and not more than that.

In other words ... ... but what is 1?

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Imagining Numbers

Definition of i

i is a number defined so that i2 = −1. Note: Despite numerous humorous t-shirts, mugs and beach blankets, it is not the case that √−1 = i. In the Real number system, the square root function is defined only on non-negative numbers. Additionally, if it were the case that √−1 = i, then we would have

√ −1 = i (−1)

1 2

= i

• (−1)

1 2

2 = i2

• (−1)2 1

2

= i2 [1]

1 2

= i2 1 = −1

which would be ... problematic.

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Imagining Numbers

Some rules for i: i has magnitude 1. Like vectors, i can be scaled using Real numbers, via scalar

• multiplication. E.g., 3i, where |3i| = 3.

We define a new number set, the Imaginary numbers, as I = {ai : a ∈ R, i2 = −1}. This set is unfortunately named, since it implies that to conceive of it requires proprietary implementation of imagination over previously defined number sets. Adding two Imaginary numbers works the way we would hope: For all a, b ∈ R, ai + bi = (a + b)i. I and R share exactly one number. I ∩ R = {0}. Negatives work the way we would hope: For all a ∈ R, ai + (−a)i = 0. i.e., the additive inverse of ai is (−a)i. To multiply two Imaginary numbers, multiply the scalars and square the i. For example (3i)(2i) = 6i2 = −6.

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Back to Equations

Once Imaginary numbers have been defined, and we can solve equations like x2 = −9, we quickly want to use this tool to solve harder quadratic equations that previously seemed to have no solution. Here’s an example. Solve: x2 − 6x + 10 = (1) x2 − 6x = −10 (2) x2 − 6x + 9 = −1 (3) (x − 3)2 = −1 (4) This yields two possibilities: x − 3 = i or x − 3 = −i. We would love to add 3 to both sides. But we don’t have any way to add non-zero Real numbers to non-zero Imaginary numbers. So we define a way! And in so doing, define a new universe of discourse.

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Complex Numbers

Definition of C

The set of Complex numbers is defined as C = {a + bi, a, b ∈ R, i2 = −1}. Notes on C: It is a fusion of the Real and Imaginary numbers, and a superset of both. In a + bi, a is called the Real part and bi is called the Imaginary part. Adding, multiplying and negating (therefore subtracting) all work as we would hope, honouring the definitions of these operations in the Real and Imaginary number sets. For example (3 + 7i)(−2 + 5i) = (3)(−2) + (3)(5i) + (−2)(7i) + (7i)(5i) = −6 + 15i − 14i + 35i2 = −6 − 35 + i = −41 + i

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Complex Numbers

Notes on C (cont’d): The magnitude of a Complex number z = a + bi is defined as |z| = √ a2 + b2. Note this is perfectly consistent for numbers that are strictly Real or strictly Imaginary.

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Multiplicative Inverses

Complex conjugates

Given a Complex number z = a + bi, we define the conjugate of z as z = a − bi. Complex conjugates have the property that when multiplied together, we always get a Real number: (z)(z) = (a + bi)(a − bi) = a2 − (bi)2 = a2 − b2i2 = a2 − b2(−1) = a2 + b2 = |z|2 ∈ R So the multiplicative inverse of z ∈ C, z = 0 is z−1 = 1 |z|2 z, or z |z|2 .

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Multiplicative Inverses (cont’d)

Example 1 - Demonstration of Multiplicative Inverse

Given z = 3 + 5i, we get z = 3 − 5i, and zz−1 = (3 + 5i)

• 3 − 5i

32 + (−5)2

• =

(3 + 5i)(3 − 5i) 34 = 34 34 = 1 Just like an additive inverse gives us a way to subtract, a multiplicative inverse gives us a way to “divide” Complex numbers. Technically there is no division with Complex numbers, though as witnessed above we often see lazy notation.

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Multiplicative Inverses (cont’d)

Example 2 - Demonstration of “Division”

Given x = 7 − 3i and y = 5 + 4i, x ÷ y, or x y , is technically not defined, but essentially means the same thing as xy−1.

x y = xy −1 = xy |y|2 (a more convenient way to express the above) = 7 − 3i 5 + 4i × 5 − 4i 5 − 4i = 35 − 28i − 15i + 12i2 52 + 42 = 1 41(23 − 43i)

This result may be written as 23−43i

42

, although to emphasize the absence

• f division with Complex numbers, it is better to write it as multiplication.

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Powers of i

This is a great time to look at a cool property of i having to do with

• exponents. Consider the following progression, which assumes that the

exponent rules we already know apply to i (they do - proof is left as an exercise): i0 = 1 i1 = i i2 = −1 i3 = i2 × i = −i i4 = i3 × i = −i × i = −i2 = −(−1) = 1 i5 = i4 × i = 1 × i = i i6 = i5 × i = i × i = i2 = −1 i7 = i6 × i = −1 × i = −i i8 = i7 × i = −i × i = 1 i9 = i8 × i = 1 × i = i i10 = i9 × i = i × i = i2 = −1

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Powers of i

Notes on powers of i: The powers of i repeat in cycles of 4, so that any non-negative integer power of i can be expressed either as

◮ 1 (if the exponent gives remainder 0 when divided by 4) or ◮ i (if the exponent gives remainder 1 when divided by 4) or ◮ −1 (if the exponent gives remainder 2 when divided by 4) or ◮ −i (if the exponent gives remainder 3 when divided by 4).

This can be summarized concisely as follows, where n can be any non-negative integer i4n = 1 i4n+1 = i i4n+2 = −1 i4n+3 = −i

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Negative Integer Exponents

We have looked at positive integer exponents, and discussed 0 in the

• exponent. But what about negative integers?

The good news is we have already defined z−1, so we can apply the definition to i−1. Note that i = 0 + i, so i = 0 − i = −i. Note that ii = |i|2 = 1. Therefore i−1 = 1

ii

• i
• = −i.

Good news!

◮ It means the cycle we saw just before actually goes backwards as well,

and

◮ We can convert any negative Integer power of i to an expression with a

positive exponent, as follows.

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Negative Integer Exponents (cont’d)

Let x > 0, where x ∈ Z. Then i−x =

• i−1x

= (−i)x (since i−1 = −i) = (−1 × i)x = (−1)xix We have so far only considered integer values for x, but this result actually holds for any Real value of x, although when x is not an integer things do get more interesting. Very handy!

Example

i−18 = (−1)18i18 = (1)(−1) (since 18 ÷ 4 leaves remainder 2, and i2 = −1) = i

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Rational Exponents

The challenge: Can we evaluate i

1 2 ?

Let’s investigate.

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Evaluating i

1 2

First, note that given two Complex numbers z1 = a1 + b1i and z2 = a2 + b2i, then z1 = z2 if and only if a1 = a2 and b1 = b2. Let i

1 2 = a + bi. Then

• i

1 2

2 = (a + bi)2 i = a2 + 2abi + b2i2 0 + 1i = a2 − b2 + 2abi This leads to the following two equations involving a and b: a2 − b2 = (1) 2ab = 1 (2) Solving this system tells us that two possibilities for i

1 2 are:

i

1 2 = ±

1 √ 2 + 1 √ 2 i

• Squaring each of these confirms their suitability.

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Evaluating (a + bi)

1 2

A similar but much more heavily involved process yields this result:

◮ If z = a + bi then ◮ z

1 2 = ±

• |z| + a

2 + [sign(b)] × i

• |z| − a

2

• ◮ Trust me!

This, combined with quadratic formula, allows us to solve quadratic equations with Complex coefficients.

◮ For example, it can be shown that the solution to

x2 − (5 − i)x + (8 − i) = 0 is

◮ x = 3 − 2i or x = 2 + i Presented by: Rich Dlin Complex Numbers For High School Students 28 / 29

Other Topics in the Accompanying File

Given time constraints today, not all topics have been presented Attendees of this presentation receive a PDF of this presentation. Attendees of this presentation also receive a PDF designed to be used by high school teachers with high school students. The file is complete with worked examples and exercises with full solutions. Topics include:

◮ A deeper look at solving equations ◮ The formal definition of i ◮ The formal definition of Complex numbers ◮ Operations with Complex numbers ◮ Graphing Complex numbers ◮ Complex numbers in polar form ◮ BONUS: Taylor series expansions and Euler’s Identity: eiπ + 1 = 0.

Thanks for attending!

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