Complex Numbers For High School Students

Transcription

1 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

2 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. Presented by: Rich Dlin Complex Numbers For High School Students 2 / 29

3 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 or 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! Presented by: Rich Dlin Complex Numbers For High School Students 3 / 29

4 Identities Consider the following equations in x: and x + y = x 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. Presented by: Rich Dlin Complex Numbers For High School Students 4 / 29

5 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 = This is actually extremely significant. Presented by: Rich Dlin Complex Numbers For High School Students 5 / 29

6 Solving Equations Let s solve some equations. But let s really think while we do. 3a 7 = 42 (1) 3a = (2) 3a + 0 = 49 (3) 3a = 49 (4) 1 3 3a = 1 49 (5) 3 1 a = 49 3 a = 49 3 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 = (Interesting thought: What are we proving when we check the answer?) We did it by using some fundamental numeric concepts, not the least of which is the power of identity elements and inverses. Presented by: Rich Dlin Complex Numbers For High School Students 6 / 29 (6) (7)

7 Solving Harder Equations Here s one that, in Ontario, is usually seen first in grade 10. Baby s First Quadratic Equation 3x 2 7 = 41 3x 2 = 48 x 2 = 16 (Ok... maybe not baby s first...) Presented by: Rich Dlin Complex Numbers For High School Students 7 / 29

8 Solving Harder Equations So we have x 2 = 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)! Presented by: Rich Dlin Complex Numbers For High School Students 8 / 29

9 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

10 Thinking More Deeply About Equations Now let s turn to a somewhat mystifying equation: Is there a solution? x 2 = 1 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? Presented by: Rich Dlin Complex Numbers For High School Students 10 / 29

11 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? Presented by: Rich Dlin Complex Numbers For High School Students 11 / 29

12 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. Presented by: Rich Dlin Complex Numbers For High School Students 12 / 29

13 Imagining Numbers Definition of 3 Three is a counting concept used to represent a quantity of objects that is one more than one more than one, and not more than that. In other words but what is 1? Presented by: Rich Dlin Complex Numbers For High School Students 13 / 29

14 Imagining Numbers Definition of i i is a number defined so that i 2 = 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 which would be... problematic. ( 1) 1 2 [ ] 2 = i ( 1) 1 2 = i 2 [ ( 1) 2] 1 2 = i 2 [1] 1 2 = i 2 1 = 1 Presented by: Rich Dlin Complex Numbers For High School Students 14 / 29

15 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, i 2 = 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) = 6i 2 = 6. Presented by: Rich Dlin Complex Numbers For High School Students 15 / 29

16 Back to Equations Once Imaginary numbers have been defined, and we can solve equations like x 2 = 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: x 2 6x + 10 = 0 (1) x 2 6x = 10 (2) x 2 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. Presented by: Rich Dlin Complex Numbers For High School Students 16 / 29

17 Complex Numbers Definition of C The set of Complex numbers is defined as C = {a + bi, a, b R, i 2 = 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) = i 14i + 35i 2 = i = 41 + i Presented by: Rich Dlin Complex Numbers For High School Students 17 / 29

18 Complex Numbers Notes on C (cont d): The magnitude of a Complex number z = a + bi is defined as z = a 2 + b 2. Note this is perfectly consistent for numbers that are strictly Real or strictly Imaginary. Presented by: Rich Dlin Complex Numbers For High School Students 18 / 29

19 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) = a 2 (bi) 2 = a 2 b 2 i 2 = a 2 b 2 ( 1) = a 2 + b 2 = z 2 R So the multiplicative inverse of z C, z 0 is z 1 = 1 z z, or z 2 z 2. Presented by: Rich Dlin Complex Numbers For High School Students 19 / 29

20 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) ( 5) 2 (3 + 5i)(3 5i) = 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. Presented by: Rich Dlin Complex Numbers For High School Students 20 / 29

21 Multiplicative Inverses (cont d) Example 2 - Demonstration of Division Given x = 7 3i and y = 5 + 4i, x y, or x, is technically not defined, y 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 i 15i + 12i = = 1 (23 43i) 41 This result may be written as 23 43i 42, although to emphasize the absence of division with Complex numbers, it is better to write it as multiplication. Presented by: Rich Dlin Complex Numbers For High School Students 21 / 29

22 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): i 0 = 1 i 1 = i i 2 = 1 i 3 = i 2 i = i i 4 = i 3 i = i i = i 2 = ( 1) = 1 i 5 = i 4 i = 1 i = i i 6 = i 5 i = i i = i 2 = 1 i 7 = i 6 i = 1 i = i i 8 = i 7 i = i i = 1 i 9 = i 8 i = 1 i = i i 10 = i 9 i = i i = i 2 = 1 Presented by: Rich Dlin Complex Numbers For High School Students 22 / 29

23 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 i 4n = 1 i 4n+1 = i i 4n+2 = 1 i 4n+3 = i Presented by: Rich Dlin Complex Numbers For High School Students 23 / 29

24 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. Presented by: Rich Dlin Complex Numbers For High School Students 24 / 29

25 Negative Integer Exponents (cont d) Let x > 0, where x Z. Then i x = ( i 1) x = ( i) x (since i 1 = i) = ( 1 i) x = ( 1) x i x 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) 18 i 18 = (1)( 1) (since 18 4 leaves remainder 2, and i 2 = 1) = i Presented by: Rich Dlin Complex Numbers For High School Students 25 / 29

26 Rational Exponents The challenge: Can we evaluate i 1 2? Let s investigate. Presented by: Rich Dlin Complex Numbers For High School Students 26 / 29

27 Evaluating i 1 2 First, note that given two Complex numbers z 1 = a 1 + b 1 i and z 2 = a 2 + b 2 i, then z 1 = z 2 if and only if a 1 = a 2 and b 1 = b 2. Let i 1 2 = a + bi. Then ( ) i = (a + bi) 2 i = a 2 + 2abi + b 2 i i = a 2 b 2 + 2abi This leads to the following two equations involving a and b: a 2 b 2 = 0 (1) 2ab = 1 (2) Solving this system tells us that two possibilities for i 1 2 ( i = ± ) i 2 are: Squaring each of these confirms their suitability. Presented by: Rich Dlin Complex Numbers For High School Students 27 / 29

28 Evaluating (a + bi) 1 2 A similar but much more heavily involved process yields this result: If z = a ( + bi then ) z 1 z + a z a 2 = ± + [sign(b)] i 2 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 x 2 (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

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: e iπ + 1 = 0. Thanks for attending! Presented by: Rich Dlin Complex Numbers For High School Students 29 / 29