Math Circles Complex Numbers, Lesson 2 Solutions Wednesday, March 28, Rich Dlin. Rich Dlin Math Circles / 24

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1 Math Circles 018 Complex Numbers, Lesson Solutions Wednesday, March 8, 018 Rich Dlin Rich Dlin Math Circles / 4

2 Warmup and Review Here are the key things we discussed last week: The numbers 1 and 0 are special, because they are identities (multiplicative and additive, respectively). When two numbers combine in an operation to result in the identity for the operation, those numbers are called inverses. For example 8 and 8 are additiive inverses of each other, while 9 and 1 9 are multiplicative inverses of eachother. Subtration and division are convenience operatations that can just as easily be represented by addition and multiplication using inverses. The number represent by the symbol i is defined so that i = 1. Although i is not actually the same as 1, we can usually safely pretend that it is. We just need to be careful. Rich Dlin Math Circles 018 / 4

3 Warmup and Review All numbers are imaginary, but some are Real, and some are Imaginary. If a R, then ai I. If we find ourselves in a situation where it would be nice to add a Real number to an Imaginary number, we can do so, and the result is called a Complex number, which in standard form looks like z = a + bi. For z = a + bi, a, b R. a is called the real part of z, and b is called the imaginary part of z. We write a = Re(z), and b = Im(z). We define a useful property of complex numbers as the modulus (or magnitude). The formula is z = a + b, which is not coincidentally reminiscent of Pyhtagorean Theorem. Rich Dlin Math Circles / 4

4 Warmup and Review Warmup Questions 1 Evaluate ( + i)(3 5i) Evaluate (3 i) 3 Evaluate (1 + 3i) 4 4 Determine z if we know that z z + 8 = 0. Let s look at the solutions to these together on the board... Rich Dlin Math Circles / 4

5 Complex Conjugates and Multiplicative Inverses Complex conjugates Given a Complex number z = a + bi, we define the conjugate of z as z = a bi. Properties of Conjugates If z, w C then 1. z + w = z + w. zw = z w 3. z = z 4. z + z = (Re(z)) R 5. z z = i(im(z)) I Let s Let s prove some of these together on the board... Rich Dlin Math Circles / 4

6 Complex Conjugates and Multiplicative Inverses (cont d) Complex conjugates have the property that when multiplied together, we always get a Real number, and a useful one at that: (z)(z) = (a + bi)(a bi) = a (bi) = a b i = a b ( 1) = a + b = z ( R) Note that: If z 0 then z 0. Since ( z ) R, then for z 0, 1 z z z = zz z = 1, which is the multiplicative identity in C. the multiplicative inverse of z C, z 0 is z 1 = 1 z z, or z z. Rich Dlin Math Circles / 4

7 Just like an additive inverse gives us a way to subtract, a multiplicative inverse gives us a way to divide. Technically there is no division with Complex numbers, though we often see and accept lazy notation. Rich Dlin Math Circles / 4 Complex Conjugates and Multiplicative Inverses (cont d) Example - Demonstration of Multiplicative Inverse Given z = 3 + 5i, we get z = 3 5i, so ( ) z zz 1 = z z [ ] 3 5i = (3 + 5i) 3 + ( 5) (3 + 5i)(3 5i) = 34 = = 1

8 Multiplicative Inverses (cont d) Example - 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 a more convenient way to express the above = 7 3i 5 + 4i 5 4i a more convenient way to think of the calculation 5 4i 35 8i 15i + 1i = = 1 (3 43i) 41 For convenience, this result may be written as 3 43i. 41 Rich Dlin Math Circles / 4

9 Try these examples 1 Given z = 4 5i, determine z 1. If x = 3 + 7i and y = + 4i, write x y in standard form. Let s look at the solutions to these together on the board... Rich Dlin Math Circles / 4

10 Exponents To think about exponents, we need some definitions. Let z C. We define: z 0 = 1 z 1 = z z k+1 = z k z, k N Note that this definition does not help us for non-natural exponents. We ll get to that. Example Find a Real solution to ( 6z ) ( i z 11 ) i z 6 = 0 Let s solve that intense monster together on the board... Rich Dlin Math Circles / 4

11 Rational Exponents The challenge: Can we evaluate i 1? Let s investigate. Rich Dlin Math Circles / 4

12 Evaluating i 1 First, note that given two Complex numbers z 1 = a 1 + b 1 i and z = a + b i, then z 1 = z if and only if a 1 = a and b 1 = b. Let i 1 = a + bi. Then ( ) i 1 = (a + bi) i = a + abi + b i 0 + 1i = a b + abi This leads to the following two equations involving a and b: a b = 0 (1) ab = 1 () Solving this system tells us that two possibilities for i 1 are ± 1 (1 + i). Squaring each of these confirms their suitability. (Is it weird that 1 = sin π 4?) Rich Dlin Math Circles / 4

13 Evaluating (a + bi) 1 Clearly then, if z = a + bi then ( ) z 1 z + a b z a = ± + i b This, combined with quadratic formula, allows us to solve quadratic equations with Complex coefficients. Let s find complex solutions to x (5 i)x + (8 i) = 0 is x = 3 i or x = + i Rich Dlin Math Circles / 4

14 Do the exercises in section Specifically make sure you try questions 4 and 5. Check your answers to 5. 1 Show that if z = a + bi 0 then z 1 z = 1. as required. z 1 z = a bi a + b a + bi 1 = a + b a + b = 1 Rich Dlin Math Circles / 4

15 Given z 1 = 4 + 3i and z = 5 i 1 Determine z 1 Determine z 1 z 1 z 1 = = 5 i 5 i 5 + i 5 + ( 1) = 5 + i 6 z 1 z 1 = 4 + 3i 5 + i i + 15i + 3i = i + 15i 3 = = i 6 Rich Dlin Math Circles / 4 6

16 Given z 1 = 4 + 3i and z = 5 i 1 Determine z 1 z z 1 z = 4 + 3i 5 i = 4 + 3i 5 + i 5 i 5 + i 0 + 4i + 15i + 3i = 5 i 0 + 4i + 15i 3 = = i Explain why your answers for (b) and (c) are the same. They are the same because the division z 1 z 1 is defined as the multiplication z 1. Rich Dlin Math Circles / 4

17 3 Evaluate each of the following. 1 (5 7i) + (6 + 5i) = 11 i (3 + 4i) (11 9i) = i 3 (3 + i) (7 4i) = 4 + 6i 4 (7 i) (3 + i) = 19 13i 10 4 Evaluate each of the following by expanding fully then using powers of i to simplify. 1 (1 + i) 3 (1 + i) 3 = (1 + i)(1 + i)(1 + i) = (1 + i)(1 + 4i + 4i ) = 1 + 4i + 4i + i + 8i + 8i 3 = 1 + 6i + 1i + 8i 3 = 1 + 6i + 1( 1) + 8( i) = 11 i Rich Dlin Math Circles / 4

18 4 Evaluate each of the following by expanding fully then using powers of i to simplify. 1 ( i) 4 ( i) 4 = ( i)( i)( i)( i) = ( i)( i)(4 4i + i ) = ( i)(8 8i + i 4i + 4i i 3 ) = ( i)(8 1i + 6i i 3 ) = 16 4i + 1i i 3 8i + 1i 6i 3 + i 4 = 16 3i + 4i 8i 3 + i 4 = 16 3i + 4( 1) 8( i) + 1 = 7 4i Rich Dlin Math Circles / 4

19 5 Solve each equation for x C. Simplify your answer as much as possible. 1 x (1 4i)x + (3 11i) = 0 Using quadratic formula, a = 1, b = (1 4i) = 1 + 4i, c = 3 11i. So x = 1 4i ± ( 1 + 4i) 4(1)(3 11i) (1) = 1 4i ± 1 8i + 16i i = 1 4i ± 1 8i i = 1 4i ± i Now determine i, using the formula. Rich Dlin Math Circles / 4

20 Note that i = ( 7) + 36 = 05 = 45. Also note that 36 > 0. Therefore, ( ) 7 36i i i = ± + i ( ) = ± + i ( ) 18 7 = ± + i ( ) = ± 9 + i 36 = ±(3 + 6i) Rich Dlin Math Circles / 4

21 Now we can continue. Either or 1 4i + (3 + 6i) x = = 4 + i = + i x = 1 4i (3 + 6i) = 10i = 1 5i Rich Dlin Math Circles / 4

22 x (8 + 1i)x (10 15i) = 0 Using quadratic formula, a =, b = (8 + 1i) = 8 1i, c = (10 15i) = i. So x = 8 + 1i ± ( 8 1i) 4()( i) () = 8 + 1i ± i + 144i i 4 = 8 + 1i ± i i 4 = 8 + 1i ± 0 + 7i 4 Now determine 0 + 7i, using the formula. Note that 0 + 7i = 7. Also note that 7 > 0. Rich Dlin Math Circles 018 / 4

23 Therefore, ( ) 0 + 7i i i = ± + i ( ) 7 7 = ± + i ( ) = ± 36 + i 36 = ±(6 + 6i) Rich Dlin Math Circles / 4

24 Now we can continue. Either or x = 8 + 1i + (6 + 6i) 4 = i 4 = 7 + 9i 8 + 1i (6 + 6i) x = 4 = + 6i 4 = 1 + 3i Rich Dlin Math Circles / 4

25 3 x (6 i)x + (8 6i) = 0 Using quadratic formula, a = 1, b = (6 i) = 6 + i, c = 8 6i. So x = 6 i ± ( 6 + i) 4(1)(8 6i) (1) = 6 i ± 36 4i + 4i 3 + 4i = 6 i ± 36 4i i = 6 i ± 0 = 3 i Rich Dlin Math Circles / 4