1 OCF Operations With Complex Numbers MCR3U - Santowski

2 (A) Review A complex number is a number that has two components: a real part and an imaginary part (ex = 2 + 3i )A complex number is a number that has two components: a real part and an imaginary part (ex = 2 + 3i ) In general, we can write any complex number in the form of a + bi. We designate any complex number by the letter z. As such z = a + bi. In general, we can write any complex number in the form of a + bi. We designate any complex number by the letter z. As such z = a + bi. Complex numbers have conjugates which means that the two complex numbers have the same real component and the imaginary components are “negative opposites”. If we designate z = a + bi, then we designate the conjugate as = a - biComplex numbers have conjugates which means that the two complex numbers have the same real component and the imaginary components are “negative opposites”. If we designate z = a + bi, then we designate the conjugate as = a - bi The same algebra rules that we learned for polynomials apply for complex numbers - the concepts of like terms and the distributive rule apply. For example, the terms 7i and i 2 are like terms; the terms 6 and 6i are unlike termsThe same algebra rules that we learned for polynomials apply for complex numbers - the concepts of like terms and the distributive rule apply. For example, the terms 7i and i 2 are like terms; the terms 6 and 6i are unlike terms Remember that i 2 = -1 and that i =  (-1)Remember that i 2 = -1 and that i =  (-1)

3 (B) Operations With Complex Numbers (1) Addition and Subtracting(1) Addition and Subtracting ex 1.Add (5 - 2i) and (-3 + 7i)ex 1.Add (5 - 2i) and (-3 + 7i) ex 2.Find the sum of (-4 - 3i) and its conjugateex 2.Find the sum of (-4 - 3i) and its conjugate ex 3.Simplify (7 - 4i) - (2 + 3i)ex 3.Simplify (7 - 4i) - (2 + 3i) ex 4.Subtract 3 - 5i from its conjugateex 4.Subtract 3 - 5i from its conjugate

4 (B) Operations With Complex Numbers (2) Multiplying(2) Multiplying ex 1. Simplify (i) i 7 (ii) (i 2 ) 5 (iii) i 75ex 1. Simplify (i) i 7 (ii) (i 2 ) 5 (iii) i 75 ex 2. Multiply 4 + 3i by 2 - iex 2. Multiply 4 + 3i by 2 - i ex 3. Expand and simplify (-3 - 2i) 2ex 3. Expand and simplify (-3 - 2i) 2 ex 4. Find the product of 5 - 2i and its conjugateex 4. Find the product of 5 - 2i and its conjugate ex 5. Find the product of every complex number and its conjugateex 5. Find the product of every complex number and its conjugate

5 (B) Operations With Complex Numbers (3) Dividing(3) Dividing A prerequisite skill is the idea of “rationalizing the denominator” - in other words, we have something in the denominator that we can algebraically “remove” or change.A prerequisite skill is the idea of “rationalizing the denominator” - in other words, we have something in the denominator that we can algebraically “remove” or change. Specifically, we do not want a term containing i in the denominator, so we must “remove it” using algebraic concepts (recall i 2 = -1 and recall the product of conjugates)Specifically, we do not want a term containing i in the denominator, so we must “remove it” using algebraic concepts (recall i 2 = -1 and recall the product of conjugates)

6 (B) Operations With Complex Numbers – Examples with Division ex 1. Simplify 5/2iex 1. Simplify 5/2i 5/2i = 5/2i * i/i multiply the fraction by i/i  why?5/2i = 5/2i * i/i multiply the fraction by i/i  why? 5i/2i 2 = 5i/(2(-1)) = 5i/-2 = -2.5i or -2 ½ i5i/2i 2 = 5i/(2(-1)) = 5i/-2 = -2.5i or -2 ½ i ex 2. Simplify (2 + 3i)/(1 – 2i)ex 2. Simplify (2 + 3i)/(1 – 2i) Multiply the “fraction” by the conjugate of its denominator  why??Multiply the “fraction” by the conjugate of its denominator  why?? = (2 + 3i) / (1 – 2i) * (1 + 2i)/(1 + 2i)= (2 + 3i) / (1 – 2i) * (1 + 2i)/(1 + 2i) = (2 + 3i)(1 + 2i) / (1 + 2i)(1 – 2i)= (2 + 3i)(1 + 2i) / (1 + 2i)(1 – 2i) = (2 + 3i + 4i + 6i 2 ) / (1 + 2i – 2i – 4i 2 )= (2 + 3i + 4i + 6i 2 ) / (1 + 2i – 2i – 4i 2 ) = (2 + 7i – 6) / (1 + 4)= (2 + 7i – 6) / (1 + 4) = (-4 + 7i) / 5= (-4 + 7i) / 5 Or -4/5 + 7i/5Or -4/5 + 7i/5

7 (B) Operations With Complex Numbers – Examples with Substitution Evaluate f(2 - i) if f(x) = 2x 2 - 8x - 2Evaluate f(2 - i) if f(x) = 2x 2 - 8x - 2 f(2 - i) = 2(2 - i)(2 - i) - 8(2 - i) - 2f(2 - i) = 2(2 - i)(2 - i) - 8(2 - i) - 2 f(2 - i) = 2(4 - 4i + i 2 ) i - 2f(2 - i) = 2(4 - 4i + i 2 ) i - 2 f(2 - i) = 8 - 8i + 2(-1) if(2 - i) = 8 - 8i + 2(-1) i f(2 - i) = -12f(2 - i) = -12 which means that 2 - i is not a factor of f(x)which means that 2 - i is not a factor of f(x)

8 (C) Web Links Link #1 : /complex/complex.html#Content then connect to Complex arithmeticLink #1 : /complex/complex.html#Content then connect to Complex arithmetic /complex/complex.html#Content /complex/complex.html#Content Link #2 : x2.htm from PurpleMathLink #2 : x2.htm from PurpleMath x2.htm x2.htm

9 (D) Homework Handout from MHR, page 150, Q1-7 eolHandout from MHR, page 150, Q1-7 eol Nelson text, p336, Q1,3,5,6-10,13Nelson text, p336, Q1,3,5,6-10,13