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OCF.1.7 - Operations With Complex OCF.1.7 - Operations With Complex NumbersNumbers

MCR3U - SantowskiMCR3U - Santowski

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(A) Review(A) Review

• A complex number is a number that has two components: A complex number is a number that has two components: a real part and an imaginary part (ex = 2 + 3a real part and an imaginary part (ex = 2 + 3ii ) )

•   In general, we can write any complex number in the In general, we can write any complex number in the form of a + bform of a + bii. We designate any complex number by the . We designate any complex number by the letter letter zz. As such z = a + b. As such z = a + bii..

• Complex numbers have conjugates which means that the Complex numbers have conjugates which means that the two complex numbers have the same real component and two complex numbers have the same real component and the imaginary components are “negative opposites”. If the imaginary components are “negative opposites”. If we designate z = a + bwe designate z = a + bii, then we designate the conjugate , then we designate the conjugate as = a - bi as = a - bi

• The same algebra rules that we learned for polynomials The same algebra rules that we learned for polynomials apply for complex numbers - the concepts of like terms apply for complex numbers - the concepts of like terms and the distributive rule apply. For example, the terms 7and the distributive rule apply. For example, the terms 7ii and and ii22 are like terms; the terms 6 and 6 are like terms; the terms 6 and 6ii are unlike terms are unlike terms

• Remember that Remember that ii22 = -1 and that = -1 and that ii = = (-1)(-1)

z

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(B) Operations With Complex Numbers(B) Operations With Complex Numbers

• (1) Addition and Subtracting (1) Addition and Subtracting   • ex 1.ex 1. Add (5 - 2Add (5 - 2ii) and (-3 + 7) and (-3 + 7ii))• ex 2.ex 2. Find the sum of (-4 - 3Find the sum of (-4 - 3ii) and its conjugate) and its conjugate• ex 3.ex 3. Simplify (7 - 4Simplify (7 - 4ii) - (2 + 3) - (2 + 3ii))• ex 4.ex 4. Subtract 3 - 5Subtract 3 - 5ii from its conjugate from its conjugate

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(B) Operations With Complex Numbers(B) Operations With Complex Numbers

• (2) Multiplying(2) Multiplying

• ex 1. Simplify (i) iex 1. Simplify (i) i7 7 (ii) (i(ii) (i22))5 5 (iii) (iii) ii7575

• ex 2. Multiply 4 + 3ex 2. Multiply 4 + 3ii by 2 - by 2 - ii• ex 3. Expand and simplify (-3 - 2ex 3. Expand and simplify (-3 - 2ii))22

• ex 4. Find the product of 5 - 2ex 4. Find the product of 5 - 2ii and its and its conjugateconjugate

• ex 5. Find the product of every complex ex 5. Find the product of every complex number and its conjugatenumber and its conjugate

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(B) Operations With Complex Numbers(B) Operations With Complex Numbers

• (3) Dividing(3) Dividing

• A prerequisite skill is the idea of “rationalizing A prerequisite skill is the idea of “rationalizing the denominator” - in other words, we have the denominator” - in other words, we have something in the denominator that we can something in the denominator that we can algebraically “remove” or change.algebraically “remove” or change.

• Specifically, we do not want a term containing Specifically, we do not want a term containing i i in the denominator, so we must “remove it” in the denominator, so we must “remove it” using algebraic concepts (recall using algebraic concepts (recall ii22 = -1 and recall = -1 and recall the product of conjugates)the product of conjugates)

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(B) Operations With Complex Numbers – Examples with (B) Operations With Complex Numbers – Examples with DivisionDivision

• ex 1. Simplify 5/2iex 1. Simplify 5/2i• 5/2i = 5/2i * i/i multiply the fraction by i/i 5/2i = 5/2i * i/i multiply the fraction by i/i why? why?• 5i/2i5i/2i22 = 5i/(2(-1)) = 5i/-2 = -2.5i or -2 ½ i = 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 Multiply the “fraction” by the conjugate of its denominator

why?? 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 + 3i + 4i + 6i22) / (1 + 2i – 2i – 4i) / (1 + 2i – 2i – 4i22))• = (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

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(B) Operations With Complex Numbers – (B) Operations With Complex Numbers – Examples with SubstitutionExamples with Substitution

• Evaluate f(2 - Evaluate f(2 - ii) if f(x) = 2x) if f(x) = 2x22 - 8x - 2 - 8x - 2• f(2 - f(2 - ii) = 2(2 - ) = 2(2 - ii)(2 - )(2 - ii) - 8(2 - ) - 8(2 - ii) - 2) - 2• f(2 - f(2 - ii) = 2(4 - 4) = 2(4 - 4ii + + ii22) - 16 + 8) - 16 + 8ii - 2 - 2• f(2 - f(2 - ii) = 8 - 8) = 8 - 8ii + 2(-1) - 18 + 8 + 2(-1) - 18 + 8ii• f(2 - f(2 - ii) = -12) = -12

• which means that 2 - which means that 2 - ii is not a factor is not a factor of f(x)of f(x)

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(C) Web Links(C) Web Links

• Link #1 : Link #1 : http://www.uncwil.edu/courses/mat1http://www.uncwil.edu/courses/mat111hb/Izs/complex/complex.html#Con11hb/Izs/complex/complex.html#Contenttent then connect to Complex arithmetic then connect to Complex arithmetic

• Link #2 : Link #2 : http://www.purplemath.com/moduleshttp://www.purplemath.com/modules/complex2.htm/complex2.htm from PurpleMath from PurpleMath

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(D) Homework(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