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Lecture 14: Complex Numbers (Section 5)

   

Added on  2020-11-09

10 Pages1389 Words264 Views
Lecture 14:Complex Numbers (Section 5)Considerx2+ 4 = 0=) x2=4This equation has no realsolution.Considerax2+ bx + c = 0We know that if= b24ac < 0,the equation has no realsolutions.For solutions to exist for allquadratic equations, we need tointroduce the complex numbers C, where we definei=p1.L14: 1 / 10DefinitionA complex number is a number of the formz = a + ib,where a, b 2 R, and idenotesp1.(That is, i2=1).We now have solutions to allquadratic equations.Examplex2+ 4x + 5 = 0We havep4 =p4p1 = 2i=)x =2 ± iL14: 2 / 10
Lecture 14: Complex Numbers (Section 5)_1
Complex Numbers (in general)Consider the complex numberz = a + ib.The realnumber a is called the realpart of z , and is writtenRe(z) = a.The realnumber b is called the purely imaginary part of z, and iswritten Im(z) = b.ExampleLet z1= 1 + 2i,z2= 6i ,z3= 4.Re(z1) = 1Im(z1) = 2Re(z2) =Im(z2) =Re(z3) =Im(z3) =L14: 3 / 10Equality of Complex NumbersLet two complex numbers bez1= a + ib,z2= c + idThen z1= z2if and only if a = c and b = d .That is, the realparts are equaland the imaginary parts are equal.ExampleIfa + 4i= 3 + ibthena = 3andb = 4.SpecialComplex NumbersLet z = a + ib1.If b = 0 then z = a is called a realnumber.2.If a = 0 then z = ib is called a purely imaginary number.L14: 4 / 10
Lecture 14: Complex Numbers (Section 5)_2
Example4,p3,1.7are realnumbers.3i ,ip2,4i3are purely imaginary numbers.L14: 5 / 10Arithmetic of Complex NumbersLet z1= a + ib,z2= c + idAddition:z1+ z2= a + ib + c + idSubtraction:z1z2= a + ib(c + id )Scalar Multiplication:If ↵ is a scalar (↵ 2 R)↵z1= ↵(a + ib)Multiplication:Here we use the usualbinomialexpansion,recalling that i2=1.z1z2= (a + ib)(c + id)L14: 6 / 10
Lecture 14: Complex Numbers (Section 5)_3

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