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Algebra lesson 1.Complex numbers 2.Matrix Algebra 3.Vector Algebra
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  Chapter 1 Complex Numbers 1.1 Introduction To have solutions for all equations bax  =  where ba ,  are integers, ,0 ≠ a  the integers were extended to the rational numbers. To have a solution for ,2 2 =  x  we went outside the rational numbers to the irrational numbers. There is no real number  x  such that 1 2 −=  x  since the square of any nonzero real number must be positive. To have a solution for ,1 2 −=  x mathematicians introduced the complex number i  with the property .1 2 −= i  We also use 1 −  to denote i . Thus the equation 01 2 =+  x  would have two roots . i ±  Also, the square root of a negative number, , 2 b −  is written in the form , bi  where b  is a real number. Furthermore, solutions of the equation ( ) 22 ba x  −=−   ( ba , both real numbers) are given by , bia x  ±=−  that is bia x  ±= . Complex numbers made their appearance in the mathematical scene in The Great Art, or in the rules of Algebra (1545) by the Italian mathematician and physician Gerolamo Cardano (1501 – 1576). Cardano studied the quadratic problem of dividing 10 into two parts such that their product is 40 and found the solutions 155  −+  and 155  −−  by standard technique. About 27 years later the engineer Raphael Bombelli (1926 – 1572) published an algebra text in which he dealt systematically with complex numbers. Definition 1.1.1    A complex number  z is any expression of the form bia z  +=   where a and b are real numbers and .1 −= i  The scalar a is called the  real part  of  z and the scalar b is called the imaginary part  of  z . Notation  If ∈+= babia z ,, , then we use Re ( )  z to denote the number a and I   ( )  zm to denote the number b i.e., Re ( )  z= a and I   ( )  zm= b . = { } ∈+ babia ,: denotes the set of all complex numbers. Remark  If ,0 = b   we write   ia 0 +  as aand ia 0 +  is identified as the  real number  a.  If ,0 = a  we write bi + 0   as   bi   and bi is called a  pure imaginary number . Every complex number is the sum of a real number and a pure imaginary number. Note. 0 stands for i 00 + .   Example 1.1.2  If ,43 i z  += then    Re ( ) ,3 =  z I   ( ) 4 =  zm (3 is the real part of  z and 4 is the imaginary part of  z ).  1.2 Operations on complex numbers Definition 1.2.1    Let ( ) ∈+=+= d cbadic zbia z ,,,, 21 . (i)    Equality : d bca z z  and 21  ==⇔= . (ii)    Addition : ( ) ( ) ( ) ( ) id bcadicbia z z  +++=+++=+   21 . (iii)   Subtraction : ( ) ( ) ( ) ( ) id bcadicbia z z  −+−=+−+=−   21 . (iv)    Multiplication: ( )( ) ( ) ( ) . 21 ibcad bd acdicbia z z  ++−=++=  (v)    Division : For ,0 2  ≠  z id cad bcd cbd acdicdicdicbiadicbia z z 2222 21   +−+++=−−⋅++=++= . (vi)   Conjugation: For , bia z  += define bia z  −= .   z  is called the conjugate  of .  z  It is also sometimes denoted by *  z . Note that ( )( ) 22 babiabia z z  +=−+= . Theorem 1.2.2  (i) ( ) ( )  ∈∀++=++ 321321321 ,,  z z z z z z z z z  . (Associative law) (ii) ∈∀+=+ 211221 ,  z z z z z z  . (Commutative law) (iii) ( ) ( )  ∈∀= 321321321 ,,  z z z z z z z z z  . (Associative law) (iv) ∈∀= 211221 ,  z z z z z z  . (Commutative law) (v) ( )  ∈∀+=+ 3213231321 ,,  z z z z z z z z z z  . (Distributive law) (vi)  ( )  ∈∀+=+ 3211313213 ,,  z z z z z z z z z z . (Distributive law) (vii)  z z z  ⇔=  is a real number.  Example 1.2.3   (i) Factorize (a) 1 2 +  x (b) ( ) 22 q p x  ++  (ii) Solve the equation .082 2 =+−  x x (iii) Let .31,42 21 i zi z  +−=+=  Compute 2122212121 ,,,,  z z z z z z z z z z  −+ . Solution . (i)(a) ( )( ) i xi xi x x  −+=−=+ 222 1  (b) ( ) ( ) ( )( ) qi p xqi p xqi p xq p x  −+++=−+=++ 22222  (ii) ( ) ( ) ( )( ) .712282228228422 2 ii x  ±=±=−±=−−±−− =  (iii)  ( ) ( ) ii z z 713412 21  +=++−=+   ( ) ( ) ii z z  +=−++=− 33412 21   ( )( ) iiiiiii z z 2141222124623142 221  +−=−+−=+−+−=+−+=   ( ) ( ) 1031,31 22222  =−+−=−−=  z zi z   ( )( ) iiiiiiii  z z z z z z −=−=+−−−−−− =−−+== 110101010121021012462 103142 2222121   Theorem 1.2.4   For any complex numbers zand w, (i) w zw z  +=+  (ii) w z zw   =  (iii) w zw z =       whenever .0 ≠ w Proof  .  Let bia z  += and . dicw  +=  Then bia z  −= and . dicw  −=  (i)    Now id bcaw z )()(  +++=+ . Then w zdicbiaid bcaw z  +=−+−=+−+=+ )()()()( . (ii)    Now ibcad bd ac zw )()(  ++−= . Then ibcad bd ac zw )()(  +−−= .  Hence zwibcad bd acidicbiaw z  =++−=−−= )()())(( . (iii)    Exercise.  Theorem 1.2.5    Let 2 ,,,, 21  ≥ n z z z n K  ,   be complex numbers. Then (i) nn  z z z z z z  +++=+++  L 2121 .... (1) (ii) nn  z z z z z z  ⋅⋅⋅⋅=⋅⋅⋅⋅ 2121 . (2) Proof  . We prove by mathematical induction. We shall only prove (i). Case 1: 2 = n  By Theorem 1.2.4, 2121  z z z z  +=+ . Case 2: 2 ≥ n.  Assume that (1) is true for some ,2 ≥= k n i.e. k k   z z z z z z  +++=+++  L 2121 ...  (3) Then ( ) 121121 ...... ++  +++=++++ k k k k   z z z z z z z z ... 1k 21  + ++++=  z z z z k   (by Theorem 1.2.4) 121  + ++++= k k   z z z z  L  (by (3)) We have shown that if (1) is true for some k n  =  then (1) is true for .1 += k n  It follows by induction that (1) is true for all positive integers 2 ≥ n. Example 1.2.6   Verify Theorem 1.2.4 for iwi z  +=−= 3,72 . Solution .  Note that iwi z  −=+= 3,72 . (i)    Now iiiw z 65)3()72(  −=++−=+ . Then iw z 65 +=+ .  Hence w ziiiw z  +=+=−++=+ 65)3()72( . (ii)    Now iii zw 1913)3)(72(  −=+−= . Then i zw 1913 += .  Hence zwiiiw z  =+=−+= 1913)3)(72( . (iii)    Now )231( 101372 iiiw z −−=+−= . Then )231( 101 iw z +−=      .  Hence      =+−= −+= w ziiiw z )231( 101372 .

Sebastiao Salgado

Jul 23, 2017

Sea Salt GW

Jul 23, 2017
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