Math144Notes

# Math144Notes - Advanced Engineering Math II Math 144...

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Advanced Engineering Math II Math 144 Lecture Notes by Stefan Waner (First printing: 2003) Department of Mathematics, Hofstra University

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2 1. Algebra and Geometry of Complex Numbers (based on §§17.1–17.3 of Zill) Definition 1.1 A complex number has the form z = (x, y), where x and y are real numbers. x is referred to as the real part of z, and y is referred to as the imaginary part of z . We write Re (z) = x , Im (z) = y. Denote the set of complex numbers by C I . Think of the set of real numbers as a subset of C I by writing the real number x as (x, 0) . The complex number (0, 1) is called i. Examples 3 = (3, 0), (0, 5), (-1, - π ), i = (0, 1). Geometric Representation of a Complex Number- in class. Definition 1.2 Addition and multiplication of complex numbers, and also multiplication by reals are given by: (x, y) + (x ' , y ' ) = ((x+x ' ), (y+y ' )) (x, y)(x ' , y ' ) = ((xx ' -yy ' ), (xy ' +x ' y)) (x, y) = ( x, y). Geometric Representation of Addition- in class. (Multiplication later) Examples 1.3 (a) 3+4 = (3, 0)+(4, 0) = (7, 0) = 7 (b) 3 4 = (3, 0)(4, 0) = (12-0, 0) = (12, 0) = 12 (c) (0, y) = y(0, 1) = yi (which we also write as iy). (d) In general, z = (x, y) = (x, 0) + (0, y) = x + iy. z = x + iy (e) Also, i 2 = (0, 1)(0, 1) = (-1, 0) = -1. i 2 = -1 (g) 4 - 3i = (4, -3) . Note In view of (d) above, from now on we shall write the complex number (x, y) as x+iy . Definitions 1.4 The complex conjugate , z , of the complex number z = x+iy given by z = x - iy. The magnitude , |z| of z = x+iy is given by |z| = x 2 +y 2 . Examples and Geometric Representation of Conjugation and Magnitude - in class. Notes 1. z + z = (x+iy) + (x-iy) = 2x = 2 Re (z) . Therefore, Re (z) = 1 2 (z+z)
3 z - z = (x+iy) - (x-iy) = 2iy = 2i Im (z) . Therefore, Im (z) = 1 2i (z-z) 2. Note that zz = (x+iy)(x-iy) = x 2 -i 2 y 2 = x 2 +y 2 = |z| 2 zz = |z| 2 3 . If z 0 , then z has a multiplicative inverse. Why? because: z · z |z| 2 = zz |z| 2 = |z| 2 |z| 2 = 1. Hence, z -1 = z |z| 2 Examples (a) 1 i = -i (b) 1 3+4i = 3-4i 25 (c) 1 1 2 (1+i) = 1 2 (1-i) (d) 1 cos ø + i sin ø = cos (-ø) + i sin (-ø) 4. There is also the Triangle Inequality: |z 1 + z 2 | |z 1 | + |z 2 |. Proof We square both sides and compare them. Write z 1 = x 1 + iy 1 and z 2 = x 2 + iy 2 . Then |z 1 + z 2 | 2 = (x 1 +x 2 ) 2 + (y 1 +y 2 ) 2 = x 1 2 + x 2 2 + 2x 1 x 2 + y 1 2 + y 2 2 + 2y 1 y 2 . On the other hand, (|z 1 | + |z 2 |) 2 = |z 1 | 2 + 2|z 1 ||z 2 | + |z 2 | 2 = x 1 2 + x 2 2 + y 1 2 + y 2 2 + 2|z 1 ||z 2 |. Subtracting, (|z 1 | + |z 2 |) 2 - |z 1 + z 2 | 2 = 2|z 1 ||z 2 | - 2(x 1 x 2 + y 1 y 2 ) = 2 [ |(x 1 ,y 1 )||(x 2 ,y 2 )| - (x 1 ,y 1 ).(x 2 ,y 2 ) ] (in vector form) = 2 [ |(x 1 ,y 1 )||(x 2 ,y 2 )| - |(x 1 ,y 1 )||(x 2 ,y 2 )| cos ] = 2 |(x 1 ,y 1 )||(x 2 ,y 2 )| (1 - cos ) 0, giving the result. Note The triangle inequality can also be seen by drawing a picture of z 1 + z 2 . 5. We now consider the polar form of these things: If z = x+iy, we can write x = r cos ø and y = r sin ø, getting z = r cos ø + ir sin ø, so This is called the polar form of z . It is important to draw pictures in order to feel comfortable with the polar representation. Here r is the magnitude of z, r = |z| , and ø is called the argument of z , denoted arg (z). To calculate ø , we can use the fact that tan ø = y/x. Thus ø is not arctan (y/x) as claimed in the book, but by: since the arctan function takes values between - π /2 and π /2. The principal value of arg (z) is the unique choice of ø such that - π < ø π . We write this as Arg (z) - π < Arg (z) ≤ π - π ≤ A

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4 Examples (a) Express z = 1+i in polar form, using the principal value (b) Same for 3 + 3 3 i (c) 6 = 6( cos 0 + i sin 0) 6.
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