Solutions Gamelin - IV 1 2 3 4 5 6 7 8 1 X X X X X X X X 2 X X X X X X X X 3 X X X X X X X X 4 X X X X X X X X 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 X

Solutions Gamelin - IV 1 2 3 4 5 6 7 8 1 X X X X X X X X 2...

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IV 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 1 X X X X X X X X X X X 2 X X X X X X X X 3 X X X X X X X 4 X X X X 5 X X X X 6 X X X X 7 X X X X X X X X X X X 8 X X X X X X X X 1
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I.1.1 Identify and sketch the set of points satisfying. (a) j z " 1 " i j = 1 (f) 0 < Im z < $ (b) 1 < j 2 z " 6 j < 2 (g) " $ < Re z < $ (c) j z " 1 j 2 + j z + 1 j 2 < 8 (h) j Re z j < j z j (d) j z " 1 j + j z + 1 j # 2 (i) Re ( iz + 2) > 0 (e) j z " 1 j < j z j (j) j z " i j 2 + j z + i j 2 < 2 Solution Let z = x + iy , where x; y 2 R . (a) Circle, centre 1 + i , radius 1 . j z " 1 " i j = 1 , j ( x " 1) + i ( y " 1) j = 1 , ( x " 1) 2 + ( y " 1) 2 = 1 2 (b) Annulus with centre 3 , inner radius 1 = 2 , outer radius 1 . 1 < j 2 z " 6 j < 2 , 1 < 2 j z " 3 j < 2 , , 1 = 2 < j z " 3 j < 1 , (1 = 2) 2 < ( x " 3) 2 + y 2 < 1 2 (c) Disk, centre 0 , radius p 3 . j x + iy " 1 j 2 + j x + iy + 1 j 2 < 8 , , ( x " 1) 2 + y 2 + ( x + 1) 2 + y 2 < 8 , x 2 + y 2 < ! p 3 " 2 (d) Interval [ " 1 ; 1] . j z " 1 j + j z + 1 j # 2 , q ( x " 1) 2 + y 2 # 2 " q ( x + 1) 2 + y 2 , , $ q ( x " 1) 2 + y 2 % 2 # $ 2 " q ( x + 1) 2 + y 2 % 2 , , q ( x + 1) 2 + y 2 # x + 1 , $ q ( x + 1) 2 + y 2 % 2 # ( x + 1) 2 , y = 0 Now, take y = 0 in the inequality, and compute the three intervals 2
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x < " 1 ; then j x " 1 j + j x + 1 j = " ( x " 1) " ( x + 1) = " 2 x ' 2 ; " 1 # x # 1 then j x " 1 j + j x + 1 j = " ( x " 1) + ( x + 1) = 2 # 2 x > 1 ; then j x " 1 j + j x + 1 j = ( x " 1) + ( x + 1) = 2 x ' 2 : (e) Halfñplane x > 1 = 2 . j z " 1 j < j z j , j z " 1 j 2 < j z j 2 , j x + iy " 1 j 2 < j x + iy j 2 , , ( x " 1) 2 + y 2 < x 2 + y 2 , x > 1 = 2 (f) Horizontal strip, 0 < y < $ . (g) Vertical strip, " $ < x < $ . (h) C n R . j Re z j < j z j , j Re ( x + iy ) j 2 < j x + iy j 2 , x 2 < x 2 + y 2 , j y j > 0 (i) Half plane y < 2 . Re ( iz + 2) > 0 , Re ( i ( x + iy ) + 2) > 0 , " y + 2 > 0 , y < 2 (j) Empty set. j z " i j 2 + j z + i j 2 < 2 , j x + iy " i j 2 + j x + iy + i j 2 < 2 , , x 2 + ( y " 1) 2 + x 2 + ( y + 1) 2 < 2 , x 2 + y 2 < 0 3
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I.1.2 Verify from the deÖnitions each of the identities (a) z + w = 1 z + 1 w (b) zw = 1 z 1 w (c) j 1 z j = j z j (d) j z j 2 = z 1 z Draw sketches to illustrate (a) and (c). Solution Substitute z = x + iy and w = u + iv , and use the deÖnitions. (a) z + w = ( x + iy ) + ( u + iv ) = ( x + u ) + ( y + v ) i = = ( x + u ) " ( y + v ) i = ( x " iy ) + ( u " iv ) = 1 z + 1 w: (b) zw = ( x + iy ) ( u + iv ) = ( xu " yv ) + ( xv + yu ) i = = ( xu " yv ) " ( xv + yu ) i = ( x " iy ) ( u " iv ) = 1 z 1 w: (c) j 1 z j = & & x + iy & & = j x " iy j = q x 2 + ( " y ) 2 = p x 2 + y 2 = j x + iy j = j z j : (d) j z j 2 = j x + iy j 2 = ! p x 2 + y 2 " 2 = x 2 + y 2 = = x 2 " i 2 y 2 = ( x + iy ) ( x " iy ) = z 1 z: 4
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I.1.3 Show that the equation j z j 2 " 2 Re (1 az ) + j a j 2 = / 2 represents a circle centered at a with radius / . Solution Let z = x + iy and a = 0 + i1 , we have j z j 2 " 2 Re (1 az ) + j a j 2 = = x 2 + y 2 " 2 Re (( 0 " i1 ) ( x + iy )) + 0 2 + 1 2 = = x 2 + y 2 " 2 Re (( 0x + 1y ) + i ( 0y " 1x )) + 0 2 + 1 2 = = x 2 + y 2 " 2 ( 0x + 1y ) + 0 2 + 1 2 = = ( x " 0 ) 2 + ( y " 1 ) 2 : Thus the equation j z j 2 " 2 Re (1 az ) + j a j 2 = / 2 , becomes ( x " 0 ) 2 + ( y " 1 ) 2 = / 2 ; which is the equation for a circle of radius / centered at ( 0; 1 ) , which in complex notation is the point a = 0 + i1 . 5
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I.1.4 Show that j z j # j Re z j + j Im z j , and sketch the set of points for which equality holds. Solution Apply triangle inequality to z = Re z + i Im z , to obtain j z j 6 j Re z j + j Im z j . Now set z = x + iy , and see then equality holds j z j = j Re z j + j Im z j , p x 2 + y 2 = p x 2 + p y 2 , ! p x 2 + y 2 " 2 = ! p x 2 + p y 2 " 2 , , x 2 + y 2 = ! p x 2 " 2 + ! p x 2 " 2 +2 p x 2 p y 2 , x 2 y 2 = 0 , x = 0 or y = 0 : Equality holds only when z is real or pure imaginary, which are all the points on the real and imaginary axis.
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  • Fall '07
  • Lim
  • Complex number, Logarithm, -2, Complex Plane, Branch point, Z Öxed

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