Tutorial%20complex%202%20solutions

Tutorial%20complex%202%20solutions - Solutions to Complex...

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Solutions to Complex Tutorial #2 Q1 (a) mapping wz α β =+ Inserting 1 , and 1, 1 zj w j zwj = += = = + produces ( ) () 15 2 (1 ) solving 1 15 3 6 j jj j j αβ = −+ + ⎬⎨ So transformation is 5 ( 2 ) 3 6 wj z =−+ + + j 5 (b) Writing , wu j v z x j y 5( ) ( 2 )( ) 3 6 uj v jxj y j + ++ + Equate the real and imaginary parts produces solving for y gives: (eliminating x) 52 3 6 ux y vx y =− − + =− + 51 0 23 uv y y + + + So, y = 0 corresponds to 2 3 + = , and corresponds to . 0 y > +< z-plane w-plane u+2v=3 (c) from the solutions of the simultaneous eqns in part b. 2 32 xv u y =−− Squaring and adding: ( 22 2 56 1 2 ) 9 x yv u u v u v + + The disk 2 z < corresponds to 4 xy + < So 2 1 2 9 4 0 53 2 0 3 5 6 5 20 25 4 5 2 5 5 u v +− < −+−< −< = = 2 z < z-plane 3/5 6/5 w-plane
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(d) The fixed points of the mapping are when z β = maps to w = , so mapping () 36 3 52 3 6 1 3 71 0 j zj z j z j j + =−+ + + ⇒ = = + Q2 The general bilinear mapping is az b w cz d + = + 3 points, 4 unknowns, so we write in terms of one of the parameters: 0, ,1 1, 0 zw j b j d w d j c b j a b
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This note was uploaded on 07/07/2009 for the course ELEC 3002 at Queensland.

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Tutorial%20complex%202%20solutions - Solutions to Complex...

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