Answer Key and Solutions Part A

Answer Key and Solutions Part A - 21. SOLUTIONS...

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Im Re 1 2 -1 -2 1 -1 -2 (0·5,0) (-0·5,0·5) (-1,-1) (2,-2) 21. SOLUTIONS 1. Cartesian Co-ordinates 1. 5, ± 2 i . 2. -3 or 2. 3. Let  z  =  a  +  bi .  Then: bi a bi a bi a i b a - - - - + - - + ) 1 ( ) 1 ( . ) 1 ( ) 1 (     ü   2 2 ) 1 ( ) 1 ( ) 1 )( 1 ( ) 1 ( b a b b i b a abi a a + - - + - - + - -    ü The result is purely imaginary when  0 ) 1 ( ) 1 ( ) 1 ( 2 2 = + - - + - b a b b a a    ü i.e.  0 2 2 = - + - b b a a i.e.  2 1 4 1 4 1 2 2 = + - + + - b b a a    i.e.  2 1 ) 2 1 ( ) 2 1 ( 2 2 = - + - b a   ü This equation represents a circle with centre  ) 2 1 , 2 1 (   ü   and radius =  2 1 .   ü 2. Polar Co-ordinates 1. (a) i w z 2 2 4 3 2 2 4 3 3 - + + = +    (b)    - 12 12 π cis     (c) 12 5 4 3 cis 2. (a) 4 2 cis z = cis w 3 = (b)     - 4 2 6 cis (c)    i 5 1 5 3 - - 3. (a) z 2  +  z  – 1 (b) ) 3 ( , 3 , 1 3 2 1 - = = - = cis z cis z z   are the associated values 2 5 2 1 4 + - = z , 2 5 2 1 5 - - = z   are the remaining values. 4. Q and R are   i and i 2 3 - + - = 5. ) 12 ( 18 - cis   or   ) 12 7 ( 18 cis 6. (a) graph (b) n n z z 2 1 = + (c) n = 8. VET Calculus 135 21.Solutions
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VET Calculus 136 21.Solutions
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7. (a) z  = 1 - i 3  = 2 cis (- 3 π ) ü   and    w  = 2 cis  6 . Then  wz  = 4 cis (- 6 )   üü Alternatively z  = 1 - i 3  and  w  = 2 cis  6  =  3  +  i  .  ( ü ) Then  wz  = 2 3  - 2 i .  ( ) (b)     Im ( z )     Re ( z )      - 3 z                wz
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Answer Key and Solutions Part A - 21. SOLUTIONS...

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