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hw1_soln - Complex Analysis Fall 2007 Homework 1 Solutions...

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Unformatted text preview: Complex Analysis Fall 2007 Homework 1: Solutions 1.1.2. (a) (2 + 3 i )(4 + i ) = (8- 3) + (12 + 2) i = 5 + 14 i (b) (8 + 6 i ) 2 = (64- 36) + (48 + 48) i = 28 + 96 i (c) 1 + 3 1 + i 2 = 1 + 3(1- i ) (1 + i )(1- i ) 2 = 1 + 3- 3 i 2 2 = 5 2- 3 2 i 2 = 25 4- 9 4 +- 15 4- 15 4 i = 4- 15 2 i 1.1.6. (a) If z = x + iy we have z + 1 2 z- 5 = ( z + 1)(2¯ z- 5) (2 z- 5)(2¯ z- 5) = 2 z ¯ z- 5 z + 2¯ z- 5 4 z ¯ z- 10( z + ¯ z ) + 25 = 2 | z | 2- 5 z + 2¯ z- 5 4 | z | 2- 10( z + ¯ z ) + 25 = 2( x 2 + y 2 )- 5 x + 2 x- 5 +- 5 yi- 2 yi 4( x 2 + y 2 )- 20 x + 25 = 2( x 2 + y 2 )- 3 x- 5 4( x 2 + y 2 )- 20 x + 25 +- 7 y 4( x 2 + y 2 )- 20 x + 25 i so that Re z + 1 2 z- 5 = 2( x 2 + y 2 )- 3 x- 5 4( x 2 + y 2 )- 20 x + 25 , Im z + 1 2 z- 5 =- 7 y 4( x 2 + y 2 )- 20 x + 25 . 1 (b) If z = x + iy then z 3 = ( x + iy ) 3 = x 3 + 3 x 2 yi + 3 xy 2 i 2 + y 3 i 3 = ( x 3- 3 xy 2 ) + (3 x 2 y- y 3 ) i so that Re z 3 = x 3- 3 xy 2 , Im z 3 = 3 x 2 y- y 3 . 1.1.18. (a) (1- i )- 1 = 1+ i (1- i )(1+ i ) = 1+ i 2 = 1 2 + 1 2 i (b) 1+ i 1- i = (1 + i )(1- i )- 1 = (1 + i ) ( 1 2 + 1 2 i ) = ( 1 2- 1 2 ) + ( 1 2- 1 2 ) i = i 1.2.2. (a) The equation z 6 + 8 = 0 is equivalent to z 6 =- 8. Since | - 8 | = 8 and arg(- 8) = π , the solutions to the latter equation are z k = 6 √ 8 cos π 6 + πk 3 + i sin π 6 + πk 3 for k = 0 , 1 , . . . , 5. (b) The equation z 3- 4 = 0 is equivalent to z 3 = 4 which, since | 4 | = 4 and arg(4) = 0, has the solutions z k = 3 √ 4 cos 2 πk 3 + i sin 2 πk 3 for k = 0 , 1 , 2. 1.2.4. Recalling that conjugation preserves the arithmetic of C , we have (8- 2 i ) 10 (4 + 6 i ) 5 = (8 + 2 i ) 10 (4- 6 i ) 5 . 1.2.6. DeMoivre’s formula and the binomial theorem give cos 6 x + i sin 6 x = (cos x + i sin x ) 6 = (cos 6 x- 15 cos 4 x sin 2 x + 15 cos 2 x sin 4 x- sin 6 x ) + i (6 cos 5 x sin x- 20 cos 3 x sin 3 x + 6 cos x sin 5 x ) ....
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hw1_soln - Complex Analysis Fall 2007 Homework 1 Solutions...

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