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380_hw_4_solutions[1]

380_hw_4_solutions[1] - z 1 and z 2 sin z 1 z 2 = sin z 1...

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Math 380 Solutions to hw 4 1. (Exercise 3.3.1) (a) z 5 + (2 + 2 i ) z 4 + 2 iz 3 = z 3 z 2 + 2(1 + i ) z + 2 i ± : Note that (1 + i ) 2 = 2 i; so we get z 5 + (2 + 2 i ) z 4 + 2 iz 3 = z 3 ( z + 1 + i ) 2 : (b) z 4 16 = ( z 2 + 4)( z 2 4) = ( z + 2 i )( z 2 i )( z + 2)( z 2) : (c) Let p ( z ) = 1 + z + z 2 + ± ± ± + z 6 : Note that ( z 1) p ( z ) = z 7 1 : So the roots of p are all the seventh roots of unity, except z = 1 : Thus p ( z ) = 6 Y k =1 ² z e 2 7 ³ : 2. 3.1.14. Let R ( z ) be a rational function with a pole of order m at z 0 : This means we can express R ( z ) = p ( z ) ( z z 0 ) m q ( z ) ((1)) where p and q z 0 : We must show that R 0 ( z ) can be expressed in the form R ( z ) = p 1 ( z ) ( z z 0 ) m +1 q 1 ( z ) where p 1 and q 1 z 0 : The quotient rule applied to (1) yields R 0 ( z ) = p 0 ( z ) ( z z 0 ) m q ( z ) p ( z ) h m ( z z 0 ) m 1 q ( z ) + ( z z 0 ) m q 0 ( z ) i ( z z 0 ) 2 m q ( z ) 2 = p 1 ( z ) ( z z 0 ) m +1 q 1 ( z ) where p 1 ( z ) = ( z z 0 )[ p 0 ( z ) q ( z ) p ( z ) q 0 ( z )] mp ( z ) q ( z ) and q 1 ( z ) = [ q ( z )] 2 : Note that p 1 and q 1 are polynomials with p 1 ( z 0 ) = mp ( z 0 ) q ( z 0 ) 6 = 0 and q 1 ( z 0 ) = [ q ( z 0 )] 2 6 = 0 : 3. (a) Verify that cos 2 ( z ) + sin 2 ( z ) = 1 for all complex numbers z: cos 2 z + sin 2 z = ´ e iz + e iz 2 µ 2 + ´ e iz e iz 2 i µ 2 = e 2 iz + 2 + e 2 iz · 4 e i 2 z 2 + e 2 iz 4 = 4 4 = 1 : 1
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(b) Verify that for all complex numbers
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Unformatted text preview: z 1 and z 2 , sin ( z 1 + z 2 ) = sin z 1 cos z 2 + cos z 1 sin z 2 : The easiest way to do this is to start with the right hand side of the equation: sin z 1 cos z 2 + cos z 1 sin z 2 = e iz 1 & e & iz 1 2 i ± e iz 2 + e & iz 2 2 + e iz 1 + e & iz 1 2 ± e iz 2 & e & iz 2 2 i = e i ( z 1 + z 2 ) + e i ( z 1 & z 2 ) & e i ( z 2 & z 1 ) & e & i ( z 1 + z 2 ) 4 i + e i ( z 1 + z 2 ) & e i ( z 1 & z 2 ) + e i ( z 2 & z 1 ) & e & i ( z 1 + z 2 ) 4 i = 2 e i ( z 1 + z 2 ) & 2 e & i ( z 1 + z 2 ) 4 i = e i ( z 1 + z 2 ) & e & i ( z 1 + z 2 ) 2 i = sin ( z 1 + z 2 ) : 2...
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380_hw_4_solutions[1] - z 1 and z 2 sin z 1 z 2 = sin z 1...

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