mat67-Homework1_Solutions

# mat67-Homework1_Solutions - MAT067 University of...

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Unformatted text preview: MAT067 University of California, Davis Winter 2007 Solutions to Homework Set 1 1. Express the following complex numbers in the form x + yi for x, y ∈ R : (a) (2 + 3 i ) + (4 + i ) Solution : By direct computation, (2 + 3 i ) + (4 + i ) = (2 + 4) + (3 + 1) i = 6 + 4 i. (b) (2 + 3 i ) 2 (4 + i ) Solution : By direct computation, (2 + 3 i ) 2 (4 + i ) =[(2 + 3 i )(2 + 3 i )](4 + i ) =(4 + 6 i + 6 i + 9 i 2 )(4 + i ) =(4 + 12 i − 9)(4 + i ) =( − 5 + 12 i )(4 + i ) = − 20 − 5 i + 48 i + 12 i 2 = − 32 + 43 i. (c) 2 + 3 i 4 + i Solution : By direct computation, 2 + 3 i 4 + i = 2 + 3 i 4 + i · 4 − i 4 − i = 8 − 2 i + 12 i − 3 i 2 16 − 4 i + 4 i − i 2 = 11 + 10 i 17 = 11 17 + 10 17 i. (d) 1 i + 3 1 + i Solution : By direct computation, 1 i + 3 1 + i = 1 i · − i − i + 3 1 + i · 1 − i 1 − i = − i + 3 − 3 i 2 = − i + 3 2 − 3 2 i = 3 2 − 5 2 i. (e) ( − i )- 1 Solution : By direct computation, ( − i )- 1 = 1 − i = 1 − i · i i = i − i 2 = i − ( − 1) = i. 2 2. Compute the real and imaginary parts of the following expressions, where z is the complex number x + yi and x, y ∈ R : (a) 1 z 2 Solution : Set z = x + yi for x, y ∈ R . Then, by direct computation, 1 z 2 = 1 z 2 · ( z ) 2 ( z ) 2 = ( z ) 2 ( z z ) 2 = ( z ) 2 ( | z | 2 ) 2 = ( z ) 2 | z | 4 = ( x − yi ) 2 ( radicalbig x 2 + y 2 ) 4 = ( x 2 − y 2 ) − 2 xyi ( x 2 + y 2 ) 2 so that Re parenleftbigg 1 z 2 parenrightbigg = ( x 2 − y 2 ) ( x 2 + y 2 ) 2 and Im parenleftbigg 1 z 2 parenrightbigg = − 2 xy ( x 2 + y 2 ) 2 . (b) 1 3 z + 2 Solution : Set z = x + yi for x, y ∈ R . Then, by direct computation, 1 3 z + 2 = 1 (3 x + 2) + 3 yi · (3 x + 2) − 3 yi (3 x + 2) − 3 yi = (3 x + 2) − 3 yi (3 x + 2) 2 + 9 y 2 so that Re parenleftbigg 1 3 z + 2 parenrightbigg = (3 x + 2) (3 x + 2) 2 + 9 y 2 and Im parenleftbigg 1 3 z + 2 parenrightbigg = − 3 y (3 x + 2) 2 + 9 y 2 . (c) z + 1 2 z − 5 Solution : Set z = x + yi for x, y ∈ R . Then, by direct computation, z + 1 2 z − 5 = ( x + 1) + yi (2 x − 5) + 2 yi · (2 x − 5) − 2 yi (2 x − 5) − 2 yi = ( x + 1)(2 x − 5) + 2 y 2 + [ − 2 y ( x + 1) + y (2 x − 5)] i (2 x − 5) 2 + 4 y 2 so that Re parenleftbigg z + 1 2 z − 5 parenrightbigg = 2 x 2 + 2 y 2 − 3 x − 5 (2 x − 5) 2 + 4 y 2 and Im parenleftbigg z + 1 2 z − 5 parenrightbigg = − 7 y (2 x − 5) 2 + 4 y 2 . (d) z 3 Solution : Set z = x + yi for x, y ∈ R . Then, by direct computation, z 3 =[( x + yi )( x + yi )]( x + yi ) =[( x 2 − y 2 ) + 2 xyi ]( x + yi ) =( x 2 − y 2 ) x − 2 xy 2 + [( x 2 − y 2 ) y + 2 x 2 y ] i = x 3 − 3 xy 2 + (3 x 2 y − y 3 ) i so that Re ( z 3 ) = x 3 − 3 xy 2 and Im ( z 3 ) = 3 x 2 y − y 3 . 3 3. Solve the following equations for z a complex number: (a) z 5 − 2 = 0 Solution : Set z 5 = x + yi = re θi for x, y, r, θ ∈ R . Then x = 2 and y = 0 so that r = radicalbig x 2 + y 2 = √ 2 2 + 0 2 = 2. Moreover, cos( θ ) = x/r = 2 / 2 = 1 and sin( θ ) =...
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## This homework help was uploaded on 04/08/2008 for the course MATH 67 taught by Professor Schilling during the Fall '07 term at UC Davis.

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mat67-Homework1_Solutions - MAT067 University of...

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