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M461solasstA

# M461solasstA - MAT 461 Assignment A Solutions(1 Sketch the...

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MAT 461 Assignment A - Solutions (1) Sketch the set of points detemined by the given condition. a. filled in circle, radius 5, center 1 - 2 i . The circle should pass through the origin. b. line y = x c. circle radius 1, center - 2 i d. all below the line y=-x (2) (a) and (c) (3) (cos π 20 + i sin π 20 ) 10 = cos(10 π 20 ) + i sin(10 π 20 ) (de Moivre’s formula) = cos π 2 + i sin π 2 = 0 + i = i (4) | z 2 - 2 z + 4 | | z | 2 + 2 | z | + 4 = 2 2 + 2 * 2 + 4 = 12 | 3 z + 10 | | 3 | z | - 10 | = | 6 - 10 | = 4 vextendsingle vextendsingle vextendsingle vextendsingle z 2 - 2 z + 4 3 x + 10 vextendsingle vextendsingle vextendsingle vextendsingle 12 4 = 3 (5) - 8 - i 8 3 = 16 e i - 2 π 3 (16 e i ( - 2 π 3 + 2 πk )) 1 4 = 2 e i ( - π 6 + πk 2 ) The roots are 2 e - i π 6 , 2 e i 2 π 6 , 2 e i 5 π 6 , 2 e i 8 π 6 or 3 - i, 1 + i 3 , - 3 + i, - 1 - i 3 (6) ( - 1 + i 3)( - i ) (2 3 + i 2) = (2 e i 2 π 3 )( e - i π 2 ) 4 e i π 6 = 1 2 e i 0 = 1 2 1

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(7) a. lim z 2 i z 2 + 4 z - 2 i = lim z 2 i ( z - 2 i )( z + 2 i ) z - 2 i = lim z 2 i z + 2 i = 4 i b. lim z 0 | 3 + z | 2 - 9 z = lim z 0 (3 + z ) (3 + z ) - 9 z = lim z 0 (3 + z )(3 + ¯ z ) - 9 z = lim z 0 9 + 3 z + 3¯ z + z ¯ z - 9 z = lim z 0 3 + ¯ z + z z Using z = x + iy , if we check this limit along the line x = 0, we get lim y 0 3 - y +
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M461solasstA - MAT 461 Assignment A Solutions(1 Sketch the...

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