M461solasstA

M461solasstA - MAT 461 Assignment A - Solutions (1) Sketch...

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Unformatted text preview: MAT 461 Assignment A - Solutions (1) Sketch the set of points detemined by the given condition. a. filled in circle, radius 5, center 1 - 2i. The circle should pass through the origin. b. line y = x c. circle radius 1, center -2i d. all below the line y=-x (2) (a) and (c) (3) (cos + i sin )10 = cos(10 ) + i sin(10 ) 20 20 20 20 = cos + i sin 2 2 = 0+i=i (de Moivre's formula) (4) |z 2 - 2z + 4| = |3z + 10| = 2 z - 2z + 4 3x + 10 (5) -2 -8 - i8 3 = 16ei 3 -2 - 1 k (16ei( 3 + 2k)) 4 = 2ei( 6 + 2 ) The roots are or (6) 2 (-1 + i 3)(-i) (2ei 3 )(e-i 2 ) = 4ei 6 (2 3 + i2) 1 i0 1 e = = 2 2 2e-i 6 , 2ei 6 , 2ei 6 , 2ei 6 3 - i, 1 + i 3, - 3 + i, -1 - i 3 2 5 8 |z|2 + 2|z| + 4 22 + 2 2 + 4 = 12 |3|z| - 10| |6 - 10| = 4 12 =3 4 1 (7) a. z2 + 4 (z - 2i)(z + 2i) = lim z2i z - 2i z2i z - 2i = lim z + 2i lim z2i = 4i b. (3 + z)(3 + z) - 9 |3 + z|2 - 9 = lim z0 z0 z z (3 + z)(3 + z ) - 9 = lim z0 z 9 + 3z + 3 + z - 9 z z = lim z0 z 3 z = lim 3 + z + z0 z lim Using z = x + iy, if we check this limit along the line x = 0, we get y0 lim 3 - y + -3y =0 y 3x =6 x Along the line y = 0, we get x0 lim 3 + x + Since these are different, the limit does not exist. c. ( 1 - 2)2 (z - 2)2 z2 2 = lim z 1 2 z z0 3( ) 3z 2 z z 2 (1 - 2z) = lim z0 3 1 = 3 lim (8) z x y 1 = 2 = 2 +i 2 2 z |z| x +y x + y2 When u = 1 4 on the image, we get x x2 + y 2 4x 2 x - 4x + 4 + y 2 (x - 2)2 + y 2 1 4 = x2 + y 2 = 4 (complete the square) 2 = 2 = 2 Repeat for the 3 other sides of the square. You get four circles - radius 1 - center 1; radius 1 - center i; radius 2 - center 2; radius 2 - center 2i. The preimage of the square is the region inside both the big circles, but outside both the small circles. 3 ...
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This note was uploaded on 05/11/2008 for the course MAT 461 taught by Professor Ibrahim during the Fall '07 term at ASU.

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M461solasstA - MAT 461 Assignment A - Solutions (1) Sketch...

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