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M1sols

# M1sols - Math 185 Solutions to Midterm I 1 1 1 z 6 2 i 2 =...

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Math 185 – Solutions to Midterm I 1. z 6 - 1 2 + i 1 2 = 0 if and only if z 6 = 1 2 - i 1 2 . So we want to find all sixth roots of the number 1 2 - i 1 2 = e - π 4 . If z = re , the equation z 6 = e - π 4 becomes r 6 e i 6 θ = e - π 4 so r = 1 and 6 θ = - π 4 +2 for some k Z . Thus the solutions to the equation are the complex numbers of the form e i ( - π 24 + 2 6 ) for k = 0 , 1 , . . . , 5. 2. (a) The function z (1+ i ) z +1 is a composition of the map w = (1+ i ) z (tilting by π 4 +scaling by a factor of 2) and the translation w w +1. Hence it maps rectangles to rectangles. The image of S is thus the rectangle with vertices 1 , 2+ i, 2 - 2 π + i (2 π +1) , 1 - 2 π + i 2 π which are the images of the vertices 0, 1,1 + 2 πi and 2 πi respectively. (b) Since e z = e x cos ( y ) + ie x sin ( y ) The horizontal line segments y = 0 and y = 2 π are both mapped to the ray 1 r e , θ = 0 and the vertical lines segments x = 0 and x = 1 are mapped to circles centered at the origin of radii 1 and e respectively. The region is mapped to the annulus 1 r e and 0 θ 2 π .

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