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# HW7 - MATH 185 SOLUTIONS OF HW VII(1(41.5(30pts By triangle...

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MATH 185 SOLUTIONS OF HW VII (1) (41.5)(30pts) By triangle inequality, Logz z 2 | ln R | + | | R 2 < ln R + π R 2 since z = ln R + where z = Re and - π < θ < π . (N.B. | ln R | = ln R since R > 1.) Hence, C R Logz z 2 dz C R Logz z 2 dz < 2 πR ln R + π R 2 = 2 π ln R + π R Moreover, lim R →∞ 2 π (ln R + π ) R = lim R →∞ 2 π (1 /R ) 1 = 0 by L’Hospital’s rule, we can conclude that lim R →∞ C R Logz z 2 dz = 0 NOTICE. If you finish to solve the problem, you need to check WHAT YOU HAVE PROVEN. Even though you write the same sentence appear- ing the problem, it is important because it can clarify your assertion. (2) (43. 5)(30pts) First we may replace the branch - π < θ < π by - π/ 2 < θ < 3 π/ 2 since - 1 is not defined in Logz and change of the branch cut does not affect to the result in this situation. So, we can use the antiderivative z 1+ i 1+ i of z i . Hence, 1 - 1 z i dz = z 1+ i 1 + i 1 - 1 = 1 - i 2 ( (1) i - ( - 1) i ) = 1 - i 2 (1 + e - π ) (3) (46. 4) (a)(20pts) Let I 1 , I 2 , I 3 , I 4 be the line segment from - a to a , from a to a + bi , from a + bi to - a + bi , from - a + bi to - a , respectively. Since f (

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HW7 - MATH 185 SOLUTIONS OF HW VII(1(41.5(30pts By triangle...

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