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Unformatted text preview: MATH 185 SOLUTIONS OF HW VII (1) (41.5)(30pts) By triangle inequality, Logz z 2 | ln R | + | i | R 2 < ln R + R 2 since z = ln R + i where z = Re i and- < < . (N.B. | ln R | = ln R since R > 1.) Hence, Z C R Logz z 2 dz Z C R Logz z 2 dz < 2 R ln R + R 2 = 2 ln R + R Moreover, lim R 2 (ln R + ) R = lim R 2 (1 /R ) 1 = 0 by LHospitals rule, we can conclude that lim R Z 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- < < by- / 2 < < 3 / 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 ....
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This note was uploaded on 09/01/2010 for the course MATH 185 taught by Professor Lim during the Fall '07 term at University of California, Berkeley.
- Fall '07