# extra_9 - in the interior or C must also be less than one...

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Extra Problems Sheet 9 Stephen Taylor May 20, 2005 1. True or false: Z | z | =1 ¯ zdz = Z | z | =1 1 z dz This is a true statement since both integrals have the value 2 πi . ± 2. If γ is the vertical line segment from z = R > 0 to z = R + 2 πi , then show that ± ± ± ± Z γ e 3 z 1 + e z dz ± ± ± ± 2 πe 3 R e R - 1 Applying a previous homework problem we have ± ± ± ± Z γ e 3 z 1 + e z dz ± ± ± ± Z γ | e 3 z | | 1 + e z | | dz | Since e z takes its largest value when z is real, the value of | dz | is 2 π and e R - 1 < e r + 1, we achieve the desired inequality. ± 3. Find all functions f analytic in G = { z : | z | < R } satisfying f (0) = i and | f ( z ) | ≤ 1 for all z G . Since the function is bounded above in its domain it must be constant by the Maximum Modulus Principle. Since it takes on the value i and 0 we ﬁnd that f i is the only function. ± 4. Suppose that f is analytic inside and on the simple closed curve C and that | f ( z ) - 1 | < 1 for all z on C . Prove that f has no zeros inside C . Since that maximum value on the curve is less than one, the maximum value

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Unformatted text preview: in the interior or C must also be less than one. So if f has a zero, then | f ( z )-1 | < 1 → 1 < 1 which would be a contradiction. So f has no zeros inside C . 1 ± 5. Consider the function f ( z ) = ( z + 1) 2 and the closed triangular region R with vertices at the points z = 0 , 2 ,i . Find the points in R where | f ( z ) | has its maximum and minimum values. By a theory in linear programming, it suﬃces to check the vertices for extremal values. We note z = 0 gives a minimum of 1, and z = 2 gives a maximum of 9. ± 6. Prove that for any analytic function f that is never zero in a simply connected domain D there exists a single-valued branch of log f ( z ) analytic in D . 2...
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extra_9 - in the interior or C must also be less than one...

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