HW11sol - P ROBLEM SET 11 Partial solution 6 Consider the function h = f g Since f and g are analytic so is h Since real parts of f and g are

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P ROBLEM SET 11 Partial solution 6. Consider the function h = f - g . Since f and g are analytic, so is h . Since real parts of f and g are identical on the boundary B , the real part of h is equal to 0 on B . Let us denote by u the real part, and by v the imaginary part of h . By the consequence of the Maximum Modulus Principle, that was given in (4) on page 192 of the textbook, both the maximum and the minimum on D B of u are achieved on B . Therefore both the maximum and the minimum of u on B D are equal to 0 . Hence u is constant, and equal to 0 on D B . By the Cauchy-Riemann equations v y = u x = 0 and v x = - u y = 0 on D B . Hence v must be a constant on D B ; we denote it by α . Therefore f = g + h = g + on D B . 8. Let us consider first the case that f is equal to zero at some point w B . Then for all z D B , | f ( z ) | ≥ | f ( w ) | = 0 and hence | f | achieves its minimum on the boundary (which is what we wanted to prove). Now consider the case that
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HW11sol - P ROBLEM SET 11 Partial solution 6 Consider the function h = f g Since f and g are analytic so is h Since real parts of f and g are

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