sol108c1

# sol108c1 - MA108C HW 1 SOLUTIONS PAUL NELSON (1) Let f = u...

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MA108C HW 1 SOLUTIONS PAUL NELSON (1) Let f = u + iv be holomorphic and non-vanishing. Since we have not yet deﬁned the complex logarithm, we prove that log | f | is harmonic by brute-force. We have ∂x log | f | = uu x + vv x u 2 + v 2 , 2 ∂x 2 log | f | = u xx u + u 2 x + v xx v + v 2 x u 2 + v 2 - 2 ( u x u + v x v ) 2 ( u 2 + v 2 ) 2 , and 2 ∂y 2 log | f | = u yy u + u 2 y + v yy v + v 2 y u 2 + v 2 - 2 ( u y u + v y v ) 2 ( u 2 + v 2 ) 2 , so Δlog | f | = u xx u + u 2 x + v xx v + v 2 x + u yy u + u 2 y + v yy v + v 2 y u 2 + v 2 - 2 ( u x u + v x v ) 2 ( u 2 + v 2 ) 2 - 2 ( u y u + v y v ) 2 ( u 2 + v 2 ) 2 . The Cauchy-Riemann equations are u x = v y ,u y = - v x , which implies (by continuity of the derivatives) that u xx + u yy = v xx + v yy = 0, and u 2 x = v 2 y , u 2 y = v 2 x , so 1 2 ( u 2 + v 2 ) 2 Δlog | f | = ( u 2 + v 2 )( u 2 x + v 2 x ) - ( u x u + v x v ) 2 - ( - v x u + u x v ) 2 = ( u 2 + v 2 )( u 2 x + v 2 x ) - u 2 x u 2 - v 2 x v 2 - v 2 x u 2 - u 2 x v 2 = 0 .

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sol108c1 - MA108C HW 1 SOLUTIONS PAUL NELSON (1) Let f = u...

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