math185-hw3sol - MATH 185 COMPLEX ANALYSIS FALL 2007\/08 PROBLEM SET 3 SOLUTIONS The real and imaginary parts of z C are denoted by Re(z and Im(z

math185-hw3sol - MATH 185 COMPLEX ANALYSIS FALL 2007/08...

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MATH 185: COMPLEX ANALYSIS FALL 2007/08 PROBLEM SET 3 SOLUTIONS The real and imaginary parts of z C are denoted by Re( z ) and Im( z ) respectively. If Ω C , we let Ω R := { ( x, y ) R 2 | x + iy Ω } . We will use | · | to denote the complex modulus in C and k · k to denote the vector 2-norm in R 2 . For f : Ω C , we say that z Ω is a zero of f if f ( z ) = 0; we say that f is identically zero on Ω, denoted f 0, if f ( z ) = 0 for all z Ω. You may use without proof any results that we have already proved in the lectures. 1.Letf:CCbe expressed in the usual formf=u+ivwhereu, v:R2R.Define thefunction (i) f is differentiable at z = x + iy as a function from C C . (ii) F is differentiable at ( x, y ) as a function from R 2 R 2 and the Jacobian matrix J F ( x, y ) R 2 × 2 is the composition of a dilation and a rotation, ie. J F ( x, y ) = r cos θ sin θ - sin θ cos θ for some r 0 and - π/ 2 θ π/ 2. How are r and θ related to u and v ? Solution. By Theorem 2.8 in the lectures, if f satisfies (i), then the Cauchy-Riemann equa- tions u x = v y , u y = - v x must hold at z . In which case, J F ( x, y ) = u x u y v x v y = u x u y - u y u x = r cos θ sin θ - sin θ cos θ where r = q u 2 x + u 2 y and θ = arctan( u y /u x ). If F satisfies (ii), then we let u x = v y = r cos θ and u y = - v x = r sin θ . In which case, f 0 ( z ) = u x + iv x = r (cos θ - i sin θ ) . In either case, we have the equality f ( z + h ) - f ( z ) h - f 0 ( z ) = | f ( z + h ) - f ( z ) - f 0 ( z ) h | | h | = k F ( x + h 1 , y + h 2 ) - F ( x, y ) - J F ( x, y ) h k k h k where h = h 1 + ih 2 C and h = ( h 1 , h 2 ) > R 2 . The denominators are equal because | h | = p h 2 1 + h 2 2 = k h k . The numerators are equal because f 0 ( z ) h = r (cos θ - i sin θ )( h 1 + ih 2 ) = r ( h 1 cos θ + h 2 sin θ ) + ir ( h 2 cos θ - h 1 sin θ ) and J F ( x, y ) h = r cos θ sin θ - sin θ cos θ h 1 h 2 = r ( h 1 cos θ + h 2 sin θ ) r ( h 2 cos θ - h 1 sin θ ) . In the other words, the limit as h 0 in C exists iff the limit as h 0 in R 2 exists. Date : October 13, 2007 (Version 1.3). 1
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2. Let Ω C be a region containing the point z = 0 and f = u + iv be a complex-valued function on Ω. Suppose all following three conditions are satisfied: (i) f is continuous at z = 0; (ii) the partial derivatives u x , u y , v x , v y exist at z = 0; (iii) the Cauchy-Riemann equations, u x = v y , u y = - v x , are satisfied at z = 0.
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