# lecture5 - f is indeed diFerentiable at 0 and ± i only...

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However there is a partial converse. Theorem 3.4. [S&T4.6] If f ( z ) = u ( x,y ) + iv ( x,y ) is a complex function on an open set S and at z 0 = x 0 + iy 0 S the partial derivatives ∂u ∂x , ∂v ∂x , ∂u ∂y , ∂v ∂y all exist, are continuous and satisfy the C-R equations ∂u ∂x = ∂v ∂y , ∂v ∂x = ∂u ∂y then f is differentiable at z 0 . Proof. Omitted. Example. Let f ( z ) = ( Re z )( Im z )( z 2 + 1) . Use the Cauchy-Riemann equations to find all points z at which f ( z ) exists. Solution. We have f ( x + iy ) = xy (( x + iy ) 2 + 1) = x 3 y xy 3 + xy + 2 ix 2 y 2 . Hence u ( x,y ) = x 3 y xy 3 + xy, v ( x,y ) = 2 x 2 y 2 . The Cauchy-Riemann equations yield 3 x 2 y y 3 + y = 4 x 2 y,x 3 3 xy 2 + x = 4 xy 2 . If x = 0 in the second equation, then y = 0 or ± 1. Assume x negationslash = 0; then the second equation becomes x 2 + 1 = y 2 , which is impossible. Hence the only points where f can be differentiable are 0 and ± i . Since u and v have partial derivatives which are clearly continuous,
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Unformatted text preview: f is indeed diFerentiable at 0 and ± i only. ±rom the Cauchy-Riemann equations we can deduce: Theorem 3.5. [S&T4.7] If f is diFerentiable in a domain D and f ′ ( z ) = 0 for all z ∈ D then f is constant in D . Proof. We have f ′ ( z ) = ∂u ∂x + i ∂v ∂x = ∂v ∂y − i ∂u ∂y , so f ′ ( z ) = 0 implies all the partial derivatives of u and v are zero. Two points in D are joined by a step path. On each horizontal and vertical segment, u and v have zero derivatives and so are constant. Therefore f takes the same values at the two endpoints of the path and is hence constant on D . s 1...
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