Question 3.40: (a) Suppose u1(x, y) = ∂u/∂x and u2(x, y) = ∂u/∂y. Prove tha...
(a) Suppose u_1(x, y)=\partial u / \partial x \text { and } u_2(x, y)=\partial u / \partial y. Prove that f^{\prime}(z)=u_1(z, 0)-i u_2(z, 0).
(b) Show how the result in (a) can be used to solve Problems 3.7 and 3.8.
(a) From Problem 3.5, we have f^{\prime}(z)=\frac{\partial u}{\partial x}-i \frac{\partial u}{\partial y}=u_1(x, y)-i u_2(x, y). Putting y=0, this becomes f^{\prime}(x)=u_1(x, 0)-i u_2(x, 0).
Then, replacing x by z, we have as required f^{\prime}(z)=u_1(z, 0)-i u_2(z, 0).
(b) Since we are given u=e^{-x}(x \sin y-y \cos y), we have
\begin{aligned}&u_1(x, y)=\frac{\partial u}{\partial x}=e^{-x} \sin y-x e^{-x} \sin y+y e^{-x} \cos y \\ &u_2(x, y)=\frac{\partial u}{\partial y}=x e^{-z} \cos y+y e^{-z} \sin y-e^{-x} \cos y \end{aligned}so that from part (a),
f^{\prime}(z)=u_1(z, 0)-i u_2(z, 0)=0-i\left(z e^{-z}-e^{-z}\right)=-i\left(z e^{-z}-e^{-z}\right)Integrating with respect to z we have, apart from a constant, f(z)=i z e^{-z}. By separating this into real and imaginary parts, v=e^{-x}(y \sin y+x \cos y) apart from a constant.
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