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Question 3.41: Suppose A is real or, more generally, suppose Im A is harmon...

Suppose A is real or, more generally, suppose Im A is harmonic. Prove that |curl grad A|= 0.

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If A = P + Qi, we have

\operatorname{grad} A=\left(\frac{\partial}{\partial x}+i \frac{\partial}{\partial y}\right)(P+i Q)=\frac{\partial P}{\partial x}-\frac{\partial Q}{\partial y}+i\left(\frac{\partial P}{\partial y}+\frac{\partial Q}{\partial x}\right)

Then

\begin{aligned}\mid \text { curl } \operatorname{grad} A \mid &=\left|\operatorname{Im}\left[\left(\frac{\partial}{\partial x}-i \frac{\partial}{\partial y}\right)\left\{\frac{\partial P}{\partial x}-\frac{\partial Q}{\partial y}+i\left(\frac{\partial P}{\partial y}+\frac{\partial Q}{\partial x}\right)\right\}\right]\right| \\ &=\left|\operatorname{Im}\left[\frac{\partial^2 P}{\partial x^2}-\frac{\partial^2 Q}{\partial x \partial y}+i\left(\frac{\partial^2 P}{\partial x \partial y}+\frac{\partial^2 Q}{\partial x^2}\right)-i\left(\frac{\partial^2 P}{\partial y \partial x}-\frac{\partial^2 Q}{\partial y^2}\right)+\left(\frac{\partial^2 P}{\partial y^2}+\frac{\partial^2 Q}{\partial y \partial x}\right)\right]\right| \\ &=\left|\frac{\partial^2 Q}{\partial x^2}+\frac{\partial^2 Q}{\partial y^2}\right|\end{aligned}

Hence if Q = 0, i.e., A is real, or if Q is harmonic, |curl grad A|= 0.

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