摘要：
\(f(z) = \mu(x,y)+i\nu(x,y)\) 调和函数，\(\nabla^2 \mu= 0\rightarrow \nabla\cdot \nabla \mu = 0\) \(\nabla^2\nu = 0\) \(f(z) = z^2 =(x^2-y^2)+i(2xy)\) 画图有 阅读全文
摘要：
https://www.zhihu.com/question/23378396 https://blog.csdn.net/weixin_37887116/article/details/107479588 阅读全文