2.设$f,g$都在$z_{0}$处可微且$f(z_{0})=g(z_{0})=0,g‘(z_{0})\neq0$.证明:$$\lim_{z\to z_{0}}\frac{f(z)}{g(z)}=\frac{f‘(z_{0})}{g‘(z_{0})}.$$

证明    注意\begin{align*}\left|\frac{f(z)}{g(z)}-\frac{f‘(z_{0})}{g‘(z_{0})}\right|&=\left|\left(\frac{f(z)-f(z_{0})}{z-z_{0}}g‘(z_{0})-\frac{g(z)-g(z_{0})}{z-z_{0}}f‘(z_{0})\right)/\left(g‘(z_{0})\cdot\frac{g(z)-g(z_{0}}{z-z_{0}}\right)\right|\to0\end{align*}

即可.

3.设$f,g$分别是区域$D,G$上的全纯函数,如果$f(D)\subset G$,那么$g\circ f$也是$D$上的全纯函数且$$(g\circ f)‘(z)=g‘(f(z))f‘(z).$$

证明    注意$$\frac{g\circ f(z)-g\circ f(z_{0})}{z-z_{0}}=\frac{g\circ f(z)-g\circ f(z_{0})}{f(z)-f(z_{0})}\cdot\frac{f(z)-f(z_{0})}{z-z_{0}}\to g‘(f(z_{0}))f‘(z_{0})$$

4.设区域$G$和区域$D$关于实轴对称.证明：如果$f(z)$是$D$上的全纯函数,那么$\overline{f(\overline{z})}$是$G$上的全纯函数.

证明    任取$z_{0}\in G$,那么$$\frac{\overline{f(\overline{z})}-\overline{f(\overline{z_{0}})}}{z-z_{0}}=\overline{\left(\frac{f(\overline{z})-f(\overline{z_{0}})}{\overline{z}-\overline{z_{0}}}\right)},z\in G$$

 注意$\overline{z},\overline{z_{0}}\in D$,且$z\to z_{0}$时必有$\overline{z}\to\overline{z_{0}}$.从而上式极限存在,即$f$在$G$中全纯.

2.1复变函数的导数习题解答