(1)
注意，对于两个绝对收敛的级数，有

$$\sum_{n=0}^{\infty} a_n \cdot \sum_{m=0}^{\infty} b_m = \sum_{n=0}^{\infty} \left( \sum_{k=0}^n a_k b_{n-k} \right)$$

所以

$$e^{z+w} = \sum_{n=0}^{\infty} \frac{(z+w)^n}{n!} = \sum_{n=0}^{\infty} \left( \sum_{k=0}^n \frac{z^k}{k!} \cdot \frac{w^{n-k}}{(n-k)!} \right) = e^z \cdot e^w$$

(2)

$$e^{it} = \sum_{n=0}^{\infty} \frac{(it)^n}{n!} = \sum_{n=0}^{\infty} (-1)^n \frac{t^{2n}}{(2n)!} + i \sum_{n=0}^{\infty} (-1)^n \frac{t^{2n+1}}{(2n+1)!} = \cos t + i \sin t$$

(3)
只需证明 $\overline{e^{it}} = e^{-it} \quad$

$$\overline{e^{it}} = \overline{\cos t + i \sin t} = \cos t - i \sin t = \cos(-t) + i \sin(-t) = e^{-it}$$

(4)

$$|e^{it}| = \sqrt{\cos^2 t + \sin^2 t} = 1$$

(5)
令 $z = x + iy$

$$|e^z| = |e^{x+iy}| = |e^x| \cdot |e^{iy}| = e^x \cdot 1 = e^{\mathrm{Re} z} > 0$$

(6)($\Rightarrow$)
由 $e^{z-w} = e^z e^{-w} = e^w e^{-w} = 1$ 得

$$1 = |e^{z-w}| = e^{\mathrm{Re}(z-w)} \implies \mathrm{Re}(z-w) = 0$$

即 $z-w = it,\ t \in \mathbb R$，对比 $e^{z-w} = 1$ 的虚部得

$$0 = \mathrm{Im}(e^{z-w}) = \mathrm{Im}(e^{it}) = \sin t \implies t = 2k\pi,\ k \in \mathbb Z$$

($\Leftarrow$)
令 $z = w + 2k\pi i,\ k \in \mathbb Z$，则

$$e^z = e^{w + 2k\pi i} = e^w \cdot e^{2k\pi i} = e^w \cdot 1 = e^w$$

绝对收敛确保逐项可微，所以

$$\frac{d}{dz} e^z = \frac{d}{dz} \sum_{n=0}^{\infty} \frac{z^n}{n!} = \sum_{n=1}^{\infty} \frac{n z^{n-1}}{n!} = \sum_{n=0}^{\infty} \frac{z^n}{n!} = e^z$$

直接计算验证

$$\frac{e^{iz} + e^{-iz}}{2} = \frac{1}{2} \sum_{n=0}^{\infty} \frac{(iz)^n + (-iz)^n}{n!} = \sum_{n=0}^{\infty} (-1)^n \frac{z^{2n}}{(2n)!} = \cos z$$

$$\frac{e^{iz} - e^{-iz}}{2i} = \frac{1}{2i} \sum_{n=0}^{\infty} \frac{(iz)^n - (-iz)^n}{n!} = \sum_{n=0}^{\infty} (-1)^n \frac{z^{2n+1}}{(2n+1)!} = \sin z$$

由上命题直接得出

$$e^{iz} = \frac{e^{iz} + e^{-iz}}{2} + \frac{e^{iz} - e^{-iz}}{2i} \cdot i = \cos z + i \sin z$$

(1)

$$\cos^2 z + \sin^2 z = \left( \frac{e^{iz} + e^{-iz}}{2} \right)^2 + \left( \frac{e^{iz} - e^{-iz}}{2i} \right)^2 = \frac{e^{2iz} + 2 + e^{-2iz}}{4} + \frac{-e^{2iz} + 2 - e^{-2iz}}{4} = 1$$

(2)

$$\cos(-z) = \frac{e^{-iz} + e^{iz}}{2} = \frac{e^{iz} + e^{-iz}}{2} = \cos z,\quad \sin(-z) = \frac{e^{-iz} - e^{iz}}{2i} = -\frac{e^{iz} - e^{-iz}}{2i} = -\sin z$$

(3)

$$\cos(z+w) = \frac{e^{i(z+w)} + e^{-i(z+w)}}{2} = \frac{e^{iz}e^{iw} + e^{-iz}e^{-iw}}{2} = \frac{(e^{iz} + e^{-iz})(e^{iw} + e^{-iw}) - (e^{iz} - e^{-iz})(e^{iw} - e^{-iw})}{4} = \cos z \cos w - \sin z \sin w$$

(4)

$$\sin(z+w) = \frac{e^{i(z+w)} - e^{-i(z+w)}}{2i} = \frac{e^{iz}e^{iw} - e^{-iz}e^{-iw}}{2i} = \frac{(e^{iz} + e^{-iz})(e^{iw} - e^{-iw}) + (e^{iz} - e^{-iz})(e^{iw} + e^{-iw})}{4i} = \sin z \cos w + \cos z \sin w$$

(5)

$$\cos(z + 2\pi) = \frac{e^{i(z + 2\pi)} + e^{-i(z + 2\pi)}}{2} = \frac{e^{iz}e^{2\pi i} + e^{-iz}e^{-2\pi i}}{2} = \frac{e^{iz} + e^{-iz}}{2} = \cos z$$

$$\sin(z + 2\pi) = \frac{e^{i(z + 2\pi)} - e^{-i(z + 2\pi)}}{2i} = \frac{e^{iz}e^{2\pi i} - e^{-iz}e^{-2\pi i}}{2i} = \frac{e^{iz} - e^{-iz}}{2i} = \sin z$$

绝对收敛确保逐项可微，所以

$$\frac{d}{dz} \cos z = \frac{d}{dz} \sum_{n=0}^{\infty} (-1)^n \frac{z^{2n}}{(2n)!} = \sum_{n=1}^{\infty} (-1)^n \frac{2n z^{2n-1}}{(2n)!} = -\sum_{n=0}^{\infty} (-1)^n \frac{z^{2n+1}}{(2n+1)!} = -\sin z$$

$$\frac{d}{dz} \sin z = \frac{d}{dz} \sum_{n=0}^{\infty} (-1)^n \frac{z^{2n+1}}{(2n+1)!} = \sum_{n=0}^{\infty} (-1)^n \frac{(2n+1) z^{2n}}{(2n+1)!} = \sum_{n=0}^{\infty} (-1)^n \frac{z^{2n}}{(2n)!} = \cos z$$

$$\frac{e^{z} + e^{-z}}{2} = \frac{1}{2} \sum_{n=0}^{\infty} \frac{z^n + (-z)^n}{n!} = \sum_{n=0}^{\infty} \frac{z^{2n}}{(2n)!} = \cosh z$$

$$\frac{e^{z} - e^{-z}}{2} = \frac{1}{2} \sum_{n=0}^{\infty} \frac{z^n - (-z)^n}{n!} = \sum_{n=0}^{\infty} \frac{z^{2n+1}}{(2n+1)!} = \sinh z$$

(1)

$$\cosh^2 z - \sinh^2 z = \left( \frac{e^{z} + e^{-z}}{2} \right)^2 - \left( \frac{e^{z} - e^{-z}}{2} \right)^2 = \frac{e^{2z} + 2 + e^{-2z}}{4} - \frac{e^{2z} - 2 + e^{-2z}}{4} = 1$$

(2)

$$\cosh(iz) = \frac{e^{iz} + e^{-iz}}{2} = \cos z,\quad \cos(iz) = \frac{e^{iz} - e^{-iz}}{2i} = i \cdot \frac{e^{z} - e^{-z}}{2} = i \sinh z$$

(3)

$$\sinh(iz) = \frac{e^{iz} - e^{-iz}}{2} = i \cdot \frac{e^{iz} - e^{-iz}}{2i} = i \sin z,\quad \sin(iz) = \frac{e^{iz} - e^{-iz}}{2i} = i \cdot \frac{e^{z} - e^{-z}}{2} = i \sinh z$$

(4)

$$\cosh(x+iy) = \frac{e^{x+iy} + e^{-(x+iy)}}{2} = \frac{e^x e^{iy} + e^{-x} e^{-iy}}{2} = \frac{(e^x + e^{-x})(\cos y + i \sin y) + (e^x - e^{-x})(\cos y - i \sin y)}{4} = \cosh x \cos y + i \sinh x \sin y$$

(5)

$$\sinh(x+iy) = \frac{e^{x+iy} - e^{-(x+iy)}}{2} = \frac{e^x e^{iy} - e^{-x} e^{-iy}}{2} = \frac{(e^x - e^{-x})(\cos y + i \sin y) + (e^x + e^{-x})(\cos y - i \sin y)}{4} = \sinh x \cos y + i \cosh x \sin y$$

绝对收敛确保逐项可微，所以

$$\frac{d}{dz} \sinh z = \frac{d}{dz} \sum_{n=0}^{\infty} \frac{z^{2n+1}}{(2n+1)!} = \sum_{n=0}^{\infty} \frac{(2n+1) z^{2n}}{(2n+1)!} = \sum_{n=0}^{\infty} \frac{z^{2n}}{(2n)!} = \cosh z$$

$$\frac{d}{dz} \cosh z = \frac{d}{dz} \sum_{n=0}^{\infty} \frac{z^{2n}}{(2n)!} = \sum_{n=1}^{\infty} \frac{2n z^{2n-1}}{(2n)!} = \sum_{n=0}^{\infty} \frac{z^{2n+1}}{(2n+1)!} = \sinh z$$

（连续性）
取 $z_0 \in D_\theta$，令 $\theta_0 = \arg z_0 \in (\theta, \theta + 2\pi),\ r_0 = |z_0|$
注意事实：实数三角函数的反函数在其定义域内连续
此时

$$\cos \theta_0 = \frac{\mathrm{Re} z_0}{r_0},\quad \sin \theta_0 = \frac{\mathrm{Im} z_0}{r_0}$$

所以

$$\theta_0 = \arccos \frac{\mathrm{Re} z_0}{r_0} = \arcsin \frac{\mathrm{Im} z_0}{r_0}$$

即

$$\arg \bigg |_{D_\theta} (z) = \underbrace{\arccos \frac{\mathrm{Re} z}{|z|}}_{\text{continuous}} \to \arccos \frac{\mathrm{Re} z_0}{r_0} = \theta_0 \quad (z \to z_0)$$

(可微性)
先证明以下引理：对于 $w_0 \in \mathbb C$

$$E_{w_0}(w) := \begin{cases} e^{w_0} & w = w_0 \\[4pt] \dfrac{e^w - e^{w_0}}{w - w_0} & w \neq w_0 \end{cases}$$

在 $\mathbb C$ 上连续，证明如下

$$\lim_{w \to w_0} E_{w_0}(w) = \lim_{w \to w_0} \frac{e^w - e^{w_0}}{w - w_0} = \left. \frac{d}{dw} e^w \right|_{w=w_0} = e^{w_0} = E_{w_0}(w_0)$$

根据此引理
接下来，取 $f = \log |_{D_\theta}$，则对于固定的 $z_0 \in D_\theta$

$$\frac{f(z_0 + h) - f(z_0)}{h} = \frac{f(z_0 + h)}{\exp(f(z_0 + h)) - \exp(f(z_0))} = \frac{1}{E_{f(z_0)}(f(z_0 + h))}$$

由于 $f, E_{f(z_0)}$ 均在各自定义域内连续，所以

$$E_{f(z_0)}(f(z_0 + h)) \to E_{f(z_0)}(f(z_0)) = e^{f(z_0)} = z_0 \quad (h \to 0)$$

从而

$$f'(z_0) = \lim_{h \to 0} \frac{f(z_0 + h) - f(z_0)}{h} = \lim_{h \to 0} \frac{1}{E_{f(z_0)}(f(z_0 + h))} = \frac{1}{z_0}$$

直接套用上命题的结论即可，略

令 $f(z) = \displaystyle\sum_{n=1}^{\infty} (-1)^{n-1} \dfrac{z^n}{n}, \ z \in D(0,1)$，系数 $a_n = \dfrac{(-1)^{n-1}}{n}$，需证 $\mathrm{Log}(1+z) = f(z)$

由于 $f$ 在 $D(0,1)$ 上可微，且微分系数由等比级数给出

$$f'(z) = \sum_{n=1}^{\infty} (-1)^{n-1} z^{n-1} = \sum_{n=1}^{\infty} (-z)^{n-1} = \frac{1}{1+z}$$

另一边，$1+z$ 在 $D(0,1)$ 可微，由链式法则也可以给出

$$\frac{d}{dz} (\mathrm{Log}(1+z)) = \frac{1}{1+z}$$

联合得到

$$\frac{d}{dz} (\mathrm{Log}(1+z) - f(z)) = 0,\quad z \in D(0,1)$$

由于 $D(0,1)$ 为领域，且微分系数恒为零，由正则性章节的结论得到

$$\mathrm{Log}(1+z) - f(z) = C,\quad z \in D(0,1)$$

取 $z=0$，得到 $C=0$，所以

$$\mathrm{Log}(1+z) = f(z) = \sum_{n=1}^{\infty} (-1)^{n-1} \frac{z^n}{n},\quad z \in D(0,1)$$

令 $f = \log |_{D_\theta} \quad$

$$\frac{d}{dz} (z^a \bigg |_{D_\theta}) = \frac{d}{dz} (e^{a f(z)}) = e^{a f(z)} \cdot a f'(z) = z^a \cdot a \cdot \frac{1}{z} = a z^{a-1}$$

取 $\ell = \log |_{D_\theta}(a) \in \mathbb C$，则 $a^z = e^{\ell z} \quad$

$$\frac{d}{dz} (a^z) = \frac{d}{dz} (e^{\ell z}) = e^{\ell z} \cdot \ell = a^z \log a \bigg |_{D_\theta}$$

由定义，$w$ 满足

$$w^n = 1 \implies e^{n \log w} = 1 \implies n \log w = 2k\pi i,\quad k \in \mathbb Z$$

所以

$$\log w = \frac{2k\pi i}{n} \implies w = e^{2k\pi i/n},\quad k \in \mathbb Z$$

由于 $e^{2(k+n)\pi i/n} = e^{2k\pi i/n} \cdot e^{2\pi i} = e^{2k\pi i/n}$，所以只需取 $k = 0, 1, 2, \ldots, n-1$ 即可得到所有不同的解，共计 n 个