复变期中小测 · 复变函数汇总

1.
该函数奇点满足 $cos z = \frac{e ^{i z} + e ^{- i z}}{2} = - 2$ ,整理得

$(e^{i z})^{2} + 4 e^{i z} + 1 = 0,$

从中解得 $e^{i z} = - 2 \pm 3 < 0$ ，因此其全部奇点为

$z_{k 1} = - i Ln (- 2 - 3) = - i [ln (2 + 3) + i π + 2 k_{1} π i]$
$= (2 k_{1} + 1) π - i ln (2 + 3) (k_{1} \in Z),$
$z_{k 2} = - i Ln (- 2 + 3) = - i [ln (2 - 3) + i π + 2 k_{2} π i]$
$= (2 k_{2} + 1) π - i ln (2 - 3) (k_{2} \in Z) .$
2.
1. (1)
  经计算，
  
  $i^{3} = exp (3 Ln i) = exp [3 (0 + i Arg i)] = exp [\frac{3}{2} (4 k + 1) π i] (k \in Z) .$
2. (2)
  对 $\forall x \in R$ ，由于
  
  $sin (i x) = \frac{e ^{- x} - e ^{x}}{2 i} = i sinh x, cos (i x) = \frac{e ^{- x} + e ^{x}}{2} = cosh x,$
  
  因此
  
  $sin (1 + 2 i) = sin 1 cos (2 i) + cos 1 sin (2 i) = sin 1 cosh 2 + i cos 1 sinh 2.$
3.
记 $f (z) = \frac{1}{z ( z - 2 i ) ( 2 z + 2 ) ^{2}}$ . 在闭路 $C$ 所包围的区域内， $f (z)$ 具有1级极点 $z_{0} = 0, z_{2} = 2 i$ 与2级极点 $z_{1} = - 1$ ，由留数定理，

$I = \oint_{C} f (z) d z = 2 π i k = 0 \sum 2 Res [f (z), z_{k}]$
$= 2 π i {\frac{1}{( - 2 i ) \cdot 2 ^{2}} + z \to z_{1} lim \frac{d}{d z} [(z + 1)^{2} \cdot \frac{1}{z ( z - 2 i ) ( 2 z + 2 ) ^{2}}] + \frac{1}{2 i \cdot ( 2 + 4 i ) ^{2}}}$
$= 2 π i \cdot [- \frac{1}{8 i} + \frac{1}{8 i} (1 - \frac{1}{( 1 + 2 i ) ^{2}}) + \frac{1}{8 i ( 1 + 2 i ) ^{2}}] = 0.$
4.
记 $f (z) = z z ˉ = x^{2} + y^{2}$ ，其中 $x = Re (z) \in R, y = Im (z) \in R$ ，可见 $f (z)$ 在复平面上不解析，不能使用柯西积分定理. 当 $z$ 在圆 $∣ z ∣ = 1$ 上时， $f (z) = ∣ z ∣^{2} = 1$ ，记 $z = e^{i θ}$ ，则由定义，

$\oint_{C} f (z) d z = \int_{- i}^{i} f (z) ∣_{x = Re (z) \equiv 0} d z + \int_{∣ z ∣ = 1, π /2 < θ < 3 π /2} f (z) d z$
$= i \int_{- 1}^{1} y^{2} d y + \int_{π /2}^{3 π /2} i e^{i θ} d θ = \frac{2}{3} i - 2 i = - \frac{4}{3} i .$
5.
记 $g (x, y) = x^{2} - y^{2} + x y + 3 y$ ，因此

$\frac{\partial ^{2} g}{\partial x ^{2}} = \frac{\partial}{\partial x} (2 x + y) = 2, \frac{\partial ^{2} g}{\partial y ^{2}} = \frac{\partial}{\partial y} (x - 2 y + 3) = - 2,$

可得 $Δ g = \frac{\partial ^{2} g}{\partial x ^{2}} + \frac{\partial ^{2} g}{\partial y ^{2}} = 0$ . 该式对 $\forall (x, y) \in R^{2}$ 都成立，因此 $g (x, y)$ 为全平面的调和函数. 设 $f (z) = u (x, y) + i v (x, y)$ ，其中 $z = x + i y (x, y \in R)$ ，那么

$f^{'} (z) = \frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x} = \frac{\partial v}{\partial y} - i \frac{\partial v}{\partial y} .$

由题意，

$x^{2} - y^{2} + 3 y = \frac{\partial u}{\partial x} - \frac{\partial v}{\partial x} = \frac{\partial v}{\partial y} + \frac{\partial u}{\partial y},$

那么利用 $Δ v = 0$ 可得

$2 \frac{\partial ^{2} u}{\partial x \partial y} = (- 2 y + 3) + 2 x, (\frac{\partial ^{2}}{\partial x ^{2}} - \frac{\partial ^{2}}{\partial y ^{2}}) u = 2 x - (- 2 y + 3), (\frac{\partial ^{2}}{\partial x ^{2}} + \frac{\partial ^{2}}{\partial y ^{2}}) u = 0,$

即有

$\frac{\partial ^{2} u}{\partial x ^{2}} = x + y - \frac{3}{2}, \frac{\partial ^{2} u}{\partial x \partial y} = x - y + \frac{3}{2}, \frac{\partial ^{2} u}{\partial y ^{2}} = - x - y + \frac{3}{2},$

由题意，

$f^{'} (0) = \frac{\partial u}{\partial x}_{(0, 0)} - i \frac{\partial u}{\partial y}_{(0, 0)} = 0, f (0) = u (0, 0) + i v (0, 0) = i,$

即有

$u (0, 0) = 0, v (0, 0) = 1; \frac{\partial u}{\partial x}_{(0, 0)} = \frac{\partial u}{\partial y}_{(0, 0)} = 0,$

因此

$\frac{\partial u}{\partial x} = \frac{\partial u}{\partial x}_{(0, 0)} + \int_{0}^{x} \frac{\partial ^{2} u}{\partial x ^{2}}_{y = 0} d x + \int_{0}^{y} \frac{\partial ^{2} u}{\partial x \partial y} x d y = \frac{1}{2} (x^{2} + 2 x y - y^{2}) - \frac{3}{2} (x - y),$
$\frac{\partial u}{\partial y} = \frac{\partial u}{\partial y}_{(0, 0)} + \int_{0}^{x} \frac{\partial ^{2} u}{\partial x \partial y}_{y = 0} d x + \int_{0}^{y} \frac{\partial ^{2} u}{\partial y ^{2}} x d y = \frac{1}{2} (x^{2} - 2 x y - y^{2}) + \frac{3}{2} (x + y),$

将其代入原方程组，并重复一次操作，可得

$\frac{\partial v}{\partial x} = \frac{\partial u}{\partial x} - (x^{2} - y^{2} + 3 y) = - \frac{1}{2} (x^{2} - 2 x y - y^{2}) - \frac{3}{2} (x + y),$
$\frac{\partial v}{\partial y} = (x^{2} - y^{2} + 3 y) - \frac{\partial u}{\partial y} = \frac{1}{2} (x^{2} + 2 x y - y^{2}) - \frac{3}{2} (x - y),$
$u (x, y) = u (0, 0) + \int_{0}^{x} \frac{\partial u}{\partial x}_{y = 0} d x + \int_{0}^{y} \frac{\partial u}{\partial y} x d y$
$= \frac{1}{6} (x^{3} + 3 x^{2} y - 3 x y^{2} - y^{3}) - \frac{3}{4} (x^{2} - 2 x y - y^{2}),$
$v (x, y) = v (0, 0) + \int_{0}^{x} \frac{\partial v}{\partial x}_{y = 0} d x + \int_{0}^{y} \frac{\partial v}{\partial y} x d y$
$= - \frac{1}{6} (x^{3} - 3 x^{2} y - 3 x y^{2} + y^{3}) - \frac{3}{4} (x^{2} + 2 x y - y^{2}) + 1.$

综上分析， $f (z)$ 表达式为

$f (z) = \frac{1}{6} (1 - i) z^{3} - \frac{3}{4} (1 + i) z^{2} + i .$

简单解法 设 $f^{'} (z) = u (x, y) + i v (x, y)$ ，由题意， $u - v = x^{2} - y^{2} + 3 y$ ，因此

$\frac{\partial u}{\partial x} - \frac{\partial v}{\partial x} = 2 x, \frac{\partial u}{\partial y} - \frac{\partial v}{\partial y} = - 2 y + 3,$

由C-R方程， $\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}, \frac{\partial u}{\partial y} = - \frac{\partial v}{\partial x}$ ，代入消去 $v$ 可得

$\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} = 2 x, \frac{\partial u}{\partial y} - \frac{\partial u}{\partial x} = - 2 y + 3,$

因此

$\frac{\partial u}{\partial x} = x + y - \frac{3}{2}, \frac{\partial u}{\partial y} = x - y + \frac{3}{2},$

选取折线路径： $(0, 0) \to (x, 0) \to (x, y)$ ，可得

$u (x, y) = u (0, 0) + \int_{0}^{x} (u - \frac{3}{2}) d u + \int_{0}^{y} (x - v + \frac{3}{2}) d v$
$= \frac{1}{2} x^{2} - \frac{3}{2} x + x y - \frac{1}{2} y^{2} + \frac{3}{2} y + C_{1},$
$v (x, y) = u (x, y) - (x^{2} - y^{2} + 3 y) = - \frac{1}{2} x^{2} - \frac{3}{2} x + x y + \frac{1}{2} y^{2} - \frac{3}{2} y + C_{1},$

由于 $f^{'} (0) = 0$ ，故 $C_{1} = 0$ ，则

$f^{'} (z) = u (x, y) + i v (x, y) = \frac{1}{2} (1 - i) x^{2} + (1 + i) x y - \frac{1}{2} (1 - i) y^{2} - \frac{3}{2} (1 + i) x + \frac{3}{2} (1 - i) y$
$= \frac{1}{2} (1 - i) (x + i y)^{2} - \frac{3}{2} (1 + i) (x + i y) = \frac{1}{2} (1 - i) z^{2} - \frac{3}{2} (1 + i) z,$

代入 $f (0) = i$ ，积分得最终结果

$f (z) = \frac{1}{6} (1 - i) z^{3} - \frac{3}{4} (1 + i) z^{2} + i .$
6.
记 $z = x + i y (x, y \in R)$ ，那么 $∣ z ∣ < 1 - Re z$ 可表示为 $x^{2} + y^{2} < 1 - x$ ，即区域 $D : 2 x < 1 - y^{2}$ . 曲线 $C : 2 x = 1 - y^{2}$ 在 $R^{2}$ 平面内为一以 $x$ 轴（在复平面内为实轴）为对称轴、焦准距 $p = 1$ 、焦点位于原点的抛物线，因此 $D$ 在复平面内表示位于 $C$ 外包含焦点的区域（不包含曲线 $C$ ）.