指数関数

定義 2..1   $w = e^{z} = e^{x}(\cos{y} + i\sin{y})$を指数関数といい,$w = e^{z}$あるいは $w = \exp{z}$と表わす.

定理 2..1 (1)   $e^{z_{1} + z_{2}} = e^{z_{1}}e^{z_{2}}, \ (e^{z})^{n} = e^{nz} \ (nは整数)$
(2) $e^{z}$は周期$2\pi i$を持つ.

証明 (1) $z_{1} = x_{1} + i y_{1}, z_{2} = x_{2} + i y_{2}$とすると,

$\displaystyle e^{z_{1} + z_{2}}$ $\displaystyle =$ $\displaystyle e^{x_{1} + x_{2} + i(y_{1} + y_{2})}$  
  $\displaystyle =$ $\displaystyle e^{x_{1} + x_{2}}(\cos{(y_{1} + y_{2})} + i\sin(y_{1} + y_{2}))$  
  $\displaystyle =$ $\displaystyle e^{x_{1} + x_{2}}[\cos{y_{1}}\cos{y_{2}} - \sin{y_{1}}\sin{y_{2}} + i(\sin{y_{1}}\cos{y_{2}} + \cos{y_{1}}\sin{y_{2}})]$  
  $\displaystyle =$ $\displaystyle e^{x_{1} + x_{2}}[(\cos{y_{1}} + i\sin{y_{1}})(\cos{y_{2}}+i\sin{y_{2}})]$  
  $\displaystyle =$ $\displaystyle e^{x_{1}}(\cos{y_{1}} + i\sin{y_{1}})e^{x_{2}}(\cos{y_{2}}+i\sin{y_{2}})$  
  $\displaystyle =$ $\displaystyle e^{z_{1}}e^{z_{2}}$  

(2) $e^{z + c} = e^{z}$が成り立つとき,関数$e^{z}$は周期$c$を持つという. そこで, $e^{z + c} = e^{z}$とおくと $e^{z}e^{c} = e^{z}$より$e^{c} = 1$ $c = a + bi$とおくと $e^{c} = e^{a + bi} = e^{a}(\cos{b} + i\sin{b}) = 1$より $e^{a}\cos{b} = 1, e^{a}\sin{b} = 0$.これより $a = 0, \cos{b} = 1, \sin{b} = 0$.これを満たす$b$ $b = 2n\pi ( n = 0, \pm1,\pm2,\ldots)$.したがって,$e^{z}$は周期$2\pi i$を持つ.

練習問題2.3
1. 次の式を満たす複素数を求めよ.
(a)
$e^{z} = 1$
(b)
$e^{z} = i$
(c)
$e^{z} = -2$

2. 次の値を$u+ iv$の形で表せ.

(a)
$\sin{2i}$
(b)
$\sin{\left(\frac{\pi}{2} + i\right)}$
(c)
$\cos{\left(\frac{\pi}{3} - i\right)}$
(d)
$\tan{\left(\frac{\pi}{6} + 2i\right)}$
(e)
$\sin(iy)$
(f)
$\cos(iy)$

3. 次の公式を証明せよ.

(a)
$1 + \tan^{2}{z} = \frac{1}{\cos^{2}{z}}$
(b)
$\sin{(-z)} = -\sin{z}$
(c)
$\cos(-z) = \cos{z}$

4. $\sin{z}$について次のことが成り立つことを示せ.

(a)
$\sin{z}$は周期$2\pi$を持つ

(5) $\tan{z}$の周期を求めよ.