3.2 正則関数

1.

微分公式と微分法を確認しよう．

(a)

$\displaystyle (z^3 - 2z^2 + 3z)'$	$\displaystyle =$	$\displaystyle (z^3)' -2(z^2)' + 3z' \ (和の導関数は導関数の和)$
	$\displaystyle =$	$\displaystyle 3z^2 - 4z + 3 ((z^{\alpha})' = \alpha z^{\alpha - 1})$

(b)

$\displaystyle ((z^2 +i)^{3})'$	$\displaystyle =$	$\displaystyle 3(z^2 + i)^{2}(z^2 + i)' \ (合成関数の微分法より)$
	$\displaystyle =$	$\displaystyle 3(z^2 + i)^{2}(2z) = 6z(z^2 + i)^{2}$

(c)

$\displaystyle (\frac{z - i}{z + i})'$	$\displaystyle =$	$\displaystyle \frac{(z-i)'(z+i) - (z-i)(z+i)'}{(z+ i)^{2}} \ (商の微分法より)$
	$\displaystyle =$	$\displaystyle \frac{z+i - (z-i)}{(z+ i)^{2}} = \frac{2i}{(z+ i)^{2}}$

2.

(a)

$\displaystyle (\tan{z})'$	$\displaystyle =$	$\displaystyle \frac{\sin{z}}{\cos{z}} = \frac{(\sin{z})'(\cos{z}) - (\sin{z})(\cos{z})'}{(\cos{z})^{2}} \ (商の微分法より)$
	$\displaystyle =$	$\displaystyle \frac{\cos^{2}{z} + \sin^{2}{z}}{\cos^{2}{z}} = \frac{1}{\cos^{2}} = \sec^{2}{z}$

(b)

$\displaystyle (\frac{1}{\cos{z}})'$	$\displaystyle =$	$\displaystyle \frac{- (\cos{z})'}{(\cos{z})^{2}} \ (商の微分法より)$
	$\displaystyle =$	$\displaystyle \frac{\sin{z}}{\cos^{2}{z}} = \sec{z}\tan{z}$

(c)

$\displaystyle (\sqrt{z^2 + 1})'$	$\displaystyle =$	$\displaystyle ((z^2 + 1)^{1/2})' = \frac{1}{2}(z^2 + 1)^{-1/2}(z^2 + 1)' \ (合成関数の微分法より)$
	$\displaystyle =$	$\displaystyle z(z^2 + 1)^{-1/2}$

(d)

$\displaystyle (\sin^{2}{z})'$	$\displaystyle =$	$\displaystyle ((\sin{z})^{2})' = 2(\sin{z})(\sin{z})' \ (合成関数の微分法より)$
	$\displaystyle =$	$\displaystyle 2\sin{z}\cos{z}$

(e)

$\displaystyle (\log(z^2+4i))'$	$\displaystyle =$	$\displaystyle \frac{1}{z^2 + 4i}(z^2 + 4i)' \ (合成関数の微分法より)$
	$\displaystyle =$	$\displaystyle \frac{2z}{z^2 + 4i}$

(f)

$\displaystyle (i^{\cos{z}})'$	$\displaystyle =$	$\displaystyle (e^{\cos{z}\log{i}})' \ (a^{z} = e^{z\log{a}}より)$
	$\displaystyle =$	$\displaystyle e^{\cos{z}\log{i}}(\cos{z}\log{i})' \ (合成関数の微分法より)$
	$\displaystyle =$	$\displaystyle e^{\cos{z}\log{i}}(-\sin{z}\log{i}) = -i^{\cos{z}}\sin{z}\log{i}$

(g)

$\displaystyle (\sin^{-1}(z-i))'$	$\displaystyle =$	$\displaystyle \frac{1}{\sqrt{1 - (z-i)^2}}(z-i)' \ (合成関数の微分法より)$
	$\displaystyle =$	$\displaystyle \frac{1}{\sqrt{1 - (z-i)^2}}$

(h)

$\displaystyle (\log(z + \sqrt{z^2 + 1}))'$	$\displaystyle =$	$\displaystyle \frac{1}{z + \sqrt{z^2 + 1}}(z+\sqrt{z^2 + 1})' \ (合成関数の微分法より)$
	$\displaystyle =$	$\displaystyle \frac{1}{z + \sqrt{z^2 + 1}}(1 + \frac{z}{\sqrt{z^2 + 1}})$
	$\displaystyle =$	$\displaystyle \frac{1}{z + \sqrt{z^2 + 1}}(\frac{\sqrt{z^2 + 1} + z}{\sqrt{z^2 + 1}}$
	$\displaystyle =$	$\displaystyle \frac{1}{\sqrt{z^2 + 1}}$

(i)

$\displaystyle (\log(\sin^{-1}{z}))'$	$\displaystyle =$	$\displaystyle \frac{1}{\sin^{-1}{z}}(\sin^{-1}{z})' \ (合成関数の微分法より)$
	$\displaystyle =$	$\displaystyle \frac{1}{\sin^{-1}{z}}(\frac{1}{\sqrt{1 - z^2}})$
	$\displaystyle =$	$\displaystyle \frac{1}{\sin^{-1}{z}\sqrt{1 - z^2}}$

(j)

$\displaystyle (z^{z})'$	$\displaystyle =$	$\displaystyle (e^{z\log{z}})' = e^{z\log{z}}(z\log{z})'$
	$\displaystyle =$	$\displaystyle e^{z\log{z}}(\log{z} + z\frac{1}{z}) = z^{z}(\log{z} + 1)$