$\mathrm{Log} \; z$

 

Prop 1

For $t \in \mathbb{R} \setminus \left\{0 \right\}$,

$\arctan t + \arctan \dfrac{1}{t} = \begin{cases}
\dfrac{\pi}{2} & t > 0 \\
-\dfrac{\pi}{2} & t < 0
\end{cases} $

 

  Case 1 : $t > 0$

  So there exists $0 < \theta < \dfrac{\pi}{2}$  s.t.

  $\tan \theta = t$

  $\therefore$  $\arctan t = \theta$

 

  Also

  $\dfrac{1}{t} = \cot \theta = \tan (\dfrac{\pi}{2} - \theta)$

  Since $0 < \dfrac{\pi}{2} - \theta < \dfrac{\pi}{2}$,

  $\arctan \dfrac{1}{t} = \dfrac{\pi}{2} - \theta$

 

  $\therefore$  $\arctan t + \arctan \dfrac{1}{t} = \dfrac{\pi}{2} $

 

  end

 

 

  Case 2 : $t < 0$

  So there exists $-\dfrac{\pi}{2} < \theta < 0$  s.t.

  $\tan \theta = t$

  $\therefore$  $\arctan t = \theta$

 

  Also

  $\dfrac{1}{t} = \cot \theta = \tan (-\dfrac{\pi}{2} - \theta)$

  Since $-\dfrac{\pi}{2} < -\dfrac{\pi}{2} - \theta < 0$,

  $\arctan \dfrac{1}{t} = -\dfrac{\pi}{2} - \theta$

 

  $\therefore$  $\arctan t + \arctan \dfrac{1}{t} = -\dfrac{\pi}{2} $

 

  end

 

 

 

Thm 1

Let $\Omega = \mathbb{C} \setminus \left\{z \; | \; \mathrm{Im} \; z = 0, \; \mathrm{Re} \; z \leq 0 \right\}$ . Then

$\mathrm{Log} \; z = \begin{cases}
\ln \left|z \right| + i \arctan \dfrac{y}{x} & \mathrm{Re} \; z > 0\\
\ln \left|z \right| + i (-\arctan \dfrac{x}{y} + \dfrac{\pi}{2}) & \mathrm{Im} \; z > 0 \\
\ln \left|z \right| + i (-\arctan \dfrac{x}{y} - \dfrac{\pi}{2}) & \mathrm{Im} \; z < 0
\end{cases}$     ( $z = x+ iy$ ) 

on $\Omega$

 

  Case 1 : $\mathrm{Re} \; z > 0$

  So $x >0$  and  $- \dfrac{\pi}{2} < \mathrm{Arg} \; z < \dfrac{\pi}{2}$

 

  By the polar representation of $z$,

  $\tan (\mathrm{Arg} \; z) = \dfrac{y}{x}$

  Since $- \dfrac{\pi}{2} < \mathrm{Arg} \; z < \dfrac{\pi}{2}$,

  $\mathrm{Arg} \; z = \arctan \dfrac{y}{x}$

 

  $\therefore$  $\mathrm{Log} \; z = \ln \left|z \right| + i \arctan \dfrac{y}{x}$

 

  end

 

 

  Case 2 : $\mathrm{Im} \; z > 0$

  So $y > 0$  and  $0 < \mathrm{Arg} \; z < \pi$

 

  By the polar representation of $z$,

  $\cot (\mathrm{Arg} \; z) = \dfrac{x}{y}$

  $\therefore$  $\tan (\dfrac{\pi}{2} - \mathrm{Arg} \; z) = \dfrac{x}{y}$

 

  Since $-\dfrac{\pi}{2} < \dfrac{\pi}{2} - \mathrm{Arg} \; z < \dfrac{\pi}{2}$,

  $\dfrac{\pi}{2} - \mathrm{Arg} \; z = \arctan \dfrac{y}{x}$

 

  $\therefore$  $\mathrm{Log} \; z = \ln \left|z \right| + i (- \arctan \dfrac{x}{y} + \dfrac{\pi}{2})$

 

  end

 

 

  Case 3 : $\mathrm{Im} \; z < 0$

  So $y < 0$  and  $-\pi < \mathrm{Arg} \; z < 0$

 

  By the polar representation of $z$,

  $\cot (\mathrm{Arg} \; z) = \dfrac{x}{y}$

  $\therefore$  $\tan (-\dfrac{\pi}{2} - \mathrm{Arg} \; z) = \dfrac{x}{y}$

 

  Since $-\dfrac{\pi}{2} < -\dfrac{\pi}{2} - \mathrm{Arg} \; z < \dfrac{\pi}{2}$,

  $-\dfrac{\pi}{2} - \mathrm{Arg} \; z = \arctan \dfrac{y}{x}$

 

  $\therefore$  $\mathrm{Log} \; z = \ln \left|z \right| + i (- \arctan \dfrac{x}{y} - \dfrac{\pi}{2})$

 

  end

 

 

 

Thm 2

Let $\Omega = \mathbb{C} \setminus \left\{z \; | \; \mathrm{Im} \; z = 0, \; \mathrm{Re} \; z \leq 0 \right\}$ . Then

$\mathrm{Log} \; z$ is differentiable on $\Omega$

 

  Let $\mathrm{Log} \; z = u + iv$

 

  Since $\dfrac{\partial }{\partial x}(\ln \sqrt{x^{2} + y^{2}} ) = \dfrac{x}{x^{2} + y^{2}}$,

  $\dfrac{\partial u}{\partial x} = \dfrac{x}{x^{2} + y^{2}}$

 

  Since $\dfrac{\partial }{\partial x}(\arctan \dfrac{y}{x}) = -\dfrac{y}{x^{2} + y^{2}}$  and  $\dfrac{\partial }{\partial x}(- \arctan \dfrac{x}{y}) = -\dfrac{y}{x^{2} + y^{2}}$,

  $\dfrac{\partial v}{\partial x} = -\dfrac{y}{x^{2} + y^{2}}$

 

  $\therefore$  $\dfrac{\partial u}{\partial x}$,  $\dfrac{\partial v}{\partial x}$ are continuous on $\Omega$

 

  Since $\dfrac{\partial }{\partial y}(\ln \sqrt{x^{2} + y^{2}} ) = \dfrac{y}{x^{2} + y^{2}}$,

  $\dfrac{\partial u}{\partial y} = \dfrac{y}{x^{2} + y^{2}}$

 

  Since $\dfrac{\partial }{\partial y}(\arctan \dfrac{y}{x}) = \dfrac{x}{x^{2} + y^{2}}$  and  $\dfrac{\partial }{\partial y}(- \arctan \dfrac{x}{y}) = \dfrac{x}{x^{2} + y^{2}}$,

  $\dfrac{\partial v}{\partial y} = \dfrac{x}{x^{2} + y^{2}}$

 

  $\therefore$  $\dfrac{\partial u}{\partial y}$,  $\dfrac{\partial v}{\partial y}$ are continuous on $\Omega$

  $\therefore$  C-R equation of $\mathrm{Log} \; z$ is satisfied on $\Omega$

 

  By Cor 1,

  $\mathrm{Log} \; z$ is differentiable on $\Omega$

 

 

 

Cor 1

Let $\Omega = \mathbb{C} \setminus \left\{z \; | \; \mathrm{Im} \; z = 0, \; \mathrm{Re} \; z \leq 0 \right\}$ . Then

$(\mathrm{Log} \; z)' = \dfrac{1}{z}$     ( $z \in \Omega$ )

 

 

 

Cor 2

1.  $(e^{z})' = e^{z}$

2.  $(\sin z)' = \cos z$

3.  $(\cos z)' = - \sin z$

 

#  Def 2  and  Thm 2

 

 

 

Prop 2

$\mathrm{Log}(1 + z) = \displaystyle \sum_{n=1}^{\infty} (-1)^{n-1} \dfrac{1}{n} z^{n}$     ( $\left|z \right| < 1$ )

 

  Let $\Omega' = \mathbb{C} \setminus \left\{z \; | \; \mathrm{Im} \; z = 0, \; \mathrm{Re} \; z \leq -1 \right\}$

  By Cor 1,

  $(\mathrm{Log}(1+z))' = \dfrac{1}{1+z}$     ( $z \in \Omega'$ )

 

  By geometric series,

  $\dfrac{1}{1+z} = \displaystyle \sum_{n=0}^{\infty} (-1)^{n}z^{n}$       ( $\left|z \right| < 1$ )

 

  By Thm 2,

  $(\displaystyle \sum_{n=0}^{\infty} (-1)^{n} \dfrac{1}{n+1} z^{n+1})' = \displaystyle \sum_{n=0}^{\infty} (-1)^{n}z^{n}$

 

  $\therefore$  $(\mathrm{Log}(1+z))' = (\displaystyle \sum_{n=1}^{\infty} (-1)^{n-1} \dfrac{1}{n} z^{n})'$

 

  By Thm 1,

  $\mathrm{Log}(1+z) = \displaystyle \sum_{n=1}^{\infty} (-1)^{n} \dfrac{1}{n} z^{n} + \alpha$     ( $\alpha \in \mathbb{C}$ )

 

  Since $\mathrm{Log}(1) = 0$,

  $\alpha = 0$

 

  $\therefore$  $\mathrm{Log}(1 + z) = \displaystyle \sum_{n=1}^{\infty} (-1)^{n-1} \dfrac{1}{n} z^{n}$     ( $\left|z \right| < 1$ )