2.4 Inverse function of an elementary function

1. $w = \sqrt{z} = z^{1/2}$ means that $w = z^{1/2} = \sqrt{r}e^{i(\theta + 2n\pi)/2}$ . Thus, it is two-valued function. Now let $z = re^{i\theta}$ . Then

$\displaystyle w_{1} = \sqrt{r}e^{i\theta/2}, w_{2} = \sqrt{r}e^{(\theta + 2\pi)i/2} = - w_{1}$

(a) Use $\log{z} = \log_{e}{r} + i (\theta + 2n\pi) (n = 0, \pm1, \pm 2, \ldots)$ .

$\displaystyle \log{2} = \log_{e}{2} + i (0 + 2n\pi) = \log_{e}{2} + 2n\pi$

(b) Use $\log{z} = \log_{e}{r} + i (\theta + 2n\pi) (n = 0, \pm1, \pm 2, \ldots)$ .

$\displaystyle \log{(-1)} = \log_{e}{\vert-1\vert} + i (\pi + 2n\pi) = i(\pi + 2n\pi) = (2n+1)\pi i$

$\displaystyle \log{i} = \log_{e}{\vert i\vert} + i (\frac{\pi}{2} + 2n\pi) = i(\frac{\pi}{2} + 2n\pi) = (2n+\frac{1}{2})\pi i$

(d) Use $\log{z} = \log_{e}{r} + i (\theta + 2n\pi) (n = 0, \pm1, \pm 2, \ldots)$ .

$\displaystyle \log{(1+i)} = \log_{e}{\vert 1+i\vert} + i (\frac{\pi}{4} + 2n\pi) = \log_{e}{\sqrt{2}} + (2n+\frac{1}{4})\pi i$

$\displaystyle (-1)^{i} = e^{i\log{(-1)}} = e^{i [\log_{e}{\vert-1\vert} + i(\pi + 2n\pi)]} = e^{-(2n+1)\pi}$

$\displaystyle (i)^{i} = e^{i\log{(i)}} = e^{i [\log_{e}{\vert i\vert} + i(\frac{\pi}{2} + 2n\pi)]} = e^{-(2n+\frac{1}{2})\pi}$

$\displaystyle (2)^{i}$	$\displaystyle =$	$\displaystyle e^{i\log{(2)}} = e^{i [\log_{e}{\vert 2\vert} + i(0 + 2n\pi)]}$
	$\displaystyle =$	$\displaystyle e^{i [\log_{e}{\vert 2\vert}}e^{-2n\pi}$
	$\displaystyle =$	$\displaystyle e^{-2n\pi}[\cos{(\log_{e}{2})}+i \sin{(\log_{e}{2})}]$

$\displaystyle (2)^{1+i}$	$\displaystyle =$	$\displaystyle 2\cdot2^{i} = 2e^{i\log{(2)}} = 2e^{i [\log_{e}{\vert 2\vert} + i(0 + 2n\pi)]}$
	$\displaystyle =$	$\displaystyle 2e^{i [\log_{e}{\vert 2\vert}}e^{-2n\pi}$
	$\displaystyle =$	$\displaystyle 2e^{-2n\pi}[\cos{(\log_{e}{2})}+i \sin{(\log_{e}{2})}]$

Let $\sin^{-1}{z} = w$ . Then $z = \sin{w}$ . Note that the complex function $\sin{w}$ is expressed by the exponential function.

$\displaystyle z = \sin{w} = \frac{e^{iw} - e^{-iw}}{2i}$

$\displaystyle 2iz$	$\displaystyle =$	$\displaystyle e^{iw} - e^{-iw} \Rightarrow$
$\displaystyle 2ize^{iw}$	$\displaystyle =$	$\displaystyle (e^{iw})^2 - 1 \Rightarrow$
0	$\displaystyle =$	$\displaystyle (e^{iw})^2 - 2iz e^{iw} - 1 \Rightarrow$
$\displaystyle e^{iw}$	$\displaystyle =$	$\displaystyle iz \pm \sqrt{(iz)^2 + 1} \Rightarrow$
$\displaystyle iw$	$\displaystyle =$	$\displaystyle \log(iz \pm \sqrt{(iz)^2 + 1}) \Rightarrow$
$\displaystyle w$	$\displaystyle =$	$\displaystyle \frac{1}{i}\log(iz \pm \sqrt{(iz)^2 + 1})$

Note Assuming that $\sqrt{(iz)^2 + 1}$ represents two branches with $\pm$ in front of the radical symbol at the same time by divalentity, only $+$

is required.

Let $\tan^{-1}{z} = w$ . Then $z = \tan{w}$ . Note that the complex function $\tan{w}$ can be represented by exponential function.

$\displaystyle z = \tan{w} = \frac{\sin{w}}{\cos{w}} = \frac{1}{i}\frac{e^{iw} - e^{-iw}}{e^{iw} + e^{-iw}}$

$\displaystyle iz(e^{iw} + e^{-iw})$	$\displaystyle =$	$\displaystyle e^{iw} - e^{-iw} \Rightarrow$
$\displaystyle iz((e^{iw})^2 + 1)$	$\displaystyle =$	$\displaystyle (e^{iw})^2 - 1 \Rightarrow$
$\displaystyle iz + 1$	$\displaystyle =$	$\displaystyle (1- iz)(e^{iw})^2\Rightarrow$
$\displaystyle (e^{iw})^2$	$\displaystyle =$	$\displaystyle \frac{1 + iz}{1 - iz} \Rightarrow$
$\displaystyle e^{iw}$	$\displaystyle =$	$\displaystyle \sqrt{\frac{1 + iz}{1 - iz}} \Rightarrow$
$\displaystyle iw$	$\displaystyle =$	$\displaystyle \log(\sqrt{\frac{1 + iz}{1 - iz}}) \Rightarrow$
$\displaystyle w$	$\displaystyle =$	$\displaystyle \frac{1}{2i}\log(\frac{1 + iz}{1 - iz})$

$\displaystyle \cos^{-1}{(1)} = \frac{1}{i}\log(1 \pm \sqrt{1^2 - 1}) = \frac{1}{i}(\log_{e}{(1)} + i(2n\pi)) = 2n\pi$

Alternate solution Let $w = \cos^{-1}{(1)}$ . Then $1 = \cos{w} = \frac{e^{iw} + e^{-iw}}{2}$ . Solve this for $w$

. Then

$\displaystyle (e^{iw})^2 - 2e^{iw} + 1$	$\displaystyle =$	$\displaystyle 0 \Rightarrow$
$\displaystyle e^{iw}$	$\displaystyle =$	$\displaystyle 1 \Rightarrow$
$\displaystyle iw$	$\displaystyle =$	$\displaystyle \log{(1)} = \log_{e}{1} + i(2n\pi) \Rightarrow$
$\displaystyle w$	$\displaystyle =$	$\displaystyle 2n\pi$

$\displaystyle \log_{e}(2 - \sqrt{3}) = \log_{e}(\frac{(2 -\sqrt{3})(2 + \sqrt{3})}{2 + \sqrt{3}} = \log(\frac{1}{2 + \sqrt{3}}) = -\log{(2 + \sqrt{3})}$

$\displaystyle \log_{e}{(2 \pm \sqrt{3})} = \pm \log_{e}{(2 + \sqrt{3})}$

$\displaystyle \sin^{-1}{(2)}$	$\displaystyle =$	$\displaystyle \frac{1}{i}\log(2i \pm \sqrt{1 - 4})$
	$\displaystyle =$	$\displaystyle \frac{1}{i}\log(2i \pm i \sqrt{3}) = \frac{1}{i}\log(i(2 \pm \sqrt{3}))$
	$\displaystyle =$	$\displaystyle \frac{1}{i}(\log{(i)} + \log{(2 \pm \sqrt{3})}) = \frac{1}{i}[i(2n + \frac{1}{2})\pi + \log_{e}{(2 \pm \sqrt{3})} + i(2n\pi)]$
	$\displaystyle =$	$\displaystyle 2n\pi \pm i \log_{e}(2 + \sqrt{3})$

$\displaystyle \cos^{-1}{(i)}$	$\displaystyle =$	$\displaystyle \frac{1}{i}\log(i \pm \sqrt{i^2 - 1}) = \frac{1}{i}\log(i(1 \pm \sqrt{2}))$
	$\displaystyle =$	$\displaystyle \frac{1}{i}(\log{i} + \log(1 \pm \sqrt{2}))$

$\displaystyle \log(1 \pm \sqrt{2}) = \left\{\begin{array}{l} \log_{e}(1 + \sqrt{2}) + i(2n\pi) \\ \log_{e}(\sqrt{2} - 1) + i(\pi + 2n\pi) \end{array}\right.$

$\displaystyle \cos^{-1}{(i)} = \left\{\begin{array}{l} -i[(2n + \frac{1}{2})\pi... ...}) ] \\ -i[(2n + \frac{3}{2})\pi + \log_{e}(\sqrt{2} - 1) ] \end{array}\right.$