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

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

2.

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

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

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

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

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

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

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

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

3.

(a) Use $a^{z} = e^{z\log{a}}$.

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

(b) Use $a^{z} = e^{z\log{a}}$.

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

(c) Use $a^{z} = e^{z\log{a}}$.

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

$$= e^{i [\log_{e}{\vert 2\vert}}e^{-2n\pi}$$

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

(d) Use $a^{z} = e^{z\log{a}}$.

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

$$= 2e^{i [\log_{e}{\vert 2\vert}}e^{-2n\pi}$$

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

4.

(a) Show $\sin^{-1}{z} = \frac{1}{i}\log{(iz \pm \sqrt{1 - z^2})}$.

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

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

Now solve this equation for $w$.

$$2iz = e^{iw} - e^{-iw} \Rightarrow$$

$$2ize^{iw} = (e^{iw})^2 - 1 \Rightarrow$$

$$= (e^{iw})^2 - 2iz e^{iw} - 1 \Rightarrow$$ 0

$$e^{iw} = iz \pm \sqrt{(iz)^2 + 1} \Rightarrow$$

$$iw = \log(iz \pm \sqrt{(iz)^2 + 1}) \Rightarrow$$

$$w = \frac{1}{i}\log(iz \pm \sqrt{(iz)^2 + 1})$$

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.

(b) Show $\tan^{-1}{z} = \frac{1}{2i}\log{\frac{1 + iz}{1 - iz}}$.

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

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

Now solve this equation for $w$. Then

$$iz(e^{iw} + e^{-iw}) = e^{iw} - e^{-iw} \Rightarrow$$

$$iz((e^{iw})^2 + 1) = (e^{iw})^2 - 1 \Rightarrow$$

$$iz + 1 = (1- iz)(e^{iw})^2\Rightarrow$$

$$(e^{iw})^2 = \frac{1 + iz}{1 - iz} \Rightarrow$$

$$e^{iw} = \sqrt{\frac{1 + iz}{1 - iz}} \Rightarrow$$

$$iw = \log(\sqrt{\frac{1 + iz}{1 - iz}}) \Rightarrow$$

$$w = \frac{1}{2i}\log(\frac{1 + iz}{1 - iz})$$

5.

(a) Use $\cos^{-1}{z} = \frac{1}{i}\log(z \pm \sqrt{z^2 - 1})$.

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

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

$$(e^{iw})^2 - 2e^{iw} + 1 = 0 \Rightarrow$$

$$e^{iw} = 1 \Rightarrow$$

$$iw = \log{(1)} = \log_{e}{1} + i(2n\pi) \Rightarrow$$

$$w = 2n\pi$$

(b) Use $\sin^{-1}{z} = \frac{1}{i}\log(iz \pm \sqrt{1 - z^2})$.

Note

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

より

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

$$\sin^{-1}{(2)} = \frac{1}{i}\log(2i \pm \sqrt{1 - 4})$$

$$= \frac{1}{i}\log(2i \pm i \sqrt{3}) = \frac{1}{i}\log(i(2 \pm \sqrt{3}))$$

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

$$= 2n\pi \pm i \log_{e}(2 + \sqrt{3})$$

(c) Use $\cos^{-1}{z} = \frac{1}{i}\log(z \pm \sqrt{z^2 - 1})$.

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

$$= \frac{1}{i}(\log{i} + \log(1 \pm \sqrt{2}))$$

Note that

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

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