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Simplification of $\displaystyle \tan \bigg(i\ln\bigg(\frac{a+ib}{a-ib}\bigg)\bigg)$ is

what i try

$$\ln\bigg(\frac{a+ib}{a-ib}\bigg)=\theta$$

$$\frac{a+ib}{a-ib}=e^{i\theta}$$

$$\frac{a+ib+a-ib}{(a+ib)-(a-ib)}=\frac{e^{i\theta}+1}{e^{i\theta}-1}=\frac{e^{i\frac{\theta}{2}}+e^{-i\frac{\theta}{2}}}{e^{i\frac{\theta}{2}}-e^{i\frac{\theta}{2}}}$$

$$\frac{a}{b}=\cot \frac{\theta}{2}$$

How do i solve it Help me please

jacky
  • 5,194

1 Answers1

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WLOG let $a=r\cos t, b=r\sin t,\dfrac ba=\tan t$ where $r,t$ are real

$$\dfrac{a+ib}{a-ib}=\dfrac{\cos t+i\sin t}{\cos t-i\sin t}$$

Using Intuition behind euler's formula

$$\dfrac{a+ib}{a-ib}=e^{2it}$$ $$\implies\ln\dfrac{a+ib}{a-ib}=2m\pi i+2it$$ where $m$ is any integer

$$\tan\left(i\ln\dfrac{a+ib}{a-ib}\right)=-\tan2t=\dfrac{2\tan t}{\tan^2t-1}$$

See also: Inverse trigonometric function identity doubt: $\tan^{-1}x+\tan^{-1}y =-\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$, when $x<0$, $y<0$, and $xy>1$

lab bhattacharjee
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