Prove that ,$\log \left(1+\mathrm{re}^{i\theta}\right)=\frac{1}{2} \log \left(t+2 r \cos \theta+r^{2}\right)+i \tan ^{-1} \frac{r \sin \theta}{1+r \cos \theta}$

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written 3.0 years ago by

prajapatijaimin • 3.3k

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Solution:

Deduce that, $\log (1+\cos \theta+i \sin \theta)=\log \left(2 \cos \frac{\theta}{2}\right)+i \frac{\theta}{2}\\$ . $\log \left(1+r e^{i \theta}\right)=\log [1+r(\cos \theta+i \sin \theta)]=\log [(1+r \cos \theta)+i(r \sin \theta)]$

$\begin{aligned} &=\frac{1}{2} \log \left[(1+r \cos \theta)^{2}+(r \sin \theta)^{2}\right]+i \tan ^{-1} \frac{r \sin \theta}{1+r \cos \theta} \\ &=\frac{1}{2} \log \left[1+2 r \cos \theta+r^{2} \cos ^{2} \theta+r^{2} \sin ^{2} \theta\right]+i \tan ^{-1} \frac{r \sin \theta}{1+r \cos \theta} \\ &=\frac{1}{2} \log \left[1+2 r \cos \theta+r^{2}\right]+i \tan ^{-1} \frac{r \sin \theta}{1+r \cos \theta}.....(i) \end{aligned}\$

Now, $\quad \log (1+\cos \theta+i \sin \theta)=\log \left(1+e^{i \theta}\right)$

Putting, $r=1$ in (i),

$\begin{aligned} \log (1+\cos \theta+i \sin \theta) &=\frac{1}{2} \log (1+2 \cos \theta+1)+i \tan ^{-1} \frac{\sin \theta}{1+\cos \theta} \\\\ =& \frac{1}{2} \log [2(1+\cos \theta)]+i \tan ^{-1} \frac{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \cos ^{2} \frac{\theta}{2}} \\\\ &=\frac{1}{2} \log \left[2 \cdot 2 \cos ^{2} \frac{\theta}{2}\right]+i \tan ^{-1}\left(\tan \frac{\theta}{2}\right)=\frac{1}{2} \log \left[\left(2 \cos \frac{\theta}{2}\right)^{2}\right]+i \frac{\theta}{2} \\\\ &=\frac{1}{2} \cdot 2 \log \left(2 \cos \frac{\theta}{2}\right)+i \cdot \frac{\theta}{2}=\log \left(2 \cos \frac{\theta}{2}\right)+i \cdot \frac{\theta}{2} . \end{aligned}\\$

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