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Question

# If θ is the amplitude of $\frac{a+ib}{a-ib}$, than tan θ = (a) $\frac{2a}{{a}^{2}+{b}^{2}}$ (b) $\frac{2ab}{{a}^{2}-{b}^{2}}$ (c) $\frac{{a}^{2}-{b}^{2}}{{a}^{2}+{b}^{2}}$ (d) none of these

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Solution

## (b) $\frac{2ab}{{a}^{2}-{b}^{2}}$ $z=\frac{a+ib}{a-ib}×\frac{a+ib}{a+ib}\phantom{\rule{0ex}{0ex}}⇒z=\frac{{a}^{2}+{i}^{2}{b}^{2}+2abi}{{a}^{2}-{i}^{2}{b}^{2}}\phantom{\rule{0ex}{0ex}}⇒z=\frac{{a}^{2}-{b}^{2}+2abi}{{a}^{2}+{b}^{2}}\phantom{\rule{0ex}{0ex}}⇒z=\frac{{a}^{2}-{b}^{2}}{{a}^{2}+{b}^{2}}+i\frac{2ab}{{a}^{2}+{b}^{2}}\phantom{\rule{0ex}{0ex}}⇒\mathrm{Re}\left(z\right)=\frac{{a}^{2}-{b}^{2}}{{a}^{2}+{b}^{2}},\mathrm{Im}\left(z\right)=\frac{2ab}{{a}^{2}+{b}^{2}}\phantom{\rule{0ex}{0ex}}\mathrm{tan}\alpha =\left|\frac{\mathrm{Im}\left(z\right)}{\mathrm{Re}\left(z\right)}\right|\phantom{\rule{0ex}{0ex}}=\frac{2ab}{{a}^{2}-{b}^{2}}\phantom{\rule{0ex}{0ex}}\alpha ={\mathrm{tan}}^{-1}\left(\frac{2ab}{{a}^{2}-{b}^{2}}\right)\phantom{\rule{0ex}{0ex}}\mathrm{Since},z\mathrm{lies}\mathrm{in}\mathrm{the}\mathrm{first}\mathrm{quadrant}.\mathrm{Therefore},\phantom{\rule{0ex}{0ex}}\mathrm{arg}\left(z\right)=\alpha ={\mathrm{tan}}^{-1}\left(\frac{2ab}{{a}^{2}-{b}^{2}}\right)\phantom{\rule{0ex}{0ex}}\mathrm{tan}\theta =\frac{2ab}{{a}^{2}-{b}^{2}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

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