Question Number 17959 by tawa tawa last updated on 13/Jul/17
Answered by alex041103 last updated on 13/Jul/17
$${Because}\:{j}=\sqrt{−\mathrm{1}}={i}\:{we}\:{can}\:{use} \\ $$$${Euler}'{s}\:{formula} \\ $$$${e}^{{ix}} ={cos}\:{x}\:+\:{isin}\:{x} \\ $$$${z}={e}^{\mathrm{1}+{i}\frac{\pi}{\mathrm{2}}} ={e}×{e}^{{i}\frac{\pi}{\mathrm{2}}} ={e}\left({cos}\left(\pi/\mathrm{2}\right)+{isin}\left(\pi/\mathrm{2}\right)\right) \\ $$$$={ie} \\ $$$${In}\:{polar}\:{form}\left({r}\:{is}\:{the}\:{dist}.\:{to}\:{de}\:{origin}\:\right. \\ $$$$\left.{and}\:{A}\:{is}\:{the}\:{argument}\:{measured}\:{in}\:{radians}\right) \\ $$$${z}={rcos}\left({A}\right)\:+\:{irsin}\left({A}\right)= \\ $$$$={r}\left({cos}\left({A}\right)+{isin}\left({A}\right)\right)= \\ $$$$={re}^{{iA}} ={e}^{{ln}\left({r}\right)+{iA}} \\ $$$$\Rightarrow{ln}\left({z}\right)={ln}\left({r}\right)+{iA} \\ $$$${Now}\:{we}\:{solve}\:{the}\:{rest} \\ $$$$\left({i}\right)\:{z}_{\mathrm{1}} ={e}^{{ln}\left(\mathrm{10}\right)+{i}\left(\frac{\pi}{\mathrm{180}°}\mathrm{37}.\mathrm{25}°\right)} ={e}^{{ln}\left(\mathrm{10}\right)+{i}\frac{\mathrm{5}\pi}{\mathrm{24}}} \\ $$$${ln}\left({z}_{\mathrm{1}} \right)={ln}\left(\mathrm{10}\right)+{i}\frac{\mathrm{5}\pi}{\mathrm{24}} \\ $$$$\left({ii}\right){z}_{\mathrm{2}} ={e}^{{ln}\left(\mathrm{10}\right)+{i}\left(\frac{\pi}{\mathrm{180}°}\mathrm{322}.\mathrm{75}°\right)} ={e}^{{ln}\left(\mathrm{10}\right)+{i}\frac{\mathrm{1291}\pi}{\mathrm{720}}} \\ $$$${ln}\left({z}_{\mathrm{2}} \right)={ln}\left(\mathrm{10}\right)+{i}\frac{\mathrm{1291}\pi}{\mathrm{720}} \\ $$
Commented by tawa tawa last updated on 13/Jul/17
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$