Question Number 127364 by mohammad17 last updated on 29/Dec/20
$${prove}\:{that}:\:{Arg}\left({log}\left(\mathrm{3}−\mathrm{4}{i}\right)\right)=\frac{\mathrm{1}}{{i}}{log}\left(\frac{\mathrm{3}−\mathrm{4}{i}}{\mathrm{5}}\right) \\ $$$$ \\ $$$${help}\:{me}\:{sir}\:{please} \\ $$
Commented by mohammad17 last updated on 29/Dec/20
$${pleas}\:{help}\:{me}\:? \\ $$
Answered by ebi last updated on 29/Dec/20
$$ \\ $$$${z}={log}\left(\mathrm{3}−\mathrm{4}{i}\right)={log}\left({a}+{bi}\right) \\ $$$${let}\:{a}+{bi}={re}^{{i}\theta} \:\left({convert}\:{to}\:{polar}\:{form}\right) \\ $$$${where}\:{r}=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\:{and}\:\theta={arg}\left({a}+{bi}\right) \\ $$$${r}=\sqrt{\mathrm{3}^{\mathrm{2}} +\left(−\mathrm{4}\right)^{\mathrm{2}} }=\sqrt{\mathrm{25}}=\mathrm{5} \\ $$$${log}\left(\mathrm{3}−\mathrm{4}{i}\right)={log}\left(\mathrm{5}{e}^{{i}\theta} \right) \\ $$$${log}\left(\mathrm{3}−\mathrm{4}{i}\right)={log}\left(\mathrm{5}\right)+{i}\theta \\ $$$${i}\theta={log}\left(\mathrm{3}−\mathrm{4}{i}\right)−{log}\left(\mathrm{5}\right) \\ $$$${i}\theta={log}\left(\frac{\mathrm{3}−\mathrm{4}{i}}{\mathrm{5}}\right) \\ $$$$\theta=\frac{\mathrm{1}}{{i}}{log}\left(\frac{\mathrm{3}−\mathrm{4}{i}}{\mathrm{5}}\right) \\ $$$$\therefore\:{arg}\left(\mathrm{3}−\mathrm{4}{i}\right)=\frac{\mathrm{1}}{{i}}{log}\left(\frac{\mathrm{3}−\mathrm{4}{i}}{\mathrm{5}}\right) \\ $$
Commented by mohammad17 last updated on 29/Dec/20
$${you}\:{are}\:{a}\:{wonderful}\:{person}\:,{thank}\:{you}\:{sir} \\ $$