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Question Number 127364 by mohammad17 last updated on 29/Dec/20

prove that: Arg(log(3−4i))=(1/i)log(((3−4i)/5)) help me sir please

$${prove}\:{that}:\:{Arg}\left({log}\left(\mathrm{3}−\mathrm{4}{i}\right)\right)=\frac{\mathrm{1}}{{i}}{log}\left(\frac{\mathrm{3}−\mathrm{4}{i}}{\mathrm{5}}\right) \\ $$$$ \\ $$$${help}\:{me}\:{sir}\:{please} \\ $$

Commented by mohammad17 last updated on 29/Dec/20

pleas help me ?

$${pleas}\:{help}\:{me}\:? \\ $$

Answered by ebi last updated on 29/Dec/20

z=log(3−4i)=log(a+bi) let a+bi=re^(iθ) (convert to polar form) where r=(√(a^2 +b^2 )) and θ=arg(a+bi) r=(√(3^2 +(−4)^2 ))=(√(25))=5 log(3−4i)=log(5e^(iθ) ) log(3−4i)=log(5)+iθ iθ=log(3−4i)−log(5) iθ=log(((3−4i)/5)) θ=(1/i)log(((3−4i)/5)) ∴ arg(3−4i)=(1/i)log(((3−4i)/5))

$$ \\ $$$${z}={log}\left(\mathrm{3}−\mathrm{4}{i}\right)={log}\left({a}+{bi}\right) \\ $$$${let}\:{a}+{bi}={re}^{{i}\theta} \:\left({convert}\:{to}\:{polar}\:{form}\right) \\ $$$${where}\:{r}=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\:{and}\:\theta={arg}\left({a}+{bi}\right) \\ $$$${r}=\sqrt{\mathrm{3}^{\mathrm{2}} +\left(−\mathrm{4}\right)^{\mathrm{2}} }=\sqrt{\mathrm{25}}=\mathrm{5} \\ $$$${log}\left(\mathrm{3}−\mathrm{4}{i}\right)={log}\left(\mathrm{5}{e}^{{i}\theta} \right) \\ $$$${log}\left(\mathrm{3}−\mathrm{4}{i}\right)={log}\left(\mathrm{5}\right)+{i}\theta \\ $$$${i}\theta={log}\left(\mathrm{3}−\mathrm{4}{i}\right)−{log}\left(\mathrm{5}\right) \\ $$$${i}\theta={log}\left(\frac{\mathrm{3}−\mathrm{4}{i}}{\mathrm{5}}\right) \\ $$$$\theta=\frac{\mathrm{1}}{{i}}{log}\left(\frac{\mathrm{3}−\mathrm{4}{i}}{\mathrm{5}}\right) \\ $$$$\therefore\:{arg}\left(\mathrm{3}−\mathrm{4}{i}\right)=\frac{\mathrm{1}}{{i}}{log}\left(\frac{\mathrm{3}−\mathrm{4}{i}}{\mathrm{5}}\right) \\ $$

Commented by mohammad17 last updated on 29/Dec/20

you are a wonderful person ,thank you sir

$${you}\:{are}\:{a}\:{wonderful}\:{person}\:,{thank}\:{you}\:{sir} \\ $$

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