Question Number 116725 by bemath last updated on 06/Oct/20
$$\mathrm{what}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{log}\:_{\mathrm{10}} \left(−\mathrm{1}\right)\:\mathrm{in}\: \\ $$$$\mathrm{complex}\:\mathrm{number} \\ $$
Commented by bemath last updated on 06/Oct/20
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{prof} \\ $$
Commented by MJS_new last updated on 06/Oct/20
![usually we take the principal value z=re^(iθ) with r∈R^+ and −π≤θ<π [ ] ⇒ principal value of ln z =ln r +iθ and log_z_2 z_1 =((ln z_1 )/(ln z_2 ))=((ln r_1 e^(iθ_1 ) )/(ln r_2 e^(iθ_2 ) ))=((ln r_1 +iθ_1 )/(ln r_2 +iθ_2 ))= =((θ_1 θ_2 +ln r_1 ln r_2 )/(ln^2 r_2 +θ_2 ^2 ))+i((θ_1 ln r_2 −θ_2 ln r_1 )/(ln^2 r_2 +θ_2 ^2 )) with r_1 , r_2 >0](https://www.tinkutara.com/question/Q116748.png)
$$\mathrm{usually}\:\mathrm{we}\:\mathrm{take}\:\mathrm{the}\:\mathrm{principal}\:\mathrm{value} \\ $$$${z}={r}\mathrm{e}^{\mathrm{i}\theta} \:\mathrm{with}\:{r}\in\mathbb{R}^{+} \:\mathrm{and}\:−\pi\leqslant\theta<\pi\:\left[\:\right] \\ $$$$\Rightarrow\:\mathrm{principal}\:\mathrm{value}\:\mathrm{of}\:\:\mathrm{ln}\:{z}\:=\mathrm{ln}\:{r}\:+\mathrm{i}\theta \\ $$$$\mathrm{and}\:\mathrm{log}_{{z}_{\mathrm{2}} } \:{z}_{\mathrm{1}} \:=\frac{\mathrm{ln}\:{z}_{\mathrm{1}} }{\mathrm{ln}\:{z}_{\mathrm{2}} }=\frac{\mathrm{ln}\:{r}_{\mathrm{1}} \mathrm{e}^{\mathrm{i}\theta_{\mathrm{1}} } }{\mathrm{ln}\:{r}_{\mathrm{2}} \mathrm{e}^{\mathrm{i}\theta_{\mathrm{2}} } }=\frac{\mathrm{ln}\:{r}_{\mathrm{1}} \:+\mathrm{i}\theta_{\mathrm{1}} }{\mathrm{ln}\:{r}_{\mathrm{2}} \:+\mathrm{i}\theta_{\mathrm{2}} }= \\ $$$$=\frac{\theta_{\mathrm{1}} \theta_{\mathrm{2}} +\mathrm{ln}\:{r}_{\mathrm{1}} \:\mathrm{ln}\:{r}_{\mathrm{2}} }{\mathrm{ln}^{\mathrm{2}} \:{r}_{\mathrm{2}} \:+\theta_{\mathrm{2}} ^{\mathrm{2}} }+\mathrm{i}\frac{\theta_{\mathrm{1}} \mathrm{ln}\:{r}_{\mathrm{2}} \:−\theta_{\mathrm{2}} \mathrm{ln}\:{r}_{\mathrm{1}} }{\mathrm{ln}^{\mathrm{2}} \:{r}_{\mathrm{2}} \:+\theta_{\mathrm{2}} ^{\mathrm{2}} } \\ $$$$\mathrm{with}\:{r}_{\mathrm{1}} ,\:{r}_{\mathrm{2}} \:>\mathrm{0} \\ $$
Commented by MJS_new last updated on 06/Oct/20
$$\mathrm{btw}\:\mathrm{also}\:\mathrm{ln}\:\mathrm{e}^{{r}} \:={r}\:\mathrm{is}\:“\mathrm{only}''\:\mathrm{the}\:\mathrm{principal} \\ $$$$\mathrm{value}:\:\mathrm{e}^{{r}} =\mathrm{e}^{{r}} \mathrm{e}^{\mathrm{2}{n}\pi\mathrm{i}} \:\Rightarrow\:\mathrm{ln}\:\mathrm{e}^{{r}} \:={r}+\mathrm{2}{n}\pi\mathrm{i}\:\mathrm{but} \\ $$$$\mathrm{who}\:\mathrm{would}\:\mathrm{like}\:\mathrm{to}\:\mathrm{use}\:\mathrm{this}? \\ $$
Answered by MJS_new last updated on 06/Oct/20
$$\mathrm{log}_{\mathrm{10}} \:\left(−\mathrm{1}\right)\:=\frac{\mathrm{ln}\:\left(−\mathrm{1}\right)}{\mathrm{ln}\:\mathrm{10}}=\frac{\mathrm{ln}\:\mathrm{e}^{\mathrm{i}\pi} }{\mathrm{ln}\:\mathrm{10}}=\frac{\mathrm{i}\pi}{\mathrm{ln}\:\mathrm{10}} \\ $$
Commented by bemath last updated on 06/Oct/20
$$\mathrm{sir}\:\mathrm{if}\:\mathrm{log}\:_{\mathrm{10}} \left(\mathrm{0}\right)\:=\:?\: \\ $$
Commented by bemath last updated on 06/Oct/20
$$−\mathrm{1}\:=\:\mathrm{cos}\:\pi+\:{i}.\mathrm{sin}\:\pi\:,\:\mathrm{sir} \\ $$$$ \\ $$
Answered by TANMAY PANACEA last updated on 06/Oct/20
![trying Log(a+ib) a=rcosθ b=rsinθ Log(re^(iθ) )=Log(re^(i(2kπ+θ)) )=logr+i(2kπ+θ) Log(a+ib)=log((√(a^2 +b^2 )) )+i{2kπ+tan^(−1) ((b/a))} Log(a+ib)=(1/2)log(a^2 +b^2 )+i(2kπ+tan^(−1) ((b/a))) log_(10) p×x=log_e p ((logp)/(log10))×x=((logp)/(loge))→x=log_e 10 log_(10) (−1)=log_e (−1)×log_(10) e log_(10) (−1) =log_e (−1)×log_(10) e =[(1/2)ln(1+0^2 )+i(2kπ+tan^(−1) (((−1)/0))]log_(10) e =[0+i(2kπ−(π/2))]log_(10) e](https://www.tinkutara.com/question/Q116731.png)
$${trying} \\ $$$${Log}\left({a}+{ib}\right) \\ $$$${a}={rcos}\theta\:\:\:{b}={rsin}\theta \\ $$$${Log}\left({re}^{{i}\theta} \right)={Log}\left({re}^{{i}\left(\mathrm{2}{k}\pi+\theta\right)} \right)={logr}+{i}\left(\mathrm{2}{k}\pi+\theta\right) \\ $$$${Log}\left({a}+{ib}\right)={log}\left(\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\:\right)+{i}\left\{\mathrm{2}{k}\pi+{tan}^{−\mathrm{1}} \left(\frac{{b}}{{a}}\right)\right\} \\ $$$${Log}\left({a}+{ib}\right)=\frac{\mathrm{1}}{\mathrm{2}}{log}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)+{i}\left(\mathrm{2}{k}\pi+{tan}^{−\mathrm{1}} \left(\frac{{b}}{{a}}\right)\right) \\ $$$${log}_{\mathrm{10}} {p}×{x}={log}_{{e}} {p} \\ $$$$\frac{{logp}}{{log}\mathrm{10}}×{x}=\frac{{logp}}{{loge}}\rightarrow{x}={log}_{{e}} \mathrm{10} \\ $$$$\boldsymbol{{log}}_{\mathrm{10}} \left(−\mathrm{1}\right)=\boldsymbol{{log}}_{\boldsymbol{{e}}} \left(−\mathrm{1}\right)×\boldsymbol{{log}}_{\mathrm{10}} \boldsymbol{{e}} \\ $$$$\boldsymbol{{log}}_{\mathrm{10}} \left(−\mathrm{1}\right) \\ $$$$=\boldsymbol{{log}}_{\boldsymbol{{e}}} \left(−\mathrm{1}\right)×\boldsymbol{{log}}_{\mathrm{10}} \boldsymbol{{e}} \\ $$$$=\left[\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+\mathrm{0}^{\mathrm{2}} \right)+{i}\left(\mathrm{2}{k}\pi+{tan}^{−\mathrm{1}} \left(\frac{−\mathrm{1}}{\mathrm{0}}\right)\right]{log}_{\mathrm{10}} {e}\right. \\ $$$$=\left[\mathrm{0}+{i}\left(\mathrm{2}{k}\pi−\frac{\pi}{\mathrm{2}}\right)\right]{log}_{\mathrm{10}} {e} \\ $$$$ \\ $$$$ \\ $$
Commented by bemath last updated on 06/Oct/20
$$\mathrm{how}\:\mathrm{about}\:\mathrm{log}\:_{\mathrm{10}} \left(\mathrm{0}\right)\:\mathrm{sir} \\ $$
Commented by Rio Michael last updated on 06/Oct/20
$$\mathrm{now}\:\mathrm{am}\:\mathrm{sure}\:\mathrm{log}_{\mathrm{10}} \left(\mathrm{0}\right)\:\mathrm{should}\:\mathrm{be}\:\mathrm{undefined} \\ $$$$\mathrm{since}\:\:\mathrm{ln}\:\left(\mathrm{0}\right)\:\mathrm{is}\:\mathrm{undefined}. \\ $$
Commented by Dwaipayan Shikari last updated on 06/Oct/20
$${Yes}! \\ $$
Commented by bemath last updated on 06/Oct/20
$$\mathrm{ok}.\:\mathrm{thank}\:\mathrm{you}\:\mathrm{all} \\ $$
Answered by Rio Michael last updated on 06/Oct/20
$$\mathrm{i}\:\mathrm{know}\:\mathrm{ln}\left(−\mathrm{1}\right)\:=\:\pi{i} \\ $$$$\mathrm{log}\left(−\mathrm{1}\right)\:=\:\frac{\mathrm{ln}\left(−\mathrm{1}\right)}{\mathrm{ln}\:\mathrm{10}}\:=\:\frac{\pi{i}}{\mathrm{ln}\:\mathrm{10}} \\ $$
Answered by Dwaipayan Shikari last updated on 06/Oct/20
$${log}_{\mathrm{10}} \left(\mathrm{1}−{x}\right)=\frac{{log}\left(\mathrm{1}−{x}\right)}{{log}\mathrm{10}}=−\frac{\mathrm{1}}{{log}\mathrm{10}}\left({x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}}+….\right) \\ $$$${X}=\mathrm{1} \\ $$$$−\frac{\mathrm{1}}{{log}\mathrm{10}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+….\right)\rightarrow−\infty \\ $$
Commented by Dwaipayan Shikari last updated on 06/Oct/20
$${Basically}\:\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}+….\:{is}\:{undefined} \\ $$$${But}\:{generally}\:{take}\:{it}\:{as}\:{diverges}… \\ $$