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Question Number 116725 by bemath last updated on 06/Oct/20

$$\mathrm{what}\mathrm{the}\mathrm{value}\mathrm{of}\log {}_{10}(-1)\mathrm{in}$$$$\mathrm{complex}\mathrm{number}$$

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

$$\mathrm{usually}\mathrm{we}\mathrm{take}\mathrm{the}\mathrm{principal}\mathrm{value}$$$$z=r{\mathrm{e}}^{\mathrm{i}\theta }\mathrm{with}r\in {\mathbb{R}}^{+}\mathrm{and}-\pi ⩽\theta <\pi \left[\right]$$$$\Rightarrow \mathrm{principal}\mathrm{value}\mathrm{of}\ln z=\ln r+\mathrm{i}\theta$$$$\mathrm{and}{\log }_{z_2}{z}_1=\frac{\ln z_1}{\ln z_2}=\frac{\ln r_1{\mathrm{e}}^{\mathrm{i}{\theta }_1}}{\ln r_2{\mathrm{e}}^{\mathrm{i}{\theta }_2}}=\frac{\ln r_1+\mathrm{i}{\theta }_1}{\ln r_2+\mathrm{i}{\theta }_2}=$$$$=\frac{{\theta }_1{\theta }_2+\ln r_1\ln r_2}{{\ln }^2{r}_2+{\theta }_2^2}+\mathrm{i}\frac{{\theta }_1\ln r_2-{\theta }_2\ln r_1}{{\ln }^2{r}_2+{\theta }_2^2}$$$$\mathrm{with}{r}_1,r_2>0$$

Commented by MJS_new last updated on 06/Oct/20

$$\mathrm{btw}\mathrm{also}\ln {\mathrm{e}}^r=r\mathrm{is}‘‘{\mathrm{only}}^{″}\mathrm{the}\mathrm{principal}$$$$\mathrm{value}:{\mathrm{e}}^r={\mathrm{e}}^r{\mathrm{e}}^{2n\pi \mathrm{i}}\Rightarrow \ln {\mathrm{e}}^r=r+2n\pi \mathrm{i}\mathrm{but}$$$$\mathrm{who}\mathrm{would}\mathrm{like}\mathrm{to}\mathrm{use}\mathrm{this}?$$

Answered by MJS_new last updated on 06/Oct/20

$${\log }_{10}(-1)=\frac{\ln (-1)}{\ln 10}=\frac{\ln {\mathrm{e}}^{\mathrm{i}\pi }}{\ln 10}=\frac{\mathrm{i}\pi }{\ln 10}$$

Commented by bemath last updated on 06/Oct/20

$$\mathrm{sir}\mathrm{if}\log {}_{10}\left(0\right)=?$$

Commented by bemath last updated on 06/Oct/20

$$-1=\cos \pi +i.\sin \pi ,\mathrm{sir}$$$$$$

Answered by TANMAY PANACEA last updated on 06/Oct/20

$$trying$$$$Log(a+ib)$$$$a=rcos\theta b=rsin\theta$$$$Log\left({re}^{i\theta }\right)=Log\left({re}^{i(2k\pi +\theta )}\right)=logr+i(2k\pi +\theta )$$$$Log(a+ib)=log\left(\sqrt{a^2+b^2}\right)+i\{2k\pi +{tan}^{-1}\left(\frac{b}{a}\right)\}$$$$Log(a+ib)=\frac{1}{2}log(a^2+b^2)+i(2k\pi +{tan}^{-1}\left(\frac{b}{a}\right))$$$${log}_{10}p\times x={log}_{e}p$$$$\frac{logp}{log10}\times x=\frac{logp}{loge}\to x={log}_{e}10$$$${\mathit{l}\mathit{o}\mathit{g}}_{10}(-1)={\mathit{l}\mathit{o}\mathit{g}}_{\mathit{e}}(-1)\times {\mathit{l}\mathit{o}\mathit{g}}_{10}\mathit{e}$$$${\mathit{l}\mathit{o}\mathit{g}}_{10}(-1)$$$$={\mathit{l}\mathit{o}\mathit{g}}_{\mathit{e}}(-1)\times {\mathit{l}\mathit{o}\mathit{g}}_{10}\mathit{e}$$$$=[\frac{1}{2}ln(1+0^2)+i(2k\pi +{tan}^{-1}\left(\frac{-1}{0}\right)]{log}_{10}e$$$$=[0+i(2k\pi -\frac{\pi }{2})]{log}_{10}e$$$$$$$$$$

Commented by bemath last updated on 06/Oct/20

$$\mathrm{how}\mathrm{about}\log {}_{10}\left(0\right)\mathrm{sir}$$

Commented by Rio Michael last updated on 06/Oct/20

$$\mathrm{now}\mathrm{am}\mathrm{sure}{\log }_{10}\left(0\right)\mathrm{should}\mathrm{be}\mathrm{undefined}$$$$\mathrm{since}\ln \left(0\right)\mathrm{is}\mathrm{undefined}.$$

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

$$\mathrm{i}\mathrm{know}\ln (-1)=\pi i$$$$\log (-1)=\frac{\ln (-1)}{\ln 10}=\frac{\pi i}{\ln 10}$$

Answered by Dwaipayan Shikari last updated on 06/Oct/20

$${log}_{10}(1-x)=\frac{log(1-x)}{log10}=-\frac{1}{log10}(x+\frac{x^2}{2}+\frac{x^3}{3}+....)$$$$X=1$$$$-\frac{1}{log10}(1+\frac{1}{2}+\frac{1}{3}+....)\to -\mathrm{\infty }$$

Commented by Dwaipayan Shikari last updated on 06/Oct/20

$$Basically1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....isundefined$$$$Butgenerallytakeitasdiverges...$$

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