# Question-76830

Question Number 76830 by peter frank last updated on 30/Dec/19
Commented by mathmax by abdo last updated on 05/Jan/20
$${let}\:{remember}\:{that}\:{arctanz}\:=\frac{\mathrm{1}}{\mathrm{2}{i}}{ln}\left(\frac{\mathrm{1}+{iz}}{\mathrm{1}−{iz}}\right)\:\left({result}\:{proved}\right) \\$$$${iln}\left(\frac{{a}−{ib}}{{a}+{ib}}\right)\:=−{iln}\left(\frac{{a}+{ib}}{{a}−{ib}}\right)\:=−{iln}\left(\frac{\mathrm{1}+{i}\frac{{b}}{{a}}}{\mathrm{1}−{i}\frac{{b}}{{a}}}\right)\:=−{i}\left(\mathrm{2}{i}\right){arctan}\left(\frac{{b}}{{a}}\right) \\$$$$=\mathrm{2}\:{arctan}\left(\frac{{b}}{{a}}\right)\:\Rightarrow{tan}\left({iln}\left(\frac{{a}−{ib}}{{a}+{ib}}\right)\right)={tan}\left(\mathrm{2}{arctan}\left(\frac{{b}}{{a}}\right)\right) \\$$$$=\frac{\mathrm{2}\frac{{b}}{{a}}}{\mathrm{1}−\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }}\:=\frac{\frac{\mathrm{2}{b}}{{a}}}{\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }}\:=\frac{\mathrm{2}{b}}{{a}}×\frac{{a}^{\mathrm{2}} }{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }\:=\frac{\mathrm{2}{ab}}{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }\:. \\$$
Commented by peter frank last updated on 05/Jan/20
$${thank}\:{you} \\$$
Answered by MJS last updated on 31/Dec/19
$$\mathrm{ln}\:\left({p}+{q}\mathrm{i}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left({p}^{\mathrm{2}} +{q}^{\mathrm{2}} \right)\:+\mathrm{i}\:\mathrm{arctan}\:\frac{{q}}{{p}} \\$$$$\mathrm{i}\:\mathrm{ln}\:\left({p}+{q}\mathrm{i}\right)\:=−\mathrm{arctan}\:\frac{{q}}{{p}}\:+\mathrm{i}\frac{\mathrm{ln}\:\left({p}^{\mathrm{2}} +{q}^{\mathrm{2}} \right)}{\mathrm{2}}= \\$$$$\:\:\:\:\:\left[\frac{{a}−{b}\mathrm{i}}{{a}+{b}\mathrm{i}}=\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }−\frac{\mathrm{2}{ab}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\mathrm{i}\:\Rightarrow\:{p}=\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} };\:{q}=−\frac{\mathrm{2}{ab}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\:\Rightarrow\:{p}^{\mathrm{2}} +{q}^{\mathrm{2}} =\mathrm{1}\right] \\$$$$=\mathrm{arctan}\:\frac{\mathrm{2}{ab}}{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} } \\$$$$\Rightarrow\:\mathrm{tan}\:\left(\mathrm{i}\:\mathrm{ln}\:\frac{{a}−{b}\mathrm{i}}{{a}+{b}\mathrm{i}}\right)\:=\frac{\mathrm{2}{ab}}{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} } \\$$
Commented by peter frank last updated on 31/Dec/19
$${explain}\:{the}\:{first}\:{line}\: \\$$
Commented by mr W last updated on 31/Dec/19
$${p}+{qi}=\sqrt{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} }\left(\frac{{p}}{\:\sqrt{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} }}+\frac{{q}}{\:\sqrt{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} }}{i}\right) \\$$$${p}+{qi}=\sqrt{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} }\left(\mathrm{cos}\:\alpha+{i}\:\mathrm{sin}\:\alpha\right)\:{with}\:\alpha=\mathrm{tan}^{−\mathrm{1}} \frac{{q}}{{p}} \\$$$${p}+{qi}=\sqrt{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} }\:{e}^{{i}\alpha} \\$$$$\Rightarrow\mathrm{ln}\:\left({p}+{qi}\right)=\mathrm{ln}\:\left(\sqrt{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} }\:{e}^{{i}\alpha} \right) \\$$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left({p}^{\mathrm{2}} +{q}^{\mathrm{2}} \right)+{i}\alpha \\$$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left({p}^{\mathrm{2}} +{q}^{\mathrm{2}} \right)+{i}\:\mathrm{tan}^{−\mathrm{1}} \frac{{q}}{{p}} \\$$
Commented by peter frank last updated on 31/Dec/19
$${thank}\:{you}\:{sir}\:{for}\:{ellaboration} \\$$