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Question Number 76830 by peter frank last updated on 30/Dec/19

Commented by mathmax by abdo last updated on 05/Jan/20

$$letrememberthatarctanz=\frac{1}{2i}ln\left(\frac{1+iz}{1-iz}\right)\left(resultproved\right)$$$$iln\left(\frac{a-ib}{a+ib}\right)=-iln\left(\frac{a+ib}{a-ib}\right)=-iln\left(\frac{1+i\frac{b}{a}}{1-i\frac{b}{a}}\right)=-i\left(2i\right)arctan\left(\frac{b}{a}\right)$$$$=2arctan\left(\frac{b}{a}\right)\Rightarrow tan\left(iln\left(\frac{a-ib}{a+ib}\right)\right)=tan\left(2arctan\left(\frac{b}{a}\right)\right)$$$$=\frac{2\frac{b}{a}}{1-\frac{b^2}{a^2}}=\frac{\frac{2b}{a}}{\frac{a^2-b^2}{a^2}}=\frac{2b}{a}\times \frac{a^2}{a^2-b^2}=\frac{2ab}{a^2-b^2}.$$

Commented by peter frank last updated on 05/Jan/20

$$thankyou$$

Answered by MJS last updated on 31/Dec/19

$$\ln (p+q\mathrm{i})=\frac{1}{2}\ln (p^2+q^2)+\mathrm{i}\arctan \frac{q}{p}$$$$\mathrm{i}\ln (p+q\mathrm{i})=-\arctan \frac{q}{p}+\mathrm{i}\frac{\ln (p^2+q^2)}{2}=$$$$[\frac{a-b\mathrm{i}}{a+b\mathrm{i}}=\frac{a^2-b^2}{a^2+b^2}-\frac{2ab}{a^2+b^2}\mathrm{i}\Rightarrow p=\frac{a^2-b^2}{a^2+b^2};q=-\frac{2ab}{a^2+b^2}\Rightarrow p^2+q^2=1]$$$$=\arctan \frac{2ab}{a^2-b^2}$$$$\Rightarrow \tan \left(\mathrm{i}\ln \frac{a-b\mathrm{i}}{a+b\mathrm{i}}\right)=\frac{2ab}{a^2-b^2}$$

Commented by peter frank last updated on 31/Dec/19

$$explainthefirstline$$

Commented by mr W last updated on 31/Dec/19

$$p+qi=\sqrt{p^2+q^2}(\frac{p}{\sqrt{p^2+q^2}}+\frac{q}{\sqrt{p^2+q^2}}i)$$$$p+qi=\sqrt{p^2+q^2}(\cos \alpha +i\sin \alpha )with\alpha ={\tan }^{-1}\frac{q}{p}$$$$p+qi=\sqrt{p^2+q^2}{e}^{i\alpha }$$$$\Rightarrow \ln (p+qi)=\ln \left(\sqrt{p^2+q^2}{e}^{i\alpha }\right)$$$$=\frac{1}{2}\ln (p^2+q^2)+i\alpha$$$$=\frac{1}{2}\ln (p^2+q^2)+i{\tan }^{-1}\frac{q}{p}$$

Commented by peter frank last updated on 31/Dec/19

$$thankyousirforellaboration$$

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