Question Number 201509 by Calculusboy last updated on 07/Dec/23
Answered by witcher3 last updated on 09/Dec/23
![tack principal definition of Log(z)=ln∣z∣+iarg(z) z∈]−(π/2),((3π)/2)[ ln(ix+ln(cos(x))=ln∣(√(x^2 +ln^2 (cos(x)))∣+i arg(ln(cos(x)+ix) arg(ln(cos(x)+ix)∈[(π/2),π[ arg(ln(cos(x)+ix)=tan^(−1) ((x/(ln(cos(x))))+π cos(ln(cos(x))+ix)=(1/2)ln(x^2 +ln^2 (cos(x))+i(tan^(−1) ((x/(ln(cos(x))))+π) ⇒2ln(cos(x)+ix)=ln(x^2 +ln^2 (cos(x))+itan^(−1) ((x/(lncos(x))))+2π principal definition ⇒2ln(cos(x)+ix)=ln(x^2 +ln^2 (cos(x))+2itan^(−1) ((x/(ln(cos(x)))) ∣2ln(cos(x)+ix)∣^2 =ln^2 (x^2 +ln^2 (cos(x)))+4arctan^2 ((x/(ln(cos(x)))) easy from here](https://www.tinkutara.com/question/Q201596.png)
$$\mathrm{tack}\:\mathrm{principal}\:\:\mathrm{definition}\:\mathrm{of}\:\mathrm{Log}\left(\mathrm{z}\right)=\mathrm{ln}\mid\mathrm{z}\mid+\mathrm{iarg}\left(\mathrm{z}\right) \\ $$$$\left.\mathrm{z}\in\right]−\frac{\pi}{\mathrm{2}},\frac{\mathrm{3}\pi}{\mathrm{2}}\left[\:\right. \\ $$$$\mathrm{ln}\left(\mathrm{ix}+\mathrm{ln}\left(\mathrm{cos}\left(\mathrm{x}\right)\right)=\mathrm{ln}\mid\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{ln}^{\mathrm{2}} \left(\mathrm{cos}\left(\mathrm{x}\right)\right.}\mid+\mathrm{i}\:\mathrm{arg}\left(\mathrm{ln}\left(\mathrm{cos}\left(\mathrm{x}\right)+\mathrm{ix}\right)\right.\right. \\ $$$$\mathrm{arg}\left(\mathrm{ln}\left(\mathrm{cos}\left(\mathrm{x}\right)+\mathrm{ix}\right)\in\left[\frac{\pi}{\mathrm{2}},\pi\left[\:\right.\right.\right. \\ $$$$\mathrm{arg}\left(\mathrm{ln}\left(\mathrm{cos}\left(\mathrm{x}\right)+\mathrm{ix}\right)=\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{x}}{\mathrm{ln}\left(\mathrm{cos}\left(\mathrm{x}\right)\right.}\right)+\pi\right. \\ $$$$\mathrm{cos}\left(\mathrm{ln}\left(\mathrm{cos}\left(\mathrm{x}\right)\right)+\mathrm{ix}\right)=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{ln}^{\mathrm{2}} \left(\mathrm{cos}\left(\mathrm{x}\right)\right)+\mathrm{i}\left(\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{x}}{\mathrm{ln}\left(\mathrm{cos}\left(\mathrm{x}\right)\right.}\right)+\pi\right)\right. \\ $$$$\Rightarrow\mathrm{2ln}\left(\mathrm{cos}\left(\mathrm{x}\right)+\mathrm{ix}\right)=\mathrm{ln}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{ln}^{\mathrm{2}} \left(\mathrm{cos}\left(\mathrm{x}\right)\right)+\mathrm{itan}^{−\mathrm{1}} \left(\frac{\mathrm{x}}{\mathrm{lncos}\left(\mathrm{x}\right)}\right)+\mathrm{2}\pi\right. \\ $$$$\mathrm{principal}\:\mathrm{definition} \\ $$$$\Rightarrow\mathrm{2ln}\left(\mathrm{cos}\left(\mathrm{x}\right)+\mathrm{ix}\right)=\mathrm{ln}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{ln}^{\mathrm{2}} \left(\mathrm{cos}\left(\mathrm{x}\right)\right)+\mathrm{2itan}^{−\mathrm{1}} \left(\frac{\mathrm{x}}{\mathrm{ln}\left(\mathrm{cos}\left(\mathrm{x}\right)\right.}\right)\right. \\ $$$$\mid\mathrm{2ln}\left(\mathrm{cos}\left(\mathrm{x}\right)+\mathrm{ix}\right)\mid^{\mathrm{2}} =\mathrm{ln}^{\mathrm{2}} \left(\mathrm{x}^{\mathrm{2}} +\mathrm{ln}^{\mathrm{2}} \left(\mathrm{cos}\left(\mathrm{x}\right)\right)\right)+\mathrm{4arctan}^{\mathrm{2}} \left(\frac{\mathrm{x}}{\mathrm{ln}\left(\mathrm{cos}\left(\mathrm{x}\right)\right.}\right) \\ $$$$\mathrm{easy}\:\mathrm{from}\:\mathrm{here} \\ $$$$ \\ $$$$ \\ $$
Commented by Calculusboy last updated on 10/Dec/23
$$\boldsymbol{{thanks}}\:\boldsymbol{{sir}} \\ $$
Commented by witcher3 last updated on 10/Dec/23
$$\mathrm{withe}\:\mathrm{Pleasur} \\ $$$$ \\ $$