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Question Number 62232 by aliesam last updated on 18/Jun/19

Commented by maxmathsup by imad last updated on 18/Jun/19

$$wehave\mid \alpha +\beta e^{i\theta }\mid =\mid \alpha +\beta cos\theta +i\beta sin\theta \mid =\sqrt{{(\alpha +\beta cos\theta )}^2+{\beta }^2{sin}^2\theta }\Rightarrow$$$$ln(\mid \alpha +{ie}^{i\theta }\mid )=\frac{1}{2}ln\{{(\alpha +\beta cos\theta )}^2+{\beta }^2{sin}^2\theta \}=\frac{1}{2}ln\{{\alpha }^2+2\alpha \beta cos\theta +{\beta }^2\}\Rightarrow$$$$I=\frac{1}{2}{\int }_0^{2\pi }ln({\alpha }^2+{\beta }^2+2\alpha \beta cos\theta )d\theta$$$$=\frac{1}{2}{\int }_0^{2\pi }\{ln({\alpha }^2+{\beta }^2)+ln(1+\frac{2\alpha \beta }{{\alpha }^2+{\beta }^2}cos\theta )\}d\theta$$$$=\pi ln({\alpha }^2+{\beta }^2)+\frac{1}{2}{\int }_0^{2\pi }ln(1+\frac{2\alpha \beta }{{\alpha }^2+{\beta }^2}cos\theta )d\theta letfind$$$$f\left(t\right)={\int }_0^{2\pi }ln(1+tcos\theta )d\theta \Rightarrow f^{{}^{\prime }}\left(t\right)={\int }_0^{2\pi }\frac{cos\theta }{1+tcos\theta }d\theta$$$$=\frac{1}{t}{\int }_0^{2\pi }\frac{1+tcos\theta -1}{1+tcos\theta }d\theta =\frac{2\pi }{t}-\frac{1}{t}{\int }_0^{2\pi }\frac{d\theta }{1+tcos\theta }$$$$changement{e}^{i\theta }=zgive{\int }_0^{2\pi }\frac{d\theta }{1+tcos\theta }={\int }_{\mid z\mid =1}\frac{1}{1+t\frac{z+z^{-1}}{2}}\frac{dz}{iz}$$$$={\int }_{\mid z\mid =1}\frac{2dz}{iz(2+tz+{tz}^{-1})}={\int }_{\mid z\mid =1}\frac{-2idz}{2z+{tz}^2+t}={\int }_{\mid z\mid =1}\frac{-2idz}{{tz}^2+2z+t}$$$$letw\left(z\right)=\frac{-2i}{{tz}^2+2z+t}polesofw?$$$${\mathrm{\Delta }}^{{}^{\prime }}=1-t^{2}case1\mid t\mid <1\Rightarrow z_1=-1+\sqrt{1-t^2}and{z}_2=-1-\sqrt{1-t^2}$$$$\mid z_1\mid -1=\mid -1+\sqrt{1-t^2}\mid -1=1-\sqrt{1-t^2}-1<0\Rightarrow \mid z_1\mid <1$$$$\mid z_2\mid -1=\mid -1-\sqrt{1-t^2}\mid -1=1+\sqrt{1-t^2}-1>0\left(toeliminatefromresidus\right)$$$${\int }_{\mid z\mid =1}w-z)dz=2i\pi Res(w,z_1)butw\left(z\right)=\frac{-2i}{t(z-z_1)(z-z_2)}\Rightarrow$$$$Res(w,z_1)=\frac{-2i}{t(z_1-z_2)}=\frac{-2i}{t2\sqrt{1-t^2}}=\frac{-i}{t\sqrt{1-t^2}}\Rightarrow$$$${\int }_{\mid z\mid =1}w\left(z\right)dz=2i\pi \left(\frac{-i}{t\sqrt{1-t^2}}\right)=\frac{2\pi }{t\sqrt{1-t^2}}\Rightarrow$$$$f^{{}^{\prime }}\left(t\right)=\frac{2\pi }{t}-\frac{2\pi }{t^2\sqrt{1-t^2}}\Rightarrow f\left(t\right)=2\pi ln\mid t\mid -2\pi \int \frac{dt}{t^2\sqrt{1-t^2}}+c$$$$\int \frac{dt}{t^2\sqrt{1-t^2}}=\int \frac{1}{t^2}{(1-t^2)}^{\frac{-1}{2}}dt{=}_{byparts}-\frac{1}{t}{(1-t^2)}^{-\frac{1}{2}}-\int -\frac{1}{t}(-\frac{1}{2})(-2t){(1-t^2)}^{-\frac{3}{2}}dt$$$$=-\frac{1}{t\sqrt{1-t^2}}+\int \frac{1}{{(1-t^2)}^{\frac{3}{2}}}dtcyangementt=sin\alpha give$$$$\int \frac{dt}{{(1-t^2)}^{\frac{3}{2}}}=\int \frac{1}{{cos}^3\alpha }cos\alpha d\alpha =\int \frac{d\alpha }{{cos}^2\alpha }=\int \frac{2d\alpha }{1+cos\left(2\alpha \right)}$$$${=}_{tan\left(\alpha \right)=u}\int \frac{2du}{(1+u^2)(1+\frac{1-u^2}{1+u^2})}=\int \frac{2du}{1+u^2+1-u^2}=\int du=u+c$$$$=tan\left(\alpha \right)+c=tan\left(arcsint\right)+c\Rightarrow$$$$f\left(t\right)=2\pi ln\mid t\mid -2\pi \{\frac{-1}{t\sqrt{1-t^2}}+tan\left(arcsint\right)\}+c\Rightarrow$$$$I=\pi ln({\alpha }^2+{\beta }^2)+\frac{1}{2}f\left(\frac{2\alpha \beta }{{\alpha }^2+{\beta }^2}\right)+c$$$$I=\pi ln({\alpha }^2+{\beta }^2)+\pi ln\mid \frac{2\alpha \beta }{{\alpha }^2+{\beta }^2}\mid -\pi \{-\frac{1}{{\frac{2\alpha \beta }{{\alpha }^2+{\beta }^2}\sqrt{1-(\frac{2\alpha \beta }{{\alpha }^2+{\beta }^{(}}})}^2}+tan\left(arcsin\left(\frac{2\alpha \beta }{{\alpha }^2+{\beta }^2}\right)\right\}$$$$+cresttofindthevalueofc....becontinued...$$$$$$$$$$

Commented by aliesam last updated on 18/Jun/19

$$thankssirbrilliantsol$$

Commented by maxmathsup by imad last updated on 18/Jun/19

$$\mid \frac{2\alpha \beta }{{\alpha }^2+{\beta }^2}\mid <1forthatitaket=\frac{2\alpha \beta }{{\alpha }^2+{\beta }^2}withoutstudyingthecase\mid t\mid >1..$$

Commented by maxmathsup by imad last updated on 18/Jun/19

$$youarewelcomesir.$$

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