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Question Number 138628 by mohammad17 last updated on 15/Apr/21

$$findtheregioninwhichthefunction$$$$$$$$f\left(z\right)=\frac{log(z-2i)}{z^2+1}isanalytic?$$$$$$$$helpmesir$$

Commented by mohammad17 last updated on 16/Apr/21

$$?????$$

Answered by mathmax by abdo last updated on 17/Apr/21

$$\mathrm{let}\mathrm{z}=\mathrm{x}+\mathrm{iy}\Rightarrow \mathrm{f}\left(\mathrm{z}\right)=\mathrm{f}(\mathrm{x}+\mathrm{iy})=\mathrm{u}(\mathrm{x},\mathrm{y})+\mathrm{iv}(\mathrm{x},\mathrm{y})$$$$\mathrm{f}(\mathrm{x}+\mathrm{iy})=\frac{\log (\mathrm{x}+\mathrm{iy}-2\mathrm{i})}{{(\mathrm{x}+\mathrm{iy})}^2+1}=\frac{\log (\mathrm{x}+\mathrm{i}(\mathrm{y}-2))}{{\mathrm{x}}^2+2\mathrm{ixy}-{\mathrm{y}}^2+1}$$$$=\frac{\log \left(\sqrt{{\mathrm{x}}^2+{(\mathrm{y}-2)}^2}{\mathrm{e}}^{\mathrm{iarctan}\left(\frac{\mathrm{y}-2}{\mathrm{x}}\right)}\right)}{{\mathrm{x}}^2-{\mathrm{y}}^2+1+2\mathrm{ixy}}=\frac{1}{2}\frac{\log ({\mathrm{x}}^2+{(\mathrm{y}-2)}^2)}{{\mathrm{x}}^2-{\mathrm{y}}^2+1+2\mathrm{ixy}}$$$$+\frac{\mathrm{iarctan}\left(\frac{\mathrm{y}-2}{\mathrm{x}}\right)}{{\mathrm{x}}^2-{\mathrm{y}}^2+1+2\mathrm{ixy}}$$$$=\frac{\log ({\mathrm{x}}^2+{(\mathrm{y}-2)}^2)({\mathrm{x}}^2-{\mathrm{y}}^2+1-2\mathrm{ixy})}{{({\mathrm{x}}^2-{\mathrm{y}}^2+1)}^2+{4\mathrm{x}}^2{\mathrm{y}}^2}$$$$+\mathrm{i}\frac{\arctan \left(\frac{\mathrm{y}-2}{\mathrm{x}}\right)({\mathrm{x}}^2-{\mathrm{y}}^2+1-2\mathrm{ixy})}{{({\mathrm{x}}^2-{\mathrm{y}}^2+1)}^2+{4\mathrm{x}}^2{\mathrm{y}}^2}$$$$=\frac{({\mathrm{x}}^2-{\mathrm{y}}^2+1)\log ({\mathrm{x}}^2+{(\mathrm{y}-2)}^2)}{{({\mathrm{x}}^2-{\mathrm{y}}^2+1)}^2+{4\mathrm{x}}^2{\mathrm{y}}^2}-2\mathrm{i}\frac{\mathrm{xylog}({\mathrm{x}}^2+{(\mathrm{y}-2)}^2)}{{({\mathrm{x}}^2-{\mathrm{y}}^2+1)}^2+{4\mathrm{x}}^2{\mathrm{y}}^2}$$$$+\mathrm{i}\frac{({\mathrm{x}}^2-{\mathrm{y}}^2+1)\arctan \left(\frac{\mathrm{y}-2}{\mathrm{x}}\right)}{{({\mathrm{x}}^2-{\mathrm{y}}^2+1)}^2+{4\mathrm{x}}^2{\mathrm{y}}^2}+\frac{2\mathrm{xyarctan}\left(\frac{\mathrm{y}-2)}{\mathrm{x}}\right)}{{({\mathrm{x}}^2-{\mathrm{y}}^2+1)}^2+{4\mathrm{x}}^2{\mathrm{y}}^2}$$$$\Rightarrow \mathrm{u}(\mathrm{x},\mathrm{y})=\frac{({\mathrm{x}}^2-{\mathrm{y}}^2+1)\log ({\mathrm{x}}^2+{(\mathrm{y}-2)}^2)+2\mathrm{xyarctan}\left(\frac{\mathrm{y}-2}{\mathrm{x}}\right)}{{({\mathrm{x}}^2-{\mathrm{y}}^2+1)}^2+{4\mathrm{x}}^2{\mathrm{y}}^2}$$$$\mathrm{and}\mathrm{v}(\mathrm{x},\mathrm{y})=\frac{({\mathrm{x}}^2-{\mathrm{y}}^2+1)\arctan \left(\frac{\mathrm{y}-2}{\mathrm{x}}\right)-2\mathrm{xylog}({\mathrm{x}}^2+{(\mathrm{y}-2)}^2)}{{({\mathrm{x}}^2-{\mathrm{y}}^2+1)}^2+{4\mathrm{x}}^2{\mathrm{y}}^2}$$$$\mathrm{after}\mathrm{we}\mathrm{apply}\mathrm{cauchy}\mathrm{conditions}\frac{\partial \mathrm{u}}{\partial \mathrm{x}}=\frac{\partial \mathrm{v}}{\partial \mathrm{y}}\mathrm{and}\frac{\partial \mathrm{u}}{\partial \mathrm{y}}=-\frac{\partial \mathrm{v}}{\partial \mathrm{x}}$$$$.....$$

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