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

find the region in which the function f(z)=((log(z−2i))/(z^2 +1)) is analytic ? help me sir

$${find}\:{the}\:{region}\:{in}\:{which}\:{the}\:{function}\: \\ $$$$ \\ $$$${f}\left({z}\right)=\frac{{log}\left({z}−\mathrm{2}{i}\right)}{{z}^{\mathrm{2}} +\mathrm{1}}\:{is}\:{analytic}\:? \\ $$$$ \\ $$$${help}\:{me}\:{sir}\: \\ $$

Commented by mohammad17 last updated on 16/Apr/21

?????

$$????? \\ $$

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

let z=x+iy ⇒f(z)=f(x+iy) =u(x,y)+iv(x,y) f(x+iy)=((log(x+iy−2i))/((x+iy)^2 +1)) =((log(x+i(y−2)))/(x^2 +2ixy−y^2 +1)) =((log((√(x^2 +(y−2)^2 ))e^(iarctan(((y−2)/x))) ))/(x^2 −y^2 +1+2ixy)) =(1/2)((log(x^2 +(y−2)^2 ))/(x^2 −y^2 +1 +2ixy)) +((iarctan(((y−2)/x)))/(x^2 −y^2 +1+2ixy)) =((log(x^2 +(y−2)^2 )(x^2 −y^2 +1−2ixy))/((x^2 −y^2 +1)^2 +4x^2 y^2 )) +i ((arctan(((y−2)/x))(x^2 −y^2 +1−2ixy))/((x^2 −y^2 +1)^2 +4x^2 y^2 )) =(((x^2 −y^2 +1)log(x^2 +(y−2)^2 ))/((x^2 −y^2 +1)^2 +4x^2 y^2 ))−2i((xylog(x^2 +(y−2)^2 ))/((x^2 −y^2 +1)^2 +4x^2 y^2 )) +i(((x^2 −y^2 +1)arctan(((y−2)/x)))/((x^2 −y^2 +1)^2 +4x^2 y^2 )) +((2xyarctan(((y−2))/x)))/((x^2 −y^2 +1)^2 +4x^2 y^2 )) ⇒u(x,y)=(((x^2 −y^2 +1)log(x^2 +(y−2)^2 )+2xyarctan(((y−2)/x)))/((x^2 −y^2 +1)^2 +4x^2 y^2 )) and v(x,y)=(((x^2 −y^2 +1)arctan(((y−2)/x))−2xylog(x^2 +(y−2)^2 ))/((x^2 −y^2 +1)^2 +4x^2 y^2 )) after we apply cauchy conditions (∂u/∂x)=(∂v/∂y) and (∂u/∂y)=−(∂v/∂x) .....

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

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