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

f i n d t h e r e g i o n i n w h i c h t h e f u n c t i o n

f (z) = \frac{l o g (z - 2 i)}{z^{2} + 1} i s a n a l y t i c ?

h e l p m e s i r

Commented by mohammad17 last updated on 16/Apr/21

? ? ? ? ?

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

let z = x + iy \Rightarrow f (z) = f (x + iy) = u (x, y) + iv (x, y)

f (x + iy) = \frac{\log (x + iy - 2 i)}{{(x + iy)}^{2} + 1} = \frac{\log (x + i (y - 2))}{x^{2} + 2 ixy - y^{2} + 1}

= \frac{\log (\sqrt{x^{2} + {(y - 2)}^{2}} e^{iarctan (\frac{y - 2}{x})})}{x^{2} - y^{2} + 1 + 2 ixy} = \frac{1}{2} \frac{\log (x^{2} + {(y - 2)}^{2})}{x^{2} - y^{2} + 1 + 2 ixy}

+ \frac{iarctan (\frac{y - 2}{x})}{x^{2} - y^{2} + 1 + 2 ixy}

= \frac{\log (x^{2} + {(y - 2)}^{2}) (x^{2} - y^{2} + 1 - 2 ixy)}{{(x^{2} - y^{2} + 1)}^{2} + {4 x}^{2} y^{2}}

+ i \frac{\arctan (\frac{y - 2}{x}) (x^{2} - y^{2} + 1 - 2 ixy)}{{(x^{2} - y^{2} + 1)}^{2} + {4 x}^{2} y^{2}}

= \frac{(x^{2} - y^{2} + 1) \log (x^{2} + {(y - 2)}^{2})}{{(x^{2} - y^{2} + 1)}^{2} + {4 x}^{2} y^{2}} - 2 i \frac{xylog (x^{2} + {(y - 2)}^{2})}{{(x^{2} - y^{2} + 1)}^{2} + {4 x}^{2} y^{2}}

+ i \frac{(x^{2} - y^{2} + 1) \arctan (\frac{y - 2}{x})}{{(x^{2} - y^{2} + 1)}^{2} + {4 x}^{2} y^{2}} + \frac{2 xyarctan (\frac{y - 2)}{x})}{{(x^{2} - y^{2} + 1)}^{2} + {4 x}^{2} y^{2}}

\Rightarrow u (x, y) = \frac{(x^{2} - y^{2} + 1) \log (x^{2} + {(y - 2)}^{2}) + 2 xyarctan (\frac{y - 2}{x})}{{(x^{2} - y^{2} + 1)}^{2} + {4 x}^{2} y^{2}}

and v (x, y) = \frac{(x^{2} - y^{2} + 1) \arctan (\frac{y - 2}{x}) - 2 xylog (x^{2} + {(y - 2)}^{2})}{{(x^{2} - y^{2} + 1)}^{2} + {4 x}^{2} y^{2}}

after we apply cauchy conditions \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} and \frac{\partial u}{\partial y} = - \frac{\partial v}{\partial x}

\dots . .