g ( ζ ) = { R e } ζ − 2 { I m } ζ {\textstyle g(\zeta )=\operatorname {\{} Re\}\zeta -2\operatorname {\{} Im\}\zeta }
= 1 2 π ∫ 0 2 π ζ + z ζ − z ( { R e } ζ − { I m } 2 ζ ) d θ = {\textstyle ={\frac {1}{2\pi }}\int _{0}^{2\pi }{\frac {\zeta +z}{\zeta -z}}\left(\operatorname {\{} Re\}\zeta -\operatorname {\{} Im\}2\zeta \right)d\theta =}
= {\textstyle =} Re 1 2 π ∫ 0 2 π ζ + z ζ − z ( ζ + ζ ¯ 2 − 2 ζ − ζ ¯ 2 i ) i ζ i ζ d θ = {\textstyle \operatorname {Re} {\frac {1}{2\pi }}\int _{0}^{2\pi }{\frac {\zeta +z}{\zeta -z}}\left({\frac {\zeta +{\overline {\zeta }}}{2}}-2{\frac {\zeta -{\overline {\zeta }}}{2i}}\right){\frac {i\zeta }{i\zeta }}d\theta =}
на | z | = 1 ζ ¯ {\textstyle |z|=1\quad {\overline {\zeta }}} → 1 ζ {\textstyle \rightarrow {\frac {1}{\zeta }}}
= {\textstyle =} Re 1 2 π i ∫ | ζ | = 1 ζ + z ζ − z ( ζ 2 + 1 2 ζ − ζ 2 − 1 i ζ ) d ζ ζ {\textstyle \operatorname {Re} {\frac {1}{2\pi i}}\int _{|\zeta |=1}{\frac {\zeta +z}{\zeta -z}}\left({\frac {\zeta ^{2}+1}{2\zeta }}-{\frac {\zeta ^{2}-1}{i\zeta }}\right){\frac {d\zeta }{\zeta }}}
0 , z {\textstyle 0,z} – особые точки
Возможно, лучше считать на ∞ {\textstyle \infty } .
Взять интеграл и выразить результат от x , y {\textstyle x,y} .
1 x 2 + 1 {\textstyle {\frac {1}{x^{2}+1}}}
Re i π ∫ R d x ( x 2 + 1 ) ( z − x ) {\textstyle \operatorname {Re} {\frac {i}{\pi }}\int _{\mathbb {R} }{\frac {dx}{(x^{2}+1)(z-x)}}}
– особые точки
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