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Научный форум dxdyМатематика, Физика, Computer Science, Machine Learning, LaTeX, Механика и Техника, Химия,Биология и Медицина, Экономика и Финансовая Математика, Гуманитарные науки |
 
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Страница 1 из 1 |
[ Сообщений: 10 ] |
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![\[
z - z_0
\]](https://dxdy-01.korotkov.co.uk/f/c/6/c/c6c50a197887f920603a9ad18f6a78df82.png "\[
z - z_0
\]")
![\[
w = \frac{z}{{z^2 + 1}};
\]](https://dxdy-02.korotkov.co.uk/f/5/b/c/5bcdefb20893e68a39b48b7d410377b182.png "\[
w = \frac{z}{{z^2 + 1}};
\]")
![\[
z_0 = - 3 - 2i;
\]](https://dxdy-04.korotkov.co.uk/f/b/b/c/bbc004382695e2709e77d95001e8dd8b82.png "\[
z_0 = - 3 - 2i;
\]")
и представьте 
как функцию от 
.![\[
\begin{array}{l}
\rho = \sqrt {\left( { - 3 + 1} \right)^2 + \left( { - 2 - 0} \right)^2 } = \sqrt {4 + 4} = 2\sqrt 2 ; \\
R = \sqrt {\left( { - 3 - 1} \right)^2 + \left( { - 2 + 0} \right)^2 } = \sqrt {16 + 4} = 2\sqrt 5 ; \\
\end{array}
\]](https://dxdy-03.korotkov.co.uk/f/e/1/2/e12671b4bb1862217125e6fa3036d9f082.png "\[
\begin{array}{l}
\rho = \sqrt {\left( { - 3 + 1} \right)^2 + \left( { - 2 - 0} \right)^2 } = \sqrt {4 + 4} = 2\sqrt 2 ; \\
R = \sqrt {\left( { - 3 - 1} \right)^2 + \left( { - 2 + 0} \right)^2 } = \sqrt {16 + 4} = 2\sqrt 5 ; \\
\end{array}
\]")
![\[
\begin{array}{l}
D1:z + 3 + 2i < 2\sqrt 2 ; \\
\frac{{z + 3 + 2i}}{{2\sqrt 2 }} < 1; \\
\end{array}
\]](https://dxdy-04.korotkov.co.uk/f/f/8/e/f8ee2c78e411332600e04ccce4fb8d3182.png "\[
\begin{array}{l}
D1:z + 3 + 2i < 2\sqrt 2 ; \\
\frac{{z + 3 + 2i}}{{2\sqrt 2 }} < 1; \\
\end{array}
\]")
![\[
\begin{array}{l}
w = \frac{{{\raise0.7ex\hbox{$1$} \!\mathord{\left/
{\vphantom {1 2}}\right.\kern-\nulldelimiterspace}
\!\lower0.7ex\hbox{$2$}}}}{{\left( {z - i} \right)}} + \frac{{{\raise0.7ex\hbox{$1$} \!\mathord{\left/
{\vphantom {1 2}}\right.\kern-\nulldelimiterspace}
\!\lower0.7ex\hbox{$2$}}}}{{\left( {z + i} \right)}} = \frac{{{\raise0.7ex\hbox{$1$} \!\mathord{\left/
{\vphantom {1 2}}\right.\kern-\nulldelimiterspace}
\!\lower0.7ex\hbox{$2$}}}}{{\left( {z + 3 + 2i} \right) - 3 - 3i}} + \frac{{{\raise0.7ex\hbox{$1$} \!\mathord{\left/
{\vphantom {1 2}}\right.\kern-\nulldelimiterspace}
\!\lower0.7ex\hbox{$2$}}}}{{\left( {z + 3 + 2i} \right) - 3 - i}} = - \frac{{{\raise0.7ex\hbox{$1$} \!\mathord{\left/
{\vphantom {1 2}}\right.\kern-\nulldelimiterspace}
\!\lower0.7ex\hbox{$2$}}}}{{\left( {3 + 3i} \right)\left[ {1 - \frac{{\left( {z + 3 + 2i} \right)}}{{3 + 3i}}} \right]}} - \frac{{{\raise0.7ex\hbox{$1$} \!\mathord{\left/
{\vphantom {1 2}}\right.\kern-\nulldelimiterspace}
\!\lower0.7ex\hbox{$2$}}}}{{\left( {3 + i} \right)\left[ {1 - \frac{{\left( {z + 3 + 2i} \right)}}{{3 + i}}} \right]}} = \\
= - \frac{1}{2}\sum\limits_{n = 0}^\infty {\frac{{\left( {z + 3 + 2i} \right)^n }}{{\left( {3 + 3i} \right)^{n + 1} }}} - \frac{1}{2}\sum\limits_{n = 0}^\infty {\frac{{\left( {z + 3 + 2i} \right)^n }}{{\left( {3 + i} \right)^{n + 1} }}} ; \\
\end{array}
\]](https://dxdy-03.korotkov.co.uk/f/2/4/2/242f535944615c05d1ed6ea7ac2598aa82.png "\[
\begin{array}{l}
w = \frac{{{\raise0.7ex\hbox{$1$} \!\mathord{\left/
{\vphantom {1 2}}\right.\kern-\nulldelimiterspace}
\!\lower0.7ex\hbox{$2$}}}}{{\left( {z - i} \right)}} + \frac{{{\raise0.7ex\hbox{$1$} \!\mathord{\left/
{\vphantom {1 2}}\right.\kern-\nulldelimiterspace}
\!\lower0.7ex\hbox{$2$}}}}{{\left( {z + i} \right)}} = \frac{{{\raise0.7ex\hbox{$1$} \!\mathord{\left/
{\vphantom {1 2}}\right.\kern-\nulldelimiterspace}
\!\lower0.7ex\hbox{$2$}}}}{{\left( {z + 3 + 2i} \right) - 3 - 3i}} + \frac{{{\raise0.7ex\hbox{$1$} \!\mathord{\left/
{\vphantom {1 2}}\right.\kern-\nulldelimiterspace}
\!\lower0.7ex\hbox{$2$}}}}{{\left( {z + 3 + 2i} \right) - 3 - i}} = - \frac{{{\raise0.7ex\hbox{$1$} \!\mathord{\left/
{\vphantom {1 2}}\right.\kern-\nulldelimiterspace}
\!\lower0.7ex\hbox{$2$}}}}{{\left( {3 + 3i} \right)\left[ {1 - \frac{{\left( {z + 3 + 2i} \right)}}{{3 + 3i}}} \right]}} - \frac{{{\raise0.7ex\hbox{$1$} \!\mathord{\left/
{\vphantom {1 2}}\right.\kern-\nulldelimiterspace}
\!\lower0.7ex\hbox{$2$}}}}{{\left( {3 + i} \right)\left[ {1 - \frac{{\left( {z + 3 + 2i} \right)}}{{3 + i}}} \right]}} = \\
= - \frac{1}{2}\sum\limits_{n = 0}^\infty {\frac{{\left( {z + 3 + 2i} \right)^n }}{{\left( {3 + 3i} \right)^{n + 1} }}} - \frac{1}{2}\sum\limits_{n = 0}^\infty {\frac{{\left( {z + 3 + 2i} \right)^n }}{{\left( {3 + i} \right)^{n + 1} }}} ; \\
\end{array}
\]")
![\[
\begin{array}{l}
D2:2\sqrt 2 < z + 3 + 2i < 2\sqrt 5 ; \\
\frac{{2\sqrt 2 }}{{z + 3 + 2i}} < 1;\frac{{z + 3 + 2i}}{{2\sqrt 5 }} < 1; \\
\end{array}
\]](https://dxdy-01.korotkov.co.uk/f/4/f/8/4f893b8e32192e84c0667aede1c322e182.png "\[
\begin{array}{l}
D2:2\sqrt 2 < z + 3 + 2i < 2\sqrt 5 ; \\
\frac{{2\sqrt 2 }}{{z + 3 + 2i}} < 1;\frac{{z + 3 + 2i}}{{2\sqrt 5 }} < 1; \\
\end{array}
\]")
![\[
\begin{array}{l}
w = \frac{{{\raise0.7ex\hbox{$1$} \!\mathord{\left/
{\vphantom {1 2}}\right.\kern-\nulldelimiterspace}
\!\lower0.7ex\hbox{$2$}}}}{{\left( {z + 3 + 2i} \right) - 3 - 3i}} + \frac{{{\raise0.7ex\hbox{$1$} \!\mathord{\left/
{\vphantom {1 2}}\right.\kern-\nulldelimiterspace}
\!\lower0.7ex\hbox{$2$}}}}{{\left( {z + 3 + 2i} \right) - 3 - i}} = \frac{{{\raise0.7ex\hbox{$1$} \!\mathord{\left/
{\vphantom {1 2}}\right.\kern-\nulldelimiterspace}
\!\lower0.7ex\hbox{$2$}}}}{{\left( {z + 3 + 2i} \right)\left[ {1 - \frac{{3 + 3i}}{{\left( {z + 3 + 2i} \right)}}} \right]}} - \frac{{{\raise0.7ex\hbox{$1$} \!\mathord{\left/
{\vphantom {1 2}}\right.\kern-\nulldelimiterspace}
\!\lower0.7ex\hbox{$2$}}}}{{\left( {3 + i} \right)\left[ {1 - \frac{{\left( {z + 3 + 2i} \right)}}{{3 + 3i}}} \right]}} = \\
= \frac{1}{2}\sum\limits_{n = 0}^\infty {\frac{{\left( {3 + 3i} \right)^n }}{{\left( {z + 3 + 2i} \right)^{n + 1} }}} - \frac{1}{2}\sum\limits_{n = 0}^\infty {\frac{{\left( {z + 3 + 2i} \right)^n }}{{\left( {3 + i} \right)^{n + 1} }}} ; \\
\end{array}
\]](https://dxdy-01.korotkov.co.uk/f/4/9/5/495d291d347481e86240db1216aaf29e82.png "\[
\begin{array}{l}
w = \frac{{{\raise0.7ex\hbox{$1$} \!\mathord{\left/
{\vphantom {1 2}}\right.\kern-\nulldelimiterspace}
\!\lower0.7ex\hbox{$2$}}}}{{\left( {z + 3 + 2i} \right) - 3 - 3i}} + \frac{{{\raise0.7ex\hbox{$1$} \!\mathord{\left/
{\vphantom {1 2}}\right.\kern-\nulldelimiterspace}
\!\lower0.7ex\hbox{$2$}}}}{{\left( {z + 3 + 2i} \right) - 3 - i}} = \frac{{{\raise0.7ex\hbox{$1$} \!\mathord{\left/
{\vphantom {1 2}}\right.\kern-\nulldelimiterspace}
\!\lower0.7ex\hbox{$2$}}}}{{\left( {z + 3 + 2i} \right)\left[ {1 - \frac{{3 + 3i}}{{\left( {z + 3 + 2i} \right)}}} \right]}} - \frac{{{\raise0.7ex\hbox{$1$} \!\mathord{\left/
{\vphantom {1 2}}\right.\kern-\nulldelimiterspace}
\!\lower0.7ex\hbox{$2$}}}}{{\left( {3 + i} \right)\left[ {1 - \frac{{\left( {z + 3 + 2i} \right)}}{{3 + 3i}}} \right]}} = \\
= \frac{1}{2}\sum\limits_{n = 0}^\infty {\frac{{\left( {3 + 3i} \right)^n }}{{\left( {z + 3 + 2i} \right)^{n + 1} }}} - \frac{1}{2}\sum\limits_{n = 0}^\infty {\frac{{\left( {z + 3 + 2i} \right)^n }}{{\left( {3 + i} \right)^{n + 1} }}} ; \\
\end{array}
\]")
![\[
\begin{array}{l}
D3:2\sqrt 5 < z + 3 + 2i; \\
\frac{{2\sqrt 5 }}{{z + 3 + 2i}} < 1; \\
\end{array}
\]](https://dxdy-04.korotkov.co.uk/f/b/f/0/bf0b23202bae33080400adb002bd18bf82.png "\[
\begin{array}{l}
D3:2\sqrt 5 < z + 3 + 2i; \\
\frac{{2\sqrt 5 }}{{z + 3 + 2i}} < 1; \\
\end{array}
\]")
![\[
\begin{array}{l}
w = \frac{{{\raise0.7ex\hbox{$1$} \!\mathord{\left/
{\vphantom {1 2}}\right.\kern-\nulldelimiterspace}
\!\lower0.7ex\hbox{$2$}}}}{{\left( {z + 3 + 2i} \right) - 3 - 3i}} + \frac{{{\raise0.7ex\hbox{$1$} \!\mathord{\left/
{\vphantom {1 2}}\right.\kern-\nulldelimiterspace}
\!\lower0.7ex\hbox{$2$}}}}{{\left( {z + 3 + 2i} \right) - 3 - i}} = \frac{{{\raise0.7ex\hbox{$1$} \!\mathord{\left/
{\vphantom {1 2}}\right.\kern-\nulldelimiterspace}
\!\lower0.7ex\hbox{$2$}}}}{{\left( {z + 3 + 2i} \right)\left[ {1 - \frac{{3 + 3i}}{{\left( {z + 3 + 2i} \right)}}} \right]}} + \frac{{{\raise0.7ex\hbox{$1$} \!\mathord{\left/
{\vphantom {1 2}}\right.\kern-\nulldelimiterspace}
\!\lower0.7ex\hbox{$2$}}}}{{\left( {z + 3 + 2i} \right)\left[ {1 - \frac{{3 + i}}{{\left( {z + 3 + 2i} \right)}}} \right]}} = \\
= \frac{1}{2}\sum\limits_{n = 0}^\infty {\frac{{\left( {3 + 3i} \right)^n }}{{\left( {z + 3 + 2i} \right)^{n + 1} }}} + \frac{1}{2}\sum\limits_{n = 0}^\infty {\frac{{\left( {3 + i} \right)^n }}{{\left( {z + 3 + 2i} \right)^{n + 1} }}} ; \\
\end{array}
\]](https://dxdy-01.korotkov.co.uk/f/8/8/b/88bf96f7fd61d7ea1815bd52d9bfb08c82.png "\[
\begin{array}{l}
w = \frac{{{\raise0.7ex\hbox{$1$} \!\mathord{\left/
{\vphantom {1 2}}\right.\kern-\nulldelimiterspace}
\!\lower0.7ex\hbox{$2$}}}}{{\left( {z + 3 + 2i} \right) - 3 - 3i}} + \frac{{{\raise0.7ex\hbox{$1$} \!\mathord{\left/
{\vphantom {1 2}}\right.\kern-\nulldelimiterspace}
\!\lower0.7ex\hbox{$2$}}}}{{\left( {z + 3 + 2i} \right) - 3 - i}} = \frac{{{\raise0.7ex\hbox{$1$} \!\mathord{\left/
{\vphantom {1 2}}\right.\kern-\nulldelimiterspace}
\!\lower0.7ex\hbox{$2$}}}}{{\left( {z + 3 + 2i} \right)\left[ {1 - \frac{{3 + 3i}}{{\left( {z + 3 + 2i} \right)}}} \right]}} + \frac{{{\raise0.7ex\hbox{$1$} \!\mathord{\left/
{\vphantom {1 2}}\right.\kern-\nulldelimiterspace}
\!\lower0.7ex\hbox{$2$}}}}{{\left( {z + 3 + 2i} \right)\left[ {1 - \frac{{3 + i}}{{\left( {z + 3 + 2i} \right)}}} \right]}} = \\
= \frac{1}{2}\sum\limits_{n = 0}^\infty {\frac{{\left( {3 + 3i} \right)^n }}{{\left( {z + 3 + 2i} \right)^{n + 1} }}} + \frac{1}{2}\sum\limits_{n = 0}^\infty {\frac{{\left( {3 + i} \right)^n }}{{\left( {z + 3 + 2i} \right)^{n + 1} }}} ; \\
\end{array}
\]")