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I start by writing $f(z)$ as $$\frac{1}{(1-(x+iy))^2}$$ and then I expand the bottom to get $$\frac{1}{(1-2x+x^2-y^2) + i(2y-2xy)}$$

The answer says $$w=\frac{(1+x^2-2x-y^2)-(i(2xy-2y)}{(1+x^2-2x-y^2)^2+(2xy-2y)^2}$$

How do I get to this stage?

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  • $\begingroup$ Multiply by the conjugate to make the denominator real... it's like $\frac{1}{a+bi}=\frac{a-bi}{(a+bi)(a-bi)}=\frac{a-bi}{a^2-b^2i^2}=\frac{a-bi}{a^2+b^2}$ $\endgroup$ May 20, 2015 at 19:52
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    $\begingroup$ Indeed, $$\frac1{a+ib}=\frac{a-ib}{a^2+b^2}.$$ $\endgroup$
    – Did
    May 20, 2015 at 19:52

3 Answers 3

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You can simplify this process a bit:

$$\begin{align} \frac1{(1-z)^2} & =\left(\frac1{1-z}\right)^2=\left(\frac{1-\overline z}{(1-z)(1-\overline z)}\right)^2\\ & = \left({1-\overline z\over 1-z-\overline z+|z|^2}\right)^2\\ & = \left({1-x+iy\over 1-2x+x^2+y^2}\right)^2,\quad\text{since }z+\overline z=2x\\ & = {1-2x+x^2-y^2+2iy(1-x)\over(1-2x+x^2+y^2)^2}\\ & ={1-2x+x^2-y^2\over(1-2x+x^2+y^2)^2}+{2y(1-x)\over(1-2x+x^2+y^2)^2}i. \end{align}$$

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Here is another way: $u(z) = \operatorname{re} f(z) = {1 \over 2} (f(z)+ \overline{f(z)})$, $v(z) = \operatorname{im} f(z) = {1 \over 2i} (f(z)- \overline{f(z)})$.

In this case, we have (also using $w \bar{w} = |w|^2$.) \begin{eqnarray} u(z) &=& {1 \over 2} ({1 \over (1-z)^2}+{1 \over (1-\bar{z})^2}) \\ &=& {1 \over 2} {(1-\bar{z})^2 + (1-z)^2 \over |1-z|^4 }\\ &=& {1 \over 2} {2 - 2 (z + \bar{z}) + z^2 + \bar{z}^2 \over |1-z|^4 } \\ &=& {1 \over 2} {2 - 2 \cdot 2 \operatorname{re}(z) + 2 \operatorname{re}(z^2) \over |1-z|^4 } \\ &=& { 1 - 2x +x^2-y^2\over ((1-x)^2+y^2)^2} \end{eqnarray}

Similarly: \begin{eqnarray} v(z) &=& {1 \over 2i} ({1 \over (1-z)^2}-{1 \over (1-\bar{z})^2}) \\ &=& {1 \over 2i} {(1-\bar{z})^2 - (1-z)^2 \over |1-z|^4 }\\ &=& {1 \over 2i} {2 (z - \bar{z}) + \bar{z}^2 - z^2 \over |1-z|^4 } \\ &=& {1 \over 2i} {2 \cdot 2i \operatorname{im}(z) - 2 i \operatorname{im}(z^2) \over |1-z|^4 } \\ &=& { 2y - 2xy\over ((1-x)^2+y^2)^2} \end{eqnarray}

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$$ \begin{align*} \frac{1}{(1 - (x + iy))^2} &= \frac{1}{1-2(x+iy)+(x+iy)^2} = \frac{1}{1+x^2-y^2-2x + i(2xy - 2y)} \\ &= \frac{1 +x^2-y^2-2x-i(2xy - 2y)}{\left[ (1+x^2-y^2-2x)+i(2xy-2y) \right]\left[ (1+x^2-y^2-2x)-i(2xy-2y) \right]} \\ &= \frac{(1+x^2-y^2-2x) - i(2xy - 2y)}{(1+x^2-y^2-2x)^2 + (2xy-2y)^2} \end{align*} $$

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  • $\begingroup$ where did you get the 2(x+iy) term from? $\endgroup$
    – Al jabra
    May 20, 2015 at 20:37
  • $\begingroup$ @Aljabra $$(1-(x+iy))^2 = 1 - (x+iy) - (x +iy) + (x+iy)^2 = 1 - 2(x+iy) + (x + iy)^2$$ $\endgroup$
    – aexl
    May 21, 2015 at 3:36

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