Find the image of $x+y=4$ under the mapping $w=z^{-1}$.

I have read of an algebraic method that seems to work (unsure if my answer is correct).

Let $$x+iy=z=\frac{1}{w}=\frac{1}{u+iv}=\frac{u}{u^2+v^2}-i\frac{v}{u^2+v^2}$$ Now upon equating real and imaginary components, $$x=\frac{u}{u^2+v^2}, \ \ y=-\frac{v}{u^2+v^2}$$

Now, \begin{align} x+y&=4 \\ \frac{u}{u^2+v^2}-\frac{v}{u^2+v^2}&=4 \\ u-v&=4(u^2+v^2) \\ 4u^2-u+4v^2+v&=0 \\ \left(u-\frac{1}{8}\right)^2+\left(v+\frac{1}{8}\right)^2&=\frac{1}{32} \ \ \ \ \ \ \ \text{(upon completing the square)} \end{align}

Is the method valid? It is very simple. Any advice would be really appreciated

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    $\begingroup$ I don't see any mistakes (except for a typo where you left out the minus sign on $y after the line "equating real and imaginary components.") What are you unsure of? I should perhaps say that I didn't check the arithmetic in the last line. $\endgroup$ – saulspatz Aug 2 '18 at 2:41
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    $\begingroup$ A bit of geometry allows you avoid handling formulas: The map $z\mapsto z^{-1}$ is an inversion with respect to the unit circle followed by a symmetry with respect to the $X$-axis. The closest point of the line $x+y=4$ to the origin is $(2,2)$ at a distance $4/\sqrt{2}$, by Pythagoras. The inversion of this point will be a point $(1/4,1/4)$ at a distance $\sqrt{2}/4$ from the origin. Reflexion with respect to the $X$-axis would be $(1/4,-1/4)$. The inversion plus reflexion transforms the line into the circle with diameter $(0,0)$ to $(1/4,-1/4)$. It follows that the center is $(1/8,-1/8)$ ... $\endgroup$ – user580373 Aug 2 '18 at 2:47
  • $\begingroup$ ... radius $\sqrt{2}/8$. $\endgroup$ – user580373 Aug 2 '18 at 2:47
  • $\begingroup$ Thank you! This is the explanation I was hoping for. This confirms my reasoning on why this works and not just how. $\endgroup$ – user557493 Aug 2 '18 at 2:53

The posted proof looks good.

For an alternative algebraic solution without resorting to cartesian:

  • write the condition $\,x+y=4\,$ as $\,z+\bar z - i(z - \bar z) = 8\,$;

  • substitute $\,z = 1/w\,$ and simplify to $\,8 w \bar w - (1+i)w - (1-i) \bar w = 0\,$;

  • rearrange the above to write as $\,\left|8w - (1-i)\right|^2 = 2\,$.

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    $\begingroup$ I was wondering, is the image of the line $x+y=4$ under the mapping $w=z^{-1}$ defined at the origin? $\endgroup$ – user557493 Aug 2 '18 at 8:15
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    $\begingroup$ @Bell If you work strictly in the complex plane then the image is indeed a punctured circle without the origin. But Möbius transformations (including inversion) are often defined on the extended complex plane, in which case the origin $w=0$ is the image of the point at infinity on the line $x+y=4$. $\endgroup$ – dxiv Aug 2 '18 at 15:43

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