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In Complex Variables and Applications, by Brown and Chyrchill (McGraw-Hill), there is the following exercise: $$\textit{Show that $|z-z_0|=R$ can be written as $|z|^2-2Re(z\bar{z_0})+|z_0|^2=R^2$}$$

I solved it this way:

  1. Square both sides: $|z-z_0|^2 = R^2$
  2. Define $w:=z-z_0$, then we have $w\bar{w}=|w|^2=R^2$
  3. Compute $(z-z_0)(\bar{z}-\bar{z_0}) = z\bar{z}-z\bar{z_0}-z_0\bar{z}+z_0\bar{z_0}=|z|^2-z\bar{z_0}-z_0\bar{z}+|z_0|^2 = R^2$
  4. Compute $-z_0\bar{z} = -x_0x+x_0yi-xy_0i-yy_0$ and $-z\bar{z_0}=-xx_0+xy_0i-x_0yi-yy_0$
  5. Sum them $-z_0\bar{z}-z\bar{z_0}=-2(x_0x_0+yy_0)=-2Re(z\bar{z_0})$
  6. Hence $|z|^2-2Re(z\bar{z_0})+|z_0|^2 = R^2$

But it looks quite tedious and long. I am sure there must be a shorter way to do this, can anyone think of some other way? I tried $|z-z_0|^2 = |(z-z_0)^2|=|z^2-2zz_0+z_0^2|=R^2$, but this leads me nowhere.

Thank you

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    $\begingroup$ There's no need for step 2. Step 4 can be shortened by noticing that $\overline{z} z_0 = \overline{z \overline{z}_0}$, so $z\overline{z}_0 + \overline{z} z_0 = 2\operatorname{Re} (z \overline{z}_0)$. $\endgroup$ – Daniel Fischer Dec 2 '16 at 13:14
  • $\begingroup$ Oh okay! Just to check, are you using the fact that for $z$ we have $Re(z) = \frac{z+\bar{z}}{2}$ and hence $Re(z\bar{z_0}) = \frac{z\bar{z_0}+\overline{z\bar{z_0}}}{2}$ ? $\endgroup$ – Euler_Salter Dec 2 '16 at 13:17
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    $\begingroup$ Yes. This (and the corresponding $w - \overline{w} = 2i \operatorname{Im} w$) is frequently used, one gets to use that without even consciously thinking after a while. $\endgroup$ – Daniel Fischer Dec 2 '16 at 13:23
  • $\begingroup$ Oh okay thank you! If you write down the solution with the improvement in the steps, I'll mark you answer! Just to know, in what other topics/subjects is complex analysis used apart from complex analysis itself? $\endgroup$ – Euler_Salter Dec 2 '16 at 13:24
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What you did is good. It is quite tedious because you detailed the steps.

You should remember the formula: $$\vert a - b \vert^2 = \vert a \vert^2 - 2Re(a\bar{b}) + \vert b \vert^2$$ as it is constantly used. Then the equation of a circle can be find in a much straightforward manner.

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  • $\begingroup$ Thank you! I'll try to remember it! $\endgroup$ – Euler_Salter Dec 2 '16 at 13:27

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