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In my book it is givenenter image description here

I could not understand how they have written the statement given in NOTE.

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    It's about rotation of a curve. – 2017-01-23
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    @MyGlasses how . – 2017-01-23
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    We can rotate every point $z$ (as a vector) with an angle $\theta$, not only in a line, but in every equation, $f(ze^{i\theta})$ is a rotation of $f(z)$. – 2017-01-23
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    @MyGlasses this is valid for every equation – 2017-01-23
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    Of course. If change $z$ by $ze^{i\theta}$ we rotate every point about origin with angle $\theta$, other case, If we change $f(z)$ to $e^{i\theta}f(z)$ then we rotate whole of shape with angle $\theta$. – 2017-01-23
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    @MyGlasses $\theta$ is taken from where ? For a line rotating it , we have to multipy all its equation with $e^{i\theta}$ ,but as above in my book it is only given that replace z with z$e^{i\theta}$. – 2017-01-23
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    As I said above, we can replace $z$ with $ze^{i\theta}$ and this angle is **arbitrary**. Also In line, we replace $z$ with $ze^{i\theta}$ and then replace $\overline{z}$ with $\overline{z}e^{-i\theta}$ for rotation. – 2017-01-23
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    @MyGlasses If we want to rotate it about any other point(not the origin) ,then how can we do it . – 2017-01-24

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Hint: divide by $|\alpha|^2 = \alpha \bar \alpha\ne 0$ and write equation as:

$$ \frac{z}{\alpha}+\frac{\bar z}{\bar \alpha} + \frac{r}{|\alpha|^2} = 0 \quad\iff\quad \operatorname{Re}\left(\frac{z}{\alpha}\right) = - \,\frac{r}{2\,|\alpha|^2} $$

The latter means that $\frac{z}{\alpha}$ describes the vertical line passing through $z_0 = - \,\frac{r}{2\,|\alpha|^2}\,$, which in turn means that $z$ describes the line through $\alpha \,z_0$ at angle $\frac{\pi}{2} + \arg(\alpha)\,$.

Replacing $z$ by $z\,e^{i \theta}$ is algebraically equivalent with replacing $\alpha$ by $\alpha \overline{e^{i \theta}} = \alpha e^{-i \theta}\,$, so the angle of the resulting line becomes $\frac{\pi}{2} + \arg(\alpha) - \theta\,$, which is at an angle of $\theta$ with the original line.