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The exercise I got a bit stuck with is at the end of chapter on complex numbers that does not deal with transformations in general. So the notion of general bilinear transformation is introduced for the purpose of this exercise on the spot.

I am mentioning this because the answer I am looking for should, therefore, be made by more or less brute-force caluculation, not by using general results concerning transformations of complex numbers.

What I should show is that general bilinear transformation, defined as transformation from $z$-plane to $Z$-plane via

$$z = \frac{a Z + b}{c Z + d}$$

transforms a circle given by

$$\left|\frac{z-z_1}{z-z_2}\right| = H$$

where $H$ is a real number not equal to $1$, into a circle or straight line, and then determine what conditions $z_1$, $z_2$ and $H$ should satisfy in order for the circle to be transformed into a straight line.

Now, when I substitute $(aZ+b)/(cZ+d)$ for $z$ in the given equation of the circle I get

$$\left|\frac{(a - c z_1)Z + b - d z_1}{(a - c z_2 ) Z + b - d z_2}\right| = H$$

and from here it seems immediately clear to me that when I raise both sides to the power of 2 and simplify the equation I will get a general conic section curve $Ax^2 + By^2 + Cx + Dy + \cdots =0$ in which $A$ won't be equal to $B$ (due simply to the difference between $z_1$ and $z_2$) and so it won't be an equation of the circle.

Where am I making a mistake?

Storyteller011
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  • I re-formatted the math with TeX. Please double-check that I didn't change the nature of your equations. (I fixed a subscript typo in your $R$ equation. BTW, shouldn't $H$ and $R$ be the same value?) – Blue Oct 08 '13 at 08:42
  • Thank you, Blue! Yes, R should've been H. It was a typo :) I fixed it now. – Storyteller011 Oct 08 '13 at 08:45
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    When you square and simplify, you'll get (in part) $$|a-c z_1|^2 |Z|^2 + \cdots = H^2 \left( |a - c z_2|^2 |Z|^2 +\cdots\right)$$ so that $$0 = |Z|^2\left( |a-c z_1|^2- H^2|a-c z_2|^2 \right) + \cdots = (x^2+y^2) A + \cdots$$ The "$\cdots$" stuff involves single powers of $Z$ and $\overline{Z}$ (and thus $x$ and $y$), and constants. The coefficients of $x^2$ and $y^2$ definitely match. – Blue Oct 08 '13 at 08:48
  • Thanks :) that was enough for me to figure out the rest of the exercise. You might write this comment in the answer if you want me to check off the topic. – Storyteller011 Oct 08 '13 at 14:47

1 Answers1

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Converting comment to answer, as requested ...

When you square and simplify, you'll get (in part) ...

$$\left| a - c z_1 \right|^2 \left|Z\right|^2 + \cdots = H^2 \left( \left|a-c z_2\right|^2 \left|Z\right|^2 + \cdots \right)$$

so that $$ 0 = \left|Z\right|^2\;\left( \left|a-c z_1\right|^2 - H^2 \left|a - c z_2\right|^2 \right) + \cdots = \left( x^2 + y^2 \right) A + \cdots$$

The "$\cdots$" stuff involves single powers of $Z$ and $\overline{Z}$ (and thus $x$ and $y$), and constants. The coefficients of $x^2$ and $y^2$ definitely match.

Blue
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