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I'm trying to prove that the set of $z$ that satisfies the following equation represents a circle or a straight line on $\mathbb C$.

$$(a\bar c-c\bar a)|w^2|+(a\bar d-c\bar b)w+(b\bar c-d\bar a)\bar w+b\bar d-d\bar b=0 $$ If $(a\bar c-c\bar a)=0$, I could prove that this is a straight line. (It took very long though. If you know how to show it simply, please give me some ideas on how to proceed.)

Apparently, if $(a\bar c-c\bar a)\neq 0$, we can get $$\Bigg |\displaystyle w+\frac{\bar a d-\bar c b}{\bar a c-\bar c a}\Bigg|=\Bigg |\frac{ad-bc}{\bar a c-\bar c a}\Bigg |.$$But I don't know how it works. I tried to factorize the first equation, but it didn't work. So I tried to expand the second equation by replacing the absolute value with the conjugate, but it didn't give me any good idea. Could someone give me some clues?

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Try putting the numbers into exponential form, then see if you can use the exponential definitions of the trig functions to reduce it to something that looks like $sin^2(k)+cos^2(l)=|z|/s$.

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$$(a\bar c-c\bar a)|w^2|+(a\bar d-c\bar b)w+(b\bar c-d\bar a)\bar w+b\bar d-d\bar b=0$$

Dividing by $a\bar c-c\bar a$ you get

$$w \bar w + \frac{a\bar d-c\bar b}{a\bar c-c\bar a}w+ \frac{b\bar c-d\bar a}{a\bar c-c\bar a}\bar w=-\frac{b\bar d-d\bar b}{a\bar c-c\bar a}$$

Now, note that

$$\overline{b\bar c-d\bar a}=-(a\bar d-c\bar b)$$ and $$\overline{a\bar c-c\bar a}=-(a\bar c-c\bar a)$$


$$\frac{b\bar c-d\bar a}{a\bar c-c\bar a}=\overline{\left( \frac{a\bar d-c\bar b}{a\bar c-c\bar a}\right)} \,.$$

For simplicity, let $z:= \frac{a\bar d-c\bar b}{a\bar c-c\bar a}$. Then, your equation is

$$w \bar w + zw+\overline{z} \bar w=-\frac{b\bar d-d\bar b}{a\bar c-c\bar a}$$

Adding $z \bar z$ yields

$$$w \bar w + zw+\overline{z} \bar w +z \bar z =-\frac{b\bar d-d\bar b}{a\bar c-c\bar a} + z \bar z $$


$$(w+\bar z)(\bar w +z)= -\frac{b\bar d-d\bar b}{a\bar c-c\bar a} + z \bar z$$

You should be able to finish from here.

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