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Describe the set of points $z$ in the complex plane that satisfies the following:

a.) $|z-1| + |z+1| = 7$

b.) $|z| = 3|z-1|$

For a, I know that it has the property that their distance from $1$ added to their distance from $-1$ is equal to $7$, but I do not know how to describe it any more further than that.

For b, I know that it must be a circle.

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a) Given two fixed points $F_1,F_2$, the set of points $P$ with $PF_1+PF_2=2a$ is the "geometric" definition of an .... ellipse.

b) Rewrite it as $$z\bar z=3(z-1)(\bar z-1)$$

and try to factor it as

$$(z-a)(\bar z- \bar z)= r^2$$

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  • $\begingroup$ Thanks a lot N. S., so I can safely assume that any equation of the form $PF_1+PF_2=2a$ is an ellipse, even in the complex plane? $\endgroup$ – Q.matin Jan 23 '13 at 6:56
  • $\begingroup$ @Q.matin Well, you can also write $z=x+iy$ plug it in, get rid of the roots and see that you get an ellipse. The point is that when you identify the $||$ as distance, the distance between points is "the same" in the complex and real plane. $\endgroup$ – N. S. Jan 23 '13 at 7:09
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Hint : Expand as $(z-1)(\bar{z}-1) + (z+1)(\bar{z}+1)$

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