Describe each set in the \(z\)-plane in (a) and (b) below, where \(\alpha\) is a complex number and \(k\) is a positive number such that \(2|\alpha|<k\).
\(|z-\alpha|+|z+\alpha|=k\);
\(|z-\alpha|+|z+\alpha| \leq k\).
Solution
The two fixed points \(\alpha\) and \(-\alpha\) are the foci. Thus part (a) is an ellipse, because the sum of the distances from \(z\) to the two foci is the constant \(k>2|\alpha|\).
Its center is \(0\). The major semiaxis is \[ a=\frac{k}{2}, \] the focal distance is \[ c=|\alpha|, \] and therefore the minor semiaxis is \[ b=\sqrt{a^2-c^2} =\frac12\sqrt{k^2-4|\alpha|^2}. \] The major axis lies on the line through \(\pm\alpha\).
If we rotate coordinates so that \(\alpha=|\alpha|>0\) and write \(z=x+iy\), the equation is \[ \boxed{ \frac{x^2}{(k/2)^2} +\frac{y^2}{(k^2/4-|\alpha|^2)}=1.} \]
Part (b) is the closed filled ellipse bounded by this curve, since it consists of the points for which the same sum of focal distances is at most \(k\).