1
$\begingroup$

Minimum value of $\lvert z_1-z_2\rvert $ given $\lvert z_1-i\rvert ^2=4$ and $\lvert z_2-6\rvert =\lvert z_2\rvert $

The answer is supposed to be $1$, but I keep getting $0$ when I graph the problem. I get that $z_2$ is a line parallel to the y-axis and for $z_1-i$, I get a circle that crosses the $z_2$ line.

And if $z_1-i$ crosses the line, my logic is, $z_1$ does as well, so the distance should be $0$.

So, where did I go wrong in my thinking?

$\endgroup$

1 Answer 1

Reset to default
2
$\begingroup$

$|z_1-i|^2=4$ is a circle with center at $i$ and radius $2$.

$|z_2-6|=|z_2|$ is a line that passes through $z=3$ and parallel to the imaginary axis. So their shortest distance is $1$, at which point the circle is at $2+i$.

$\endgroup$
4
  • $\begingroup$ Right, I thought 4 would be the radius. A sub-question, if you're willing. Is the circle defined by the $\lvert z_1 \rvert ^2 = c$ just the circle of $\lvert z_1-i \rvert ^2 = c $ moved by $1$ on the y axis? $\endgroup$
    – John Doe
    Commented May 9, 2015 at 18:44
  • $\begingroup$ @JohnDoe: You can of course think it that way. But more easily, it is a circle centered at the origin. $\endgroup$
    – KittyL
    Commented May 9, 2015 at 18:46
  • $\begingroup$ I meant, in the sense, that if in an exercise I was given $\lvert z_1 - x - yi\rvert ^2 = c$ and get $z_2$ defined as a circle, line, whatever, if I am searching for the minimal distance between the two complex numbers, I would not be wrong in regarding the points of $z_1$ as the circle centered at the origin with radius $c^{1/2}$. In essence, the $x$ and $yi$ parameters are meaningless? $\endgroup$
    – John Doe
    Commented May 9, 2015 at 18:52
  • $\begingroup$ @JohnDoe: That depends on where your $x+yi$ is, and also the direction of the line. In this case, your line happens to be parallel to the $y$ axis, and your circle happens to be centered on $y$ axis. For example, what if the line is $|z_2-i|=|z_2-6|$, or what if the circle is $|z-(i+i)|=1$? $\endgroup$
    – KittyL
    Commented May 9, 2015 at 18:57

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .