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The complex numbers, $z$ and $w$ satisfy the inequalities $|z-3-2i|\le2$ and $|w-7-5i|\le 1$. Find the least possible value of $|z-w|$. Thats my work till now

$$|z-3-2i| = |z-(3+2i)| |z|-|3+2i| |z|-|3+2i| \le 2 |z|^2-|3+2i|^2 \le 2^2 z\overline{z}-(13)\le 4 |z|^2 < 17$$

Kusavil
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Abmon98
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    Do you know geometric interpretation of complex numbers? Draw circles $|z-z_1| \le 2$ and $|z-z_2| \le 1$ for $z_1 = 3 + 2i$, $z_2 = 7 + 5i$, then connect their centres with a line segment. It should be easier to understand problem now ;) Just find distance between $z_1$ and $z_2$, then subtract sum of circle radiuses. – Kusavil Mar 07 '15 at 12:45
  • I only know ways of representing complex numbers till now through vectors and argand diagrams by plotting just their coordinates. where do that circle come from? – Abmon98 Mar 07 '15 at 15:41
  • @Kusavil, you could write up your comment as an answer. com nets may disappear for no reason. – abel Mar 07 '15 at 15:44

2 Answers2

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The figure is the Argand diagram that represents the problem. enter image description here

Inside the circle centered at $C_1$ are the numbers $z$ such that $|z-3-2i|\le2$, inside the other circle the points such that $|w-7-5i|\le 1$ The searched points are the red points and, since the triangle $C_1,C_2,M$ has sides that form a Pythagorean triple, the hypotenuse has length $5$ and the red segment has lenght $2$.

Emilio Novati
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We can imagine $z$ and $\omega$ are in the circle $\odot_1$(centre$(3,2)$) and $\odot_2$(centre$(7,5)$),according to $|z−3−2i|\leqslant2$ and $|w−7−5i|\leqslant1$,their radius are $2$ and $1$. So,the question becomes to calculate the shortest distance between two circles. $$d_{min}=\sqrt{(3-7)^2+(2-5)^2}-1-2=2$$

Faye Tao
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