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For all complex numbers $z_1,z_2$ satisfying $|z_1|=12$ and $|z_2-3-4i|=5$ , find the minimum value of $|z_1-z_2|$

Can we go like this :

Let $z_1 = x +iy$ therefore $|z_1| = \sqrt{x_1^2+y_1^2}$ and $z_2 = x_2+ iy_2$

$|z_2-3-4i| = \sqrt{(x_2-3)^2+ (y_2-4)^2}$ so $x_2^2-6x_2+9+y_2^2-8y_2+16 =25$ ( squaring both sides) .. Please guide if it is correct..

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    $\begingroup$ is that $z_1$ in $|z-1-z_2|$ $\endgroup$ – Maesumi Apr 11 '13 at 1:27
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Another hint:

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Hence, the maximum is $22$ and the minimum is $2$ but you should prove this.

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  • $\begingroup$ Don't you mean $2$? We are looking for the minimum distance. $\endgroup$ – Ross Millikan Apr 12 '13 at 3:39
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An algebraic method

$$ |z_1 - z_2 | \geq |z_1| - |z_2| \\ |z_2-(3+4i)| \geq |z_2| - 5 \\ |z_2|\leq 10\\ |z_1 - z_2 | \geq 12-10=2 $$

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Hint: Draw the sets $\mathcal{A} = \{z_1: |z_1| = 12\}$ and $\mathcal{B} = \{z_2: |z_2-3-4i| = 5\}$ in the complex plane (both will be circles). Now draw the set $\mathcal{B}' = \{1+z_2:|z_2-3-4i| = 5\}$, which will be "$\mathcal{B}$ shifted one unit to the right." You would like to find the two closest points of $\mathcal{A}$ and $\mathcal{B}'$. The answer will be obvious as soon as you draw these circles.

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  • $\begingroup$ nice solution Lord Soth..... $\endgroup$ – juantheron Dec 15 '13 at 11:11
  • $\begingroup$ @juantheron, I am new to complex numbers and trying to understand why $1+z_{2}$ is plotted in this scenario? $\endgroup$ – Sunil Feb 1 '17 at 6:23

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