# Minimum value of $|z-w|$ where $z,w \in \mathbb C$ such that $|z|=11$, and $|w+4+3i|=5$?

I was thinking about the problem:

What is the minimum value of $|z-w|$ where $z,w \in \mathbb C$ such that $|z|=11$, and $|w+4+3i|=5$?

My attempts: I notice that $|z-w| \geq |z|-|w|=11-|w|$. Also if i take $w=u+iv$ then $|w+4+3i|=5$ gives $|w|^2+8u+6v=0$. Now i can not proceed. Can someone point me in the right direction? Thanks in advance for your time.

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If you think of it geometrically, you'll get a sense of what you needed to do algebraically to actually prove what the answer is. $|z|=11$ means $z$ is a point of the circle centered in $0$ and radius $11$. $|w+4+3i|=5$ means $w$ is a point of the circle centered in $4+3i$ and radius $5$. This isn't an answer or a hint, just an observation. –  Git Gud Jan 10 '13 at 17:59
Thanks a lot sir.It's been useful observation. +1 from me. –  user52976 Jan 10 '13 at 18:11

Geometrically, $|z| = 11$ is the circle centered at the origin of radius $11$. If a point $w$ is inside the circle $|z| = 11$, elementary geometry shows that the the minimum distance from $w$ to this circle is $11 - |w|$ (the outward distance from $w$ to the circle).

$|w + 4 + 3i | = 5$ is the circle centered at $(-4,-3)$ of radius $5$. I claim that this circle is wholly contained inside the circle $|z| = 11$. This is evident geometrically, and to prove it observe that by the triangle inequality $|w| \leq |w + 4 + 3i| + |-(4 + 3i)| = 5 + 5 = 10$. Equality occurs when $w = -8 - 6i$, the farther of the two points on the circle that are on the line connecting the origin to the center $(-4,-3i)$. The minimum of $11 - |w|$ on this circle occurs when $|w|$ is maximized, namely at $w = -8 - 6i$ when $|w| = 10$. So the minimum distance is given by $11 - 10 = 1$.

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Let $z=a+bi$ and $w=c+di$.

Now, $|z|=11\implies \sqrt{a^2+b^2}=11\implies a^2+b^2=121$, so $z$ is constrained to this circle (shown in blue below).

On the other hand, $|w+4+3i|=5\implies \sqrt{(c+4)^2+(d+3)^2}=5\implies (c+4)^2+(d+3)^2=25$, so $w$ is constrained to this circle (in red below).

The task is to minimize $|z-w|=\sqrt{(a-c)^2+(b-d)^2}$ subject to the two constraints above. Geometrically, we are looking to find the shortest distance between any two points which lie on the blue and red circles, respectively. A little calculus reveals the minimum is $1$ and is obtained when $a=-44/5$, $b=-33/5$ and $c=-8$, $d=-6$, as shown below.

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I have little bit of problem in understanding the line " A little calculus reveals the minimum is 1". Can you explain a little bit about it,sir? –  user52976 Jan 10 '13 at 18:46
Use the method of Lagrange multipliers. Look at Example 5 here to see how. It takes a little scribbling to do by hand, but it's just basic calculus and algebra. –  JohnD Jan 10 '13 at 19:39
Thanks a lot sir.I have got it. –  user52976 Jan 11 '13 at 4:39
@user33640: The upshot of this method is that is general enough to work regardless of the two circles you are given. –  JohnD Jan 11 '13 at 14:25

let $z=11(\cos A+i\sin A), w=r(\cos B+i\sin B)$ so $|w+4+3i|=\sqrt{(r\cos B+4)^2+(r\sin B+3)^2}=\sqrt{r^2+25+2r(4\cos B+3\sin B)}$

So, $r^2+25+2r(4\cos B+3\sin B)=25\implies r=-2(4\cos B+3\sin B)$ as $r\ne0$

$|z-w|^2=(11\cos A-r\cos B)^2+(11\sin A-r\sin B)^2=11^2+r^2-22r\cos(A-B)\ge (11-r)^2$

Now, let $R\sin C=4,R\cos C=3$ where $R>0$ so that $R^2=4^2+3^2=5^2\implies R=5$ and $\sin C=\frac45,\cos C=\frac35$

so $4\cos B+3\sin B=R\sin(B+C)=5\sin(B+C)$

So, $-5\le 4\cos B+3\sin B\le 5\implies -10\le r\le 10$

$\implies|z-w|^2\ge (11-10)^2=1$ which occurs if $A=B$ and if $B+C=-\frac\pi2\implies \sin B=\sin(-\frac\pi2-C)=-\sin(\frac\pi2+C)=-\cos C=-\frac34$ and $\cos B=\cos(-\frac\pi2-C)=\cos(\frac\pi2+C)=-\sin C=-\frac45$

So, $z=-\frac{11}5(4+3i),w=-\frac{10}5(4+3i)=-(8+6i)$

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Thanks a lot sir.I have got it.+1 from me for the lucid explanation. –  user52976 Jan 10 '13 at 18:58
@user33640, my pleasure. –  lab bhattacharjee Jan 10 '13 at 19:05

$|z-w|> = |z|-|w|$

$|w+4+3i|=<|w|+|4+3i|$

$|w|>= 0$ $Then Min|w|=0$

$|z-w|>=(11-0)$

$|z-w|>=11$

$Min|z-w|=11$

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