Complex Numbers: Finding the least value of |z-w|

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Homework Statement



The complex numbers, z and w satisfy the inequalities |z-3-2i|<=2 and |w-7-5i|<=1

Find the least possible value of |z-w|


Homework Equations



No clue at all.



The Attempt at a Solution



Since its |z-w| i figured that the least possible value will only be when both are max. I tried finding the maximum distance of each complex number by using [tex]\sqrt{}(x^2+y^2)[/tex]+r and came up with a Z=[tex]\sqrt{}13[/tex]+2 and W being [tex]\sqrt{}74[/tex]+1

Both of which are incorrect as z-w gives 4 while the answer is 2
 
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Answers and Replies

  • #2
jbunniii
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The triangle inequality gives you

[tex]|z-w| \leq |z| + |w|[/tex]

This gives you an upper bound for the distance.

You can also use the triangle inequality to obtain a lower bound, as follows:

[tex]|z| = |(z - w) + w|[/tex]

[tex]|w| = |(w - z) + z|[/tex]

Now apply the triangle inequality to the right hand sides of both of the above equalities and rearrange to get a lower bound for [itex]|z - w|[/itex].
 
  • #3
LCKurtz
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They describe two circular discs. The closest points will be on the line between their centers.
 
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Could you kindly explain this a little bit:

I first get |z| by using |z|<= 2+|3+2i| and then put z= 2+|3+2i| in the formula for |z| that u have given?
 
  • #5
jbunniii
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OK, to keep the notation simpler, let [itex]z_0 = 3 + 2i[/itex] and [itex]w_0 = 7 + 5i[/itex].

Then you can write

[tex]z - w = (z - z_0) - (w - w_0) + (z_0 - w_0)[/tex]

Therefore

[tex]|z - w| = |(z - z_0) - (w - w_0) + (z_0 - w_0)|[/tex]

You can apply the triangle inequality in reverse to obtain

[tex]|(z - z_0) - (w - w_0) + (z_0 - w_0)| \geq | |z_0 - w_0| - |(z - z_0) - (w - w_0)||[/tex]

Now, [itex]|z_0 - w_0|[/itex] is just a positive constant (call it [itex]c[/itex]), so the minimization of the right hand side is easy. The task is to mimimize

[tex]|c - |(z - z_0) - (w - w_0)||[/tex]

If there are [itex]z,w[/itex] that make this zero, then the minimum is zero. Otherwise, the minimum is achieved by maximizing [itex]|(z - z_0) - (w - w_0)|[/itex], which you can easily do by using the triangle inequality again.

By the way, this procedure will give you a lower bound. You still have to justify why there are [itex]z,w[/itex] that achieve the lower bound. (Think: under what condition does the triangle inequality become an equality?)
 
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They describe two circular discs. The closest points will be on the line between their centers.


Do u mean it will be |3+2i - ( 7+5i)|?? but that gives the answer 5 while it is 2 in the book
 
  • #7
LCKurtz
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Do u mean it will be |3+2i - ( 7+5i)|?? but that gives the answer 5 while it is 2 in the book

No. |z - (3+2i)| is the distance from x + yi to 3 + 2i. If that is equal to 2 it says the point (x,y) is distance 2 from the point (3,2) in the xy plane. That describes a circle and the inequality describes the interior of that circle. Similarly for the other inequality. Draw a picture.
 
  • #8
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No. |z - (3+2i)| is the distance from x + yi to 3 + 2i. If that is equal to 2 it says the point (x,y) is distance 2 from the point (3,2) in the xy plane. That describes a circle and the inequality describes the interior of that circle. Similarly for the other inequality. Draw a picture.


hmm..I get this part. You are saying that Z and W lie on the circumference of the circle with center 3,2 and radius 2 and center 7,5 with radius 1 right?

I cannot get beyond this point. The question is asking for the minimum value of |z-w| so this means that both Z and W should be at maximum distance from the origin?
 
  • #9
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OK, to keep the notation simpler, let [itex]z_0 = 3 + 2i[/itex] and [itex]w_0 = 7 + 5i[/itex].

Then you can write

[tex]z - w = (z - z_0) - (w - w_0) + (z_0 - w_0)[/tex]

Therefore

[tex]|z - w| = |(z - z_0) - (w - w_0) + (z_0 - w_0)|[/tex]

You can apply the triangle inequality in reverse to obtain

[tex]|(z - z_0) - (w - w_0) + (z_0 - w_0)| \geq | |z_0 - w_0| - |(z - z_0) - (w - w_0)||[/tex]

Now, [itex]|z_0 - w_0|[/itex] is just a positive constant (call it [itex]c[/itex]), so the minimization of the right hand side is easy. The task is to mimimize

[tex]|c - |(z - z_0) - (w - w_0)||[/tex]

If there are [itex]z,w[/itex] that make this zero, then the minimum is zero. Otherwise, the minimum is achieved by maximizing [itex]|(z - z_0) - (w - w_0)|[/itex], which you can easily do by using the triangle inequality again.

By the way, this procedure will give you a lower bound. You still have to justify why there are [itex]z,w[/itex] that achieve the lower bound. (Think: under what condition does the triangle inequality become an equality?)




Thanks!! Although i would like to find a simpler way to do it as well.
 
  • #10
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Ok by drawing a picture the shortest distance between Z and W is coming out too be[tex]\sqrt{}74[/tex]-[tex]\sqrt{}13[/tex]-3=1.9996

I did this by calculating W by [tex]\sqrt{}74[/tex]- 1 and Z by [tex]\sqrt{}13[/tex] + 2, then subtracted the distance W from Z to get its smallest value.

I would be greatly obliged if someone can check my method, or better yet do the whole question.


Thanks in advance
 
  • #11
LCKurtz
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hmm..I get this part. You are saying that Z and W lie on the circumference of the circle with center 3,2 and radius 2 and center 7,5 with radius 1 right?

I cannot get beyond this point. The question is asking for the minimum value of |z-w| so this means that both Z and W should be at maximum distance from the origin?

Why would you think that? It asks for the minimum distance between z and w, nothing about how far from the origin they are. As I said in my first post, the points closest to each other on the circles will be on the line joining the centers. Draw a picture. Look at it. It is simple geometry. You can do it in your head.
 
  • #12
jbunniii
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Another thing you can do to simplify the problem is to translate and rotate the points so that one of them is at the origin and the other one is on the positive x axis.

The question remains the same: what is the minimum distance between two points given that they have to live within disks of a certain radius around the points. Rotating and translating the picture will not change the answer, but it can make it easier to picture (and compute) the answer.
 

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