Complex variables

[Steven G]

Characterize geometrically the following sets.
1)$$\dfrac{|z-3|}{|z-5|} = 1$$2) $$\dfrac{|z-3|}{|z-5|} = \dfrac{\sqrt{2}}{2}$$
For the first one (and 2nd one) they are just asking to describe all complex numbers where the magnitude of Z-3 = the magnitude of Z-5? So a=4 and b can be anything?


Next set of problems.
Use complex numbers to characterize geometrically the following sets.
1)$$\epsilon_1$$ ={M in R^2 / AM=BM}, where A and B are two points of R^2
2)$$\epsilon_2$$ ={M in R^2 / AM= 3BM}, where A and B are two points of R^2

To be honest I am not sure what it means when they say AM=BM. Is AM the length of line from point A to point B??


Sorry about posting two different types of problems in one post. SK, just deal with it.

 

[joshuaa]

show what you got, and where you are stuck, then we can help you

 

[Steven G]

show what you got, and where you are stuck, then we can help you

I did tell my answer for the first question and how I got it. Is it correct?

 

[Dr.Peterson]

Characterize geometrically the following sets.
1)$$\dfrac{|z-3|}{|z-5|} = 1$$2) $$\dfrac{|z-3|}{|z-5|} = \dfrac{\sqrt{2}}{2}$$
For the first one (and 2nd one) they are just asking to describe all complex numbers where the magnitude of Z-3 = the magnitude of Z-5? So a=4 and b can be anything?

You didn't define a and b! I imagine you mean, z = 4 + bi, for any real b. It's the perpendicular bisector, right?

For the second, you have the locus of points with a fixed ratio of distances from two given points. (That could be all they want.) If you don't recall what that is (I'm not sure myself), you can solve algebraically. I'd start by changing it to $$\dfrac{|z-3|^2}{|z-5|^2} = \dfrac{1}{2}$$. Then I'd let [MATH]z = x + iy[/MATH].

Use complex numbers to characterize geometrically the following sets.
1)$$\epsilon_1$$ ={M in R^2 / AM=BM}, where A and B are two points of R^2
2)$$\epsilon_2$$ ={M in R^2 / AM= 3BM}, where A and B are two points of R^2

To be honest I am not sure what it means when they say AM=BM. Is AM the length of line from point A to point B??

Apparently they are using "/" where I'd use "|" for "such that"; I'd say that AM means the distance from A to M, perhaps better called |AM| or m(AM) or d(AM) or whatever, depending on the book.

Do you see how similar these are to the first pair?

 

[Steven G]

I realized a few minutes ago that I failed to define a and b. Ouch!

 

[joshuaa]

we know [MATH]z = x + iy[/MATH]
for (a)

the condition will be satisfied when [MATH]x = 4[/MATH] and [MATH]y = 0[/MATH]
since the value of [MATH]y[/MATH] does not affect the condition, you can extend [MATH]y[/MATH] to have all values


for (b)

the condition will be satisfied when [MATH]x = \frac{15}{4}[/MATH] and [MATH]y = \frac{\sqrt{7}}{4}[/MATH] or [MATH]y = \frac{-\sqrt{7}}{4}[/MATH]

 

[Steven G]

You didn't define a and b! I imagine you mean, z = 4 + bi, for any real b. It's the perpendicular bisector, right?

For the second, you have the locus of points with a fixed ratio of distances from two given points. (That could be all they want.) If you don't recall what that is (I'm not sure myself), you can solve algebraically. I'd start by changing it to $$\dfrac{|z-3|^2}{|z-5|^2} = \dfrac{1}{2}$$. Then I'd let [MATH]z = x + iy[/MATH].


Apparently they are using "/" where I'd use "|" for "such that"; I'd say that AM means the distance from A to M, perhaps better called |AM| or m(AM) or d(AM) or whatever, depending on the book.

Do you see how similar these are to the first pair?

Yes, based on the 1st set of problems. |AM| would have been more appropriate.
Thanks!

 

[nasi112]

we know [MATH]z = x + iy[/MATH]
for (a)

the condition will be satisfied when [MATH]x = 4[/MATH] and [MATH]y = 0[/MATH]
since the value of [MATH]y[/MATH] does not affect the condition, you can extend [MATH]y[/MATH] to have all values


for (b)

the condition will be satisfied when [MATH]x = \frac{15}{4}[/MATH] and [MATH]y = \frac{\sqrt{7}}{4}[/MATH] or [MATH]y = \frac{-\sqrt{7}}{4}[/MATH]

Please let the student have a chance to DISCOVER the solution

 

[Dr.Peterson]

we know [MATH]z = x + iy[/MATH]
for (a)
the condition will be satisfied when [MATH]x = 4[/MATH] and [MATH]y = 0[/MATH]
since the value of [MATH]y[/MATH] does not affect the condition, you can extend [MATH]y[/MATH] to have all values

for (b)
the condition will be satisfied when [MATH]x = \frac{15}{4}[/MATH] and [MATH]y = \frac{\sqrt{7}}{4}[/MATH] or [MATH]y = \frac{-\sqrt{7}}{4}[/MATH]

Please let the student have a chance to DISCOVER the solution

It's okay. That's not really the solution. And hopefully Jomo knows that.

The solution to each is a set (line or curve) and is not found by finding just one point. (And if you want a point, there's a much easier one to find.)

 

[joshuaa]

Please let the student have a chance to DISCOVER the solution

lol

 

[joshuaa]

It's okay. That's not really the solution. And hopefully Jomo knows that.

The solution to each is a set (line or curve) and is not found by finding just one point. (And if you want a point, there's a much easier one to find.)

Peterson, you're kidding right?

of course I know

[MATH]\frac{|z - 3|}{|z - 5|} = 1[/MATH]
represents a vertical line and all points there satisfies the condition

And i also know

[MATH]\frac{|z-3|}{|z-5|} = \frac{\sqrt{2}}{2}[/MATH]
represents a circle and all points on the circle satisfies the condition

And i know

Jomo was joking to ask that question

--------there are impossible things to happen

one of them

Jomo needs a solution? Yes, if it was a Joke.

 

[pka]

Characterize geometrically the following sets.
1)$$\dfrac{|z-3|}{|z-5|} = 1$$2) $$\dfrac{|z-3|}{|z-5|} = \dfrac{\sqrt{2}}{2}$$

Following the directions, note that the complex plane $\mathbb{C}$ is a metric space.
The metric is the distance function. Thus $|z-3|$ is the distance $z$ is from $3$
So it seems the question is asking for the set of all points in $\mathbb{C}$ that are equally distant from $3~\&~5$.
That is the set $\{z\in\mathbb{C} : \Re(z)=4\}$ which is the perpendicular bisector of the interval $[3,5]$ SEE HERE

 

[Steven G]

Following the directions, note that the complex plane $\mathbb{C}$ is a metric space.
The metric is the distance function. Thus $|z-3|$ is the distance $z$ is from $3$
So it seems the question is asking for the set of all points in $\mathbb{C}$ that are equally distant from $3~\&~5$.
That is the set $\{z\in\mathbb{C} : \Re(z)=4\}$ which is the perpendicular bisector of the interval $[3,5]$ SEE HERE

Yes, this is how I was thinking about it!

 

[Dr.Peterson]

Yes, this is how I was thinking about it!

The second question is the interesting one; I recalled that the set of points with a fixed ratio of distance from two points was a circle, but it's not one of those things that are drilled into you (like the perpendicular bisector), so I had to prove it to be sure. I presume you've done so by now, or already had.

By the way, is it true that you were joking in asking the question? I prefer to take people seriously rather than laugh at them and then find they were serious.

 

[Steven G]

Peterson, you're kidding right?

of course I know

[MATH]\frac{|z - 3|}{|z - 5|} = 1[/MATH]
represents a vertical line and all points there satisfies the condition

And i also know

[MATH]\frac{|z-3|}{|z-5|} = \frac{\sqrt{2}}{2}[/MATH]
represents a circle and all points on the circle satisfies the condition

And i know

Jomo was joking to ask that question

--------there are impossible things to happen

one of them

Jomo needs a solution? Yes, if it was a Joke.

Do you realize that I am the weakest helper on this forum? Yes, from time to time I have questions. And yes, this was a serious post. Sometimes I need help thinking, but once I get it I have it nicely.

 

[Steven G]

Hmm. now you folks have me thinking. Why when we changed the 1 to 1/rt(2) do we now get a circle? I won't describe how I feel now.

 

[Steven G]

Here is what I understand. We want the set of all points in the complex plane such that when you multiply the distance between Z and 3 by sqrt(2) you get the same distance as the distance between Z and 5.
OK, it does work out to a circle.

 

[Dr.Peterson]

Hmm. now you folks have me thinking. Why when we changed the 1 to 1/rt(2) do we now get a circle? I won't describe how I feel now.

That's why I second-guessed my recollection that it was a circle. It's surprising. And I'm not coming up with a nice geometrical explanation, though I'm sure there is one.

My particular thought was: How could changing the focus-and-directrix form of an ellipse to two points, rather than a line and a point, again give a conic section? But it does.

 

[LCKurtz]

This thread surprised me too. Especially the fact that the value of the ratio constant doesn't matter. After a bit of searching I found this interesting site on the topic:

Locus of Points in a Given Ratio to Two Points

www.cut-the-knot.org

 

[joshuaa]

for (1)

i have chosen the two points [MATH]A(3, 0)[/MATH] and [MATH]B(0, 2)[/MATH]
then

[MATH]\frac{|z - 3|}{|z - 2i|} = 1[/MATH]
will give us a straight line

let [MATH]M[/MATH] be any point on the straight line

to find the locus of point [MATH]M[/MATH] that has a ratio of 1 ([MATH]AM = BM[/MATH]) for any distance, we just have to solve

[MATH]\frac{|z - 3|}{|z - 2i|} = 1[/MATH]
then

[MATH](x - 3)^2 + y^2 = x^2 + (y - 2)^2[/MATH]
[MATH]x^2 - 6x + 9 + y^2 = x^2 + y^2 - 4y + 4[/MATH]
[MATH]-6x + 9 = -4y + 4[/MATH]
so, geometrically, we have this straight line

[MATH]y = \frac{3}{2}x - \frac{5}{4}[/MATH]
now, you can test any point [MATH]M[/MATH] in this straight line, it will have the same distance to points [MATH]A[/MATH] and [MATH]B[/MATH] ([MATH]AM = BM[/MATH])

now let us rearrange the equation of the straight line to get the general complex form

[MATH]-6x + 4y + 5 = 0[/MATH] (it looks nicer without fractions)

let [MATH]a = -6, b = 4, c = 5[/MATH]
then the general form is

[MATH](a + bi)\overline{z} + (a - bi)z + 2c = 0[/MATH]
[MATH](-6 + 4i)\overline{z} + (-6 - 4i)z + 10 = 0[/MATH]

for (b)

we will use the same two point, [MATH]A(3, 0)[/MATH] and [MATH]B(0, 2)[/MATH]
then

[MATH]\frac{|z - 3|}{|z - 2i|} = 3[/MATH]
this will give us a circle

[MATH]x^2 - 6x + 9 + y^2 = 9x^2 + 9y^2 - 36y + 36[/MATH]
[MATH]8x^2 + 6x + 8y^2 - 36y + 27 = 0[/MATH]
[MATH](x + \frac{3}{8})^2 + (y - \frac{9}{4})^2 = (\sqrt{\frac{117}{64}})^2[/MATH]
this is a circle with center [MATH](\frac{-3}{8}, \frac{9}{4})[/MATH] and radius [MATH]\sqrt{\frac{117}{64}}[/MATH]
you can test the distance of any point [MATH]M[/MATH] on the circle to the points [MATH]A[/MATH] and [MATH]B[/MATH], it will be a distance ratio ([MATH]AM = 3BM[/MATH])

the complex form of this circle is

[MATH]|z + \frac{3}{8} - \frac{9}{4}i| = \sqrt{\frac{117}{64}}[/MATH]
And the general form

[MATH]z\overline{z} + (\frac{3}{8} + \frac{9}{4}i)z + (\frac{3}{8} - \frac{9}{4}i)\overline{z} + \frac{27}{8} = 0[/MATH]