Author Topic: subsets of C (Read 4054 times) Tweet Share

enpassant · « **on:** March 30, 2008, 09:30:17 pm »

Find the cartesian equations for (i) |z+i|+|z-i|=1 and (ii) |z+i|-|z-i|=3. :idiot2:

ed_saifa · « **Reply #1 on:** March 30, 2008, 09:37:04 pm »

let z= x + yi

Toothpaste · « **Reply #2 on:** March 30, 2008, 09:49:36 pm »

(i)
$z=x+yi$

Substitute z = x+yi into your equation:
$|x+yi+i|+|x+yi-i|=1$

Group like terms:
$|x+(y+1)i|+|x+(y-1)i|=1$

$\sqrt{x^2+(y+1)^2}+\sqrt{x^2+(y-1)^2}=1$

Bring one of the square roots over to the right hand side.
$\sqrt{x^2+(y+1)^2}=1-\sqrt{x^2+(y-1)^2}$

Square both sides - "undo the square roots"
$(\sqrt{x^2+(y+1)^2})^2=(1-\sqrt{x^2+(y-1)^2})^2$

... Quadratic expansion on the RHS by the way.
$x^2+(y+1)^2=1-2\sqrt{x^2+(y-1)^2}+x^2+(y-1)^2$

Move it around. Expand $(y-1)^2$ and $(y+1)^2$
$y^2+2y+1=1-2\sqrt{x^2+(y-1)^2}+y^2-2y+1$

$y^2+2y-1=-2\sqrt{x^2+(y-1)^2}+y^2-2y$

$4y-1=-2\sqrt{x^2+(y-1)^2}$

Square both sides.
$(4y-1)^2=(-2\sqrt{x^2+(y-1)^2})^2$

Expand.
$16y^2-8y+1=4(x^2+(y-1)^2)$

Expand $(y-1)^2$ .
$16y^2-8y+1=4(x^2+y^2-2y+1)$

$16y^2-8y+1=4x^2+4y^2-8y+4$

$12y^2-4x^2=3$

Divide through by 3 to get 1 on the RHS.

$4y^2-\frac{4x^2}{3}=1$

See coblin's post for domain.
$|z+i|+|z-i|=1$ can't exist because Ahmad says so.

(ii) I didn't show all the steps for this one because when you get the hang of it you'll be able to get to the answer faster.

$|z+i|-|z-i|=3$

$z=x+yi$

$|x+yi+i|-|x+yi-i|=3$

$\sqrt{x^2+(y+1)^2}=3+\sqrt{x^2+(y-1)^2}$

$x^2+(y+1)^2=9+6\sqrt{x^2+(y-1)^2}+x^2+(y-1)^2$

$y^2+2y+1=9+6\sqrt{x^2+(y-1)^2}+y^2-2y+1$

$4y-9=6\sqrt{x^2+(y-1)^2}$

$16y^2-72y+81=36(x^2+(y-1)^2)$

$16y^2-72y+81=36x^2+36y^2-72y+36$

$-20y^2+45=36x^2$

$\frac{4x^2}{5}+\frac{4y^2}{9}=1$

enpassant · « **Reply #3 on:** March 30, 2008, 10:45:37 pm »

Not making sense!

Toothpaste · « **Reply #4 on:** March 30, 2008, 10:46:40 pm »

Quote from: enpassant on March 30, 2008, 10:45:37 pm

Not making sense!

All of it, or certain bits?

I'll put in annotations to show you what I did.

Collin Li · « **Reply #5 on:** March 30, 2008, 10:49:16 pm »

Quote from: enpassant on March 30, 2008, 10:45:37 pm

Not making sense!

It's just pure algebraic hackwork. There are a few milestones.

1. Change the modulus into the algebraic expression (surds)

Now, we want to simplify the relationship between $x$ and $y$ :

2. Move one of the surds to the other side so that it is easier to square both sides (which is critical to simplifying the $x$ and $y$ terms).

Now we can square both sides.

3. One surd will be left over after simplifying. Isolate it and square both sides again.

4. Simplify a final time.

That is the strategy, and that is what Toothpick has done

evaporade · « **Reply #6 on:** March 30, 2008, 10:56:14 pm »

I agree it made no sense. Sketch and see.

Toothpaste · « **Reply #7 on:** March 30, 2008, 10:59:15 pm »

Quote from: evaporade on March 30, 2008, 10:56:14 pm

I agree it made no sense. Sketch and see.

Ahmad pointed that out too. (part i)

Collin Li · « **Reply #8 on:** March 30, 2008, 11:02:41 pm »

Quote from: evaporade on March 30, 2008, 10:56:14 pm

I agree it made no sense. Sketch and see.

I haven't checked the working, but I have noticed she hasn't put down any domains and ranges. There are certain restrictions.

From this line of working:

Quote

$4y-1=-2\sqrt{x^2+(y-1)^2}$

Since the RHS is negative:

$\implies 4y-1 > 0 \implies y > \frac{1}{4}$

This may be why the function might nominally exist in cartesian form, but may actually not show up, or make sense.

Ahmad · « **Reply #9 on:** March 30, 2008, 11:11:52 pm »

$1 = \sqrt{x^2 + (y+1)^2} + \sqrt{x^2 + (y-1)^2} \ge |y+1| + |y-1| \ge |(y+1) - (y-1)| = 2$ contradiction.

enpassant · « **Reply #10 on:** March 31, 2008, 08:45:48 am »

What do I say for (i) and (ii)?

Mao · « **Reply #11 on:** March 31, 2008, 07:55:55 pm »

the algebra ought to be "trusted", but if you need to imagine how it'd look, think of it this way:

$|z-c|$ means the "distance of any point z from a particular point c"

so going to (i), it basically says "distance from -i + distance from i =1", which cannot make sense, because the minimum distance between i and -i is 2

for (ii) a bit harder to imagine, "distance from -i - distance from i = 3", or "distance from -i = distance from i + 3", so naturally you have a hyperbola (well, you would think, but it doesnt actually work)

evaporade · « **Reply #12 on:** March 31, 2008, 10:10:10 pm »

are you sure of (ii)?

Mao · « **Reply #13 on:** April 01, 2008, 12:51:09 pm »

=S this is very wierd, i see nothing wrong with toothpick's workings, but:

given $\frac{4x^2}{5}+\frac{4y^2}{9}=1$

it can be shown that the verticle axis is 3 units long, with its maximum pt at (0,1.5)

so the cartesian coordinate translates to $z=\frac{3}{2}i$

when we plug this into the original equation:

$|\frac{3}{2}i+i | - | \frac{3}{2}i-i |$

$\implies |\frac{5}{2}i |-|\frac{1}{2}i|$

$\implies \frac{5}{2}-\frac{1}{2}=2\neq 3$

=S

Mao · « **Reply #14 on:** April 01, 2008, 01:46:28 pm »

a not very pedantic proof (nor handwritten)

what I mean to say is:

if $|z+i|-|z-i|=3$ is true:
$\exists z: |z+i|=3+|z-i|$

or, the distance from $-i$ is 3 units more than the distance from $i$

If we were to link $i$ and $-i$ to $z$ , then we'd have a triangle, with $|z+i|$ being the longest edge, and has the same magnitude as $|z-i|+3$

the two smaller sides would have the magnitudes of $|z-i|$ and 2.

But this is evidently false, as the sum of magnitudes of the smaller sides need to add up to larger than the longest side (or you just have 3 lines enclosing absolutely nothing),

therefore, so long as $z$ can be represented on the Argand diagram ( $z \in C$ ), the relation $|z+i|-|z-i|=3$ is false

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enpassant

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