# Possibly Difficult Graphing Problem

Discussion in 'Math' started by MrAl, Jun 21, 2016.

1. ### MrAl Thread Starter Distinguished Member

Jun 17, 2014
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Hello there,

I was recently asked a question which is shown in the attachment.
Anyone want to take a stab at this?
I dont want to post any possible solutions yet because i was hoping someone could come up with a better one that is simpler or more direct.
The main problem is to do the graph as in the attachment question part 1.2 after doing part 1.1.

Last edited: Jun 21, 2016
2. ### WBahn Moderator

Mar 31, 2012
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What's wrong with

w = (1-2x) + i2(x-1)y

Where -1 <= x <= +1 and y = +/- sqrt(x² - 1)

At least that's what I see by inspection -- could be overlooking something or have screwed something up.

3. ### wayneh Expert

Sep 9, 2010
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Someone recently made news plotting functions containing imaginary terms. Got a lot of attention. Is this related to that? Sorry, too lazy to look it up myself.

4. ### WBahn Moderator

Mar 31, 2012
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At first blush sounds like another case of mathematically illiterate reporters -- plotting complex relations has been being done for a few centuries now.

5. ### wayneh Expert

Sep 9, 2010
12,379
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No, it was a clever approach if somewhat limited. I've looked but can't find it now. Nuts, I hate when that happens.

6. ### WBahn Moderator

Mar 31, 2012
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Would be interesting to see if you track it down.

Who did it get a lot of attention from?

7. ### wayneh Expert

Sep 9, 2010
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I'll keep looking but so far it's very frustrating. Maybe it didn't get as much attention as I had thought.

8. ### Tesla23 Active Member

May 10, 2009
323
67
for $w = (z-1)^2-1$ the circle $|z| = 1$ is mapped as follows:
1. $(z-1)$ shifts it left by 1 to be centred on (-1,0)
2. squaring the points on the circle produces a heart shape which is easily sketched (just transform a few points)
3. subtracting the 1 on the RHS shifts it left by 1

The net results is shown here using Wolfram Alpha:

parametric plot (re((1-exp(i*t))^2)-1, im((1-exp(i*t))^2)), t=0..2pi

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9. ### MrAl Thread Starter Distinguished Member

Jun 17, 2014
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Hi again,

That's looks like an interesting approach, simpler than my original approach. The main question was after that graph the circle on the W plane as in part 1.2, so how would you go about graphing that (the W plane is in terms of u and v, u is the horizontal axis and v the vertical)?

10. ### MrAl Thread Starter Distinguished Member

Jun 17, 2014
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515
Hello there,

That also looks like an interesting approach also, but could you show how you arrived at those two equation parts?

11. ### Tesla23 Active Member

May 10, 2009
323
67
$z^2-2z = (z-1)^2 - 1$

$z=x+iy \text{ with } x^2+y^2=1 \text{ results in the circle } |z|=1$

12. ### MrAl Thread Starter Distinguished Member

Jun 17, 2014
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Hello,

Yes, but you basically just said exactly what you said in your first post. What you might do is show the relationship between the exponentials and how you got them from the final 'circle'. This is for the benefit of another reader too
Thanks much.

Side note: The resulting graph looks like a cardioid, but i dont think it is one unless we stretch the definition a little. A test might help, but i'll leave that for later.

13. ### WBahn Moderator

Mar 31, 2012
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w = (1-2x) + i2(x-1)y
Where -1 <= x <= +1 and y = +/- sqrt(x² - 1)

u = (1-2x)
v = 2(x-1)y = +/- 2(x-1)sqrt(x² - 1)

Now just plot u vs. v as you walk x from -1 to +1 noting that for each value of x you get one value for u and two values for v (and thus you know that the plot is symmetric about the u axis).

14. ### MrAl Thread Starter Distinguished Member

Jun 17, 2014
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Hello again,

Thanks, that looks reasonable. I had solved it analytically but i felt that was too much work and the expression too complicated to be practical.

15. ### WBahn Moderator

Mar 31, 2012
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Another way is to note that since x varies from -1 to +1, u varies from +3 to -1.

Then you can solve for x in terms of u

x = (1-u)/2

and plug that into the equation for v

v = (+/-) 2( (1-u)/2 - 1 )sqrt(((1-u)/2)² - 1)

v = (+/-) 2(-(1+u)/2)sqrt( (1 - 2u + u² - 4) /4)

v = (+/-)(u+1)sqrt(u² - 2u - 3)/2

v = (+/-)(u+1)sqrt((u-3)(u+1))/2

I did all the manipulations in my head, so there may well be a mistake in here somewhere.

I don't think, for instance, that this is going to match what Tesla23 got because I have the domain going from -1 to +3, where his plot shows the domain extending down below -1 a bit.

If I set u=0 I see a problem right away. I think I see a problem with my original formulation. I don't have time to look at it further now, but I'll try later today.

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16. ### MrAl Thread Starter Distinguished Member

Jun 17, 2014
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Hi again,

Wait a minute, i think something is wrong.
u=1-2*t
v=2*(t-1)*sqrt(t^2-1)

With t=0, we get:
u=1
v=-2*j (plus and minus actually)

which cant be right.

With t=0.5 we get:
u=0
v=-0.86602540378444*j (plus and minus really)

and the true result when u=0 is v=2.5424 approximately (plus and minus).

Also, how did the circle x^2+y^2=1 fit into that solution?
In other words, it is u and v after x and y are solutions to that circle.

17. ### WBahn Moderator

Mar 31, 2012
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$
w \; = \; z^2 \, - \, 2z
z \; = \; x \, + \, iy
w \; = \; $$x \, + \, iy$$^2 \, - \, 2$$x \, + \, iy$$
w \; = \; x^2 \, + \, i2xy \, - \, y^2 \, - \, 2x \, - \, i2y
w \; = \; $$x^2 \, - \, 2x \, - \, y^2$$ \, + \, i$$2xy \, - \, 2y$$
$

I'm pretty sure I see where I messed up. In the real part, I think I had +y² which I then combined with x² to get 1.

Continuing on by adding and subtracting x² to the real part:

$
w \; = \; $$2x^2 \, - \, 2x \, - \, \(x^2 \, + \, y^2$$ \)\, + \, i2y$$x \, - \, 1$$
w \; = \; $$2x^2 \, - \, 2x \, - \, 1$$\, + \, i2y$$x \, - \, 1$$
$

We can now get rid of the y by using

$
y \; = \; \pm \sqrt{1 \, - \, x^2}
$

Giving us

$
w \; = \; $$2x^2 \, - \, 2x \, - \, 1$$\, \pm \, i2\sqrt{1 \, - \, x^2}$$x \, - \, 1$$
$

So, parametrically,

$
u \; = \; 2x^2 \, - \, 2x \, - \, 1
v \; = \; 2\sqrt{1 \, - \, x^2}$$x \, - \, 1$$
$

as x varies from -1 to +1.

We could then solve x in terms of u and substitute into v to get v in terms of u.

But now I really do need to get out of here. I'll come back to this later.

Last edited: Jun 22, 2016
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18. ### MrAl Thread Starter Distinguished Member

Jun 17, 2014
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Hello again,

Ok thanks much. And after that we have to (presumably) solve for x and y and plug into x^2+y^2=1
and solve that for u and v.
Unless of course you see it differently.

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19. ### WBahn Moderator

Mar 31, 2012
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The variable y is gone. The constraint placed by the equation for the circle was used to eliminate it from the parametric equations at the end of my last post (assuming I did it right this time).

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20. ### WBahn Moderator

Mar 31, 2012
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I just plotted it and it agrees with Tesla23's result:

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