# Possibly Difficult Graphing Problem

Thread Starter

#### MrAl

Joined Jun 17, 2014
9,757
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.

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#### WBahn

Joined Mar 31, 2012
28,156
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.

#### wayneh

Joined Sep 9, 2010
17,201
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.

#### WBahn

Joined Mar 31, 2012
28,156
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.
At first blush sounds like another case of mathematically illiterate reporters -- plotting complex relations has been being done for a few centuries now.

#### wayneh

Joined Sep 9, 2010
17,201
No, it was a clever approach if somewhat limited. I've looked but can't find it now. Nuts, I hate when that happens.

#### WBahn

Joined Mar 31, 2012
28,156
No, it was a clever approach if somewhat limited. I've looked but can't find it now. Nuts, I hate when that happens.
Would be interesting to see if you track it down.

Who did it get a lot of attention from?

#### wayneh

Joined Sep 9, 2010
17,201
I'll keep looking but so far it's very frustrating. Maybe it didn't get as much attention as I had thought.

#### Tesla23

Joined May 10, 2009
524
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.

View attachment 108041
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

Thread Starter

#### MrAl

Joined Jun 17, 2014
9,757
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.
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)?

Thread Starter

#### MrAl

Joined Jun 17, 2014
9,757
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

View attachment 108081
Hello there,

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

#### Tesla23

Joined May 10, 2009
524
Hello there,

That also looks like an interesting approach also, but could you show how you arrived at those two equation parts?
$$z^2-2z = (z-1)^2 - 1$$

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

Thread Starter

#### MrAl

Joined Jun 17, 2014
9,757
$$z^2-2z = (z-1)^2 - 1$$

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

#### WBahn

Joined Mar 31, 2012
28,156
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)?
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).

Thread Starter

#### MrAl

Joined Jun 17, 2014
9,757
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).
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.

#### WBahn

Joined Mar 31, 2012
28,156
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.

Thread Starter

#### MrAl

Joined Jun 17, 2014
9,757
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).
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.

#### WBahn

Joined Mar 31, 2012
28,156
$$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.

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Thread Starter

#### MrAl

Joined Jun 17, 2014
9,757
$$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.
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.

#### WBahn

Joined Mar 31, 2012
28,156
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.
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).

#### WBahn

Joined Mar 31, 2012
28,156
I just plotted it and it agrees with Tesla23's result: