Help with simple complex no question from terry lee (2 Viewers)

[chocolatelatte]

- sketch the locus of z if |z|<=1 and |(1-z)/(1+z)|<=sqrt3
now i understand the answers in saying -sqrt3<=tan(theta/2)<=sqrt3 and then sketching the angles from that
But I just don't understand why you can't split the modulus and bring it to the other side so you get |(1-z)|<=sqrt3|1+z| which gives you (x+2)^2+y^2<=3 and then sketching that..... isn't that a valid method

thank youuuu DD

 

[chocolatelatte]

oh yes , that was a typing error. but does that mean you can work out the answer like that instead of using the tan(theta/2) thingo, which seems to yield a very different looking sketch when drawn?
thanks!!

 

[chocolatelatte]

also on a totally unrelated note ,
how would you find the horizontal asymptotes of y=(e^x-e^-x)/(e^x+e^-x) ?
THANK YOU SO MUCH

 

[Squar3root]

also on a totally unrelated note ,
how would you find the horizontal asymptotes of y=(e^x-e^-x)/(e^x+e^-x) ?
THANK YOU SO MUCH

that's arctan(x) and has horizontal asymptotes y = +/- 1

 

[InteGrand]

that's arctan(x) and has horizontal asymptotes y = +/- 1

* tanh(x)

 

[braintic]

that's arctan(x) and has horizontal asymptotes y = +/- 1

He is not supposed to know that, and it doesn't really develop his understanding.

Answer:

As x approaches infinity, the e^(-x) and x terms become insignificant wrt e^x.
So the function is basically equal to (e^x)/(e^x) = 1.
The numerator is smaller than the denominator due to the fact it has -e^(-x), so the function approaches 1 from below.

As x approaches minus infinity, the e^x and x terms become insignificant wrt e^(-x).
So the function is basically equal to [-e^(-x)] /(e^x) = -1.
The numerator is closer to zero than the denominator due to the fact that its terms have opposite signs so partially cancel, so the function approaches -1 from above.

 

[InteGrand]

also on a totally unrelated note ,
how would you find the horizontal asymptotes of y=(e^x-e^-x)/(e^x+e^-x) ?
THANK YOU SO MUCH

To find horizonatal asymptotes, we always look at the behaviour as .

As , the terms approach 0 and , so y = 1 is a horizontal asymptote. The numerator is always just less than the denominator as you can see for positive x, so y approaches 1 from below.

You can see that the function is an odd function (if you replace all x's with -x, you get negative the original function, so the behaviour as x approaches negative infinity will be that y approaches -1 (from above), so y = -1 is also a horizontal asymptote.

Edit: beaten to it by braintic

 

[FrankXie]

oh yes , that was a typing error. but does that mean you can work out the answer like that instead of using the tan(theta/2) thingo, which seems to yield a very different looking sketch when drawn?
thanks!!

 

[chocolatelatte]

thanks guys

I have another question from terry lee which is confusing me :'(
The complex numbers a=3+i , b=-5i , c=2+2i are represented by points A,B and C respectively. Find distances AB, AC and angle BAC , and hence, the area of triangle ABC.

OK, so I'm find with finding the lengths of AB and AC. But when it comes to angle BAC, I find it to be pi - tan^-1(3) , while the answer is simply tan^-1 (3). I don't understand why mine is the supplementary angle to that since my working seems correct.

My working:
- i drew the diagram
- found arg (BA) and arg (AC), and then found pi-arg(AC) because I wanted the angle in the triangle formed by B, A and the intercept line CA makes with the axis. then I found angle BAC by adding these two angles (exterior angle)
but that gives me pi - tan^-1 (3) - the obtuse angle (while the answer is just tan^-1(3))

Their working:
angleBAC = arg(b-a/c-a)=arg(-3-6i/-1+i)=arg((-3-6i)(-1-i)/2)=arg(-3+9i/2) which gives the acute angle BAC to be tan^-1(3)
I follow their working but I don't understand how angle BAC is the result of that particular arg subtraction.

HELP PLEASE.
THANKS

 

[FrankXie]

how can arg(-3+9i/2) be acute?

 

[porcupinetree]

May I ask, why does the locus have to be contained within the unit circle? I understand that it must be outside of (x+2)^2 + y^2 = 3, but I don't understand why the unit circle is involved. Thanks

 

[InteGrand]

May I ask, why does the locus have to be contained within the unit circle? I understand that it must be outside of (x+2)^2 + y^2 = 3, but I don't understand why the unit circle is involved. Thanks

Because we are given that .

 

[WhoStanLeee]

The Terry Lee solution is slightly different to the intersection of the two circle inequalities. Terry Lee's solution is the region bounded by the unit circle, y=√3x and y=-√3x. If these methods are equally valid, how come they bear different results?

 

[porcupinetree]

Because we are given that .

Oh of course, how silly of me to miss that

 

[FrankXie]

The Terry Lee solution is slightly different to the intersection of the two circle inequalities. Terry Lee's solution is the region bounded by the unit circle, y=√3x and y=-√3x. If these methods are equally valid, how come they bear different results?

well I don't have the Terry Lee book so I can't check out the solution. but if the information I got from the thread is precise and complete, then I am sure that the solution from the book was incorrect, the correct solution should be the intersection of the two circle inequalities, but not the intersection of the circle and the two radii of the circle. It is obvious that points like z=-1/4, z=-1/4+i/4 should be included in the locus because they satisfy both inequalities, while the solution from the book rejects these points.