complex No. Q (1 Viewer)

[Will Hunting]

Draw up a small Cartesian plane. Draw a ray at 45 degrees, it will be in the 1st quadrant, right?
1st quad, (x,y)=(+,+)
Its reflection will be -45 and in the 4th qudrant
4th quad reflection, (x,y)=(+,-)
Similar for 2nd and 3rd:
2nd quad: (x,y)=(-,+)
3rd quad relfection: (x,y)=(-,-)
So you apply this two the x and y components of each argument and can then remove the negative sign.

These are the principles upon which you've based your reasoning, however, do not illustrate the exact conditions upon which your proof relies, respective to the question. You have solved the problem assuming the half-ray emanating from the P(2,0) has 0 ≤ arg(z-2) ≤ π/2, then solved the problem again assuming the half-ray emanating from the same P(2,0) has π/2 ≤ arg(z-2) ≤ π.

Hence, you have not divided the plane into four quadrants, but rather into two regions,

Re(z) ≥ 2,
Re(z) < 2

(The interface of these two regions gives values that satify both cases)

Your proof isn't false, though it seems you may have overshot in terms of length, perhaps including something that is just as comprehensively accounted for in Sirius' or Queenie's method (It's like proving a = b, then proving -a = -b; the algebraic deductions of above methods, as well as the behaviour of the resultant graph, show everything that you've shown)

 

[Slidey]

1) I did not realise Sirius had already answered his question.
2) I did it that way because it was something new I wanted to try.
3) It's sketch, not prove.

 

[Will Hunting]

1) I did not realise Sirius had already answered his question

That's cool

2) I did it that way because it was something new I wanted to try.

Also cool, but be aware that your second case is superfluous for the purpose of answering the original question Observing the relationship between arg(z-2) and arg(z-3+2i) when the product, (z-2)[z-(3-2i)] is taken, and, additionally, noting that the graph is continuous for all x, your second case can be comfortably withdrawn.

3) It's sketch, not prove.

I was referring to your post and not the original question when I used this term. The original question is asking for the locus to be sketched, however, your post, viewed as independent of the question, can be considered reasoning toward a conclusion (i.e. the equation of the locus), or a proof.

 

[Slidey]

Also cool, but be aware that your second case is superfluous

How exactly could I not be aware of that, considering both cases returned the same result?

 

[Sirius Black]

thanx to u all-Will Hunting, Slide Rule & KFunk

I am really appreciated about ur guys' work on this question. It is actually good to see some other brilliant attempts, like the geometrical approach from Slide Rule...
anyway, thanx to u all and i am going to sketch my lovely hyperbola now

 

[Will Hunting]

How exactly could I not be aware of that, considering both cases returned the same result?

You can't be, unless you pay regard to the properties exhibited by the algebra, the trig and/or the graph itself! That way you would know its superfluity before you even wasted the time writing it! (not that it isn't tops to explore new things for yourself, which I don't, at all, discourage. All I'm about is how to make the process as speedy and efficient as can be. I'm a nut for quick kills and easy ways out )

 

[Slidey]

I just told you that I was aware of its superfuous nature; I knew why both cases were identical (a+bi=-x+yi has the same solution set as -a+bi=x+yi). You do NOT have to keep trying to explain why to me!

The primary reason I continued with case 2 was for consistency and because chronologically, I did case 2 first when writing the post (hence it made sense that when doing case 1, I do full working for it). It took all of 20 seconds more..

Have we beaten this dead horse enough?

Post script: I am not getting angry; frustrated, rather. Your addition to Queenie's solution was very nice. But this topic was over once KFunk wrapped up the range.

 

[Will Hunting]

Ha ha ha! Fluid, as ever, Slide! Way to go!

I just told you that I was aware of its superfuous nature

Doubtless, but I just told you that that awareness should have come before a prodigal step was set down, through other precursory, yet obviously neglected, observations.

Post script: I am getting angry. ROOOOOARRGGGHHHHH!!! Fee fie foe fum! I smell a Mathematical travesty on the wind!

 

[samwell]

Arg(z-2)+Arg(z-3+2i)=0
Arg(z-2) means that the line goes via the origin. so y2-y1/x2-x1 for the first case should equal the -ve the gradient of the second case.
y/x-2=-(y+2)/x+3
y(x+3)=-(x-2)(y+2)
xy+3y=-(xy+2x-2y-4)
xy+3y+xy+2x-2y-4=0
y+2xy+2x=4
I think this is the answer but the graph lookz really wrong

 

[Slidey]

Arg(z-2)+Arg(z-3+2i)=0
Arg(z-2) means that the line goes via the origin. so y2-y1/x2-x1 for the first case should equal the -ve the gradient of the second case.
y/x-2=-(y+2)/x+3
y(x+3)=-(x-2)(y+2)
xy+3y=-(xy+2x-2y-4)
xy+3y+xy+2x-2y-4=0
y+2xy+2x=4
I think this is the answer but the graph lookz really wrong

You should use your brackets better. Anything on the bottom should ALWAYS be in brackets unless it's just one variable or constant.

Anyway, your solution is equivalent to the one I gave (scroll up past Will_Hunting's bullshit ramblings), except you've made a sign error when substituting in the points. It should be x-3 not x+3.

But it's great that you've having an independent crack at it.

How does the graph look wrong, either way? You're in 4u, so you've obviously seen hyperbolae before (they have a number of forms). This hyperbola is your run of the mill version, just translated a bit. It has no oblique asymptotes (y=x+1/x), nor is it rotated by 45 degrees (x^2-y^2=r^2).

-5y+2xy+2x=4
2xy-5y=4-2x
y(2x-5)=-2(x-2)
y=-(2x-4)/(2x-5)
y=-(2x-5+1)/(2x-5)
y=-1 - 1/(2x-5)