locus question (1 Viewer)

[Lurch]

If you are given a complex No. z=t+1/t where t=r(cosa+isina) given a=(pi/4) and r varies and asked to draw the locus of the point P on an argand diagram

I get to the point :
x=(r+1/r) times 1/(sqrt(2))
y=(r-1/r) times 1/(sqrt(2))

now what I don't get is how do I go from here to the final step of getting the answer x^2 - y^2=2 how do they make the leap?

 

[CM_Tutor]

You have x = (r + 1 / r) / sqrt(2) _____ (1)
and y = (r - 1 / r) / sqrt(2) _____ (2)

(1) + (2): x + y = [(r + 1 / r) + (r - 1 / r)] / sqrt(2) = r * sqrt(2) _____ (A)
(1) - (2): x - y = [(r + 1 / r) - (r - 1 / r)] / sqrt(2) = sqrt(2) / r _____ (B)

(A) * (B) (x + y)(x - y) = r * sqrt(2) * sqrt(2) / r
x² - y² = 2

 

[Lurch]

I understand the above but my next question is why are we doing x+y, then x-y and then multiplying the two. Like what is the reason why we did that?

 

[KeypadSDM]

Intuition.

 

[CM_Tutor]

Originally posted by Lurch
I understand the above but my next question is why are we doing x+y, then x-y and then multiplying the two. Like what is the reason why we did that?

KeypadSDM is right, my maths instincts told me that was the fastest way to solve the problem.

A more general approach - you need to get the relationship between x and y, so you need to eliminate r. So, you could take one of the equations, make r the subject, and substitue this into the other, and then tidy up. This will work, it's just much messier.

What I did was to look at the equations, and note we had an A + B and an A - B. Thus, adding gives 2A and subtracting gives 2B. In this case, A = r and B = 1 / r, so multiplying will eliminate r, as required. By doing this, I avoid having to find r on its own, which is good as I really don't care what it is anyway, I only care about the relationship between x and y. Is this clearer?

 

[Lurch]

thanks CM; That clarifies it
much appreciated..