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Help with complex numbers(locus) and hyperbolic functions

  1. Feb 4, 2009 #1
    1. The problem statement, all variables and given/known data
    Question(1) : Find the Cartesian equation of Re[ z - i / z + 1 ] = 0. If the locus is a circle, give its radius and the coordinates of its center.

    3. The attempt at a solution
    Workings : So I attempted to solve the problem and my workings are as below

    ... Since Re = Real part,

    Let z = x + iy

    Re[ x + iy - i / x + iy + 1 ] = 0

    x/x+1 = 0

    x = 0

    Right here I am assuming that the locus is at all the points of the line x=0.



    NEXT, to obtain the radius and the coordinates of the center of the locus(circle),

    [modulus] z - i/ z + 1 [modulus] = 0
    [modulus] z - i [modulus] = 0
    [modulus] x + iy - i [modulus]= 0
    square root[ (x - 0)2 + (y - 1)2 ] = 0
    (x - 0)2 + (y - 1)2 = 0

    Therefore, the radius of the circle is 0 and the coordinates of the center is ( 0, 1 )

    1. The problem statement, all variables and given/known data
    Question (2) : Obtain all the real solutions of the following equation: 9 sinh 4x - 82 sinh 3x + 9 sinh 2x = 0 . Show all your derivations.

    3. The attempt at a solution
    I first subsituted [ ex - e-x / 2 ] into all the sinh available in the equation with their specific value of x.

    9[ e4x - e-4x / 2 ] - 82[ e3x - e-3x / 2 ] + 9[e2x - e-2x/ 2 ] = 0

    I multiply the whole equation by 2 and decided to multiply the integer outside of each boxes,

    9e4x - 9e-4x- 82e3x + 82e-3x + 9e2x - 9e-2x = 0

    Then I tried to separate e4x to e4 . ex and regroup the ones with ex and e-x

    [ 9e4- 82e3 + 9e2 ] ex = [ 9e4 - 82e3 + 9e2 ] e-x

    Then I multiplied both sides with ex

    [ex]2 = 1

    ex= square root + of 1 (chosed only +ve value as the question mentioned about real solutions)

    Then I applied ln

    x ln e = ln square root of 1

    x = 0

    Any comments would be a great help and much appreciated. Thank you in advance and have a nice day.

    Regards
    Charles
     
    Last edited: Feb 4, 2009
  2. jcsd
  3. Feb 4, 2009 #2

    Office_Shredder

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    For the first one: Re(a/b) =/= Re(a)/Re(b) in general

    e.g. (1+i)/(2+i) = (1+i)(2-i)/5 = (3+i)/5 which has real part 3/5, not 1/2

    Furthermore, a line is never a circle
     
  4. Feb 4, 2009 #3
    Thanks for the heads up Office_Shredder

    Well then I think I should then make my first step by multiplying a conjugate of the denominator

    (z-i)/(z+1) = 0
    [( x2 + x + y2 - y ) + i ( y-x-1)] / x2 + 2x + 1 + y2 = 0

    Then i tried completing the squares (with the Real part = 0) to get the equation for circle (x-a)2 + (y-b)2 = r2

    I end up with

    [x2 + x + (-1/2)2 ] - (-1/2)2 + [y2 - y + (1/2)2] - (1/2)2 = 0

    [x -(-)(-1/2)]2 + [y-1/2]2 = (-1/2)2 + (1/2)2

    I end up with my radius being 0 and coordinates as (1/2,1/2). There is no circles with a 0 radius so Im still lost. Maybe completing the squares wasnt the right step?
     
    Last edited: Feb 4, 2009
  5. Feb 4, 2009 #4
    I'm confused with your notation...

    Do you mean: Re[ (z-i)/(z+1)] or Re[z - (i/z) + 1] or maybe even something else...

    I'm guessing you mean the first one. In that case, you're on the right track. You made a small error in finishing the square of the x-dependent part (a minus sign). Furthermore, the radius of your circle is not zero. The squared numbers on the right hand side do not add up to 0...
     
  6. Feb 4, 2009 #5
    xepma I apologise for the confusion that I have caused. Yes I meant (z-i)/(z+1) = 0.

    After you have pointed out my error, this is what I've got and please let me know if it is the correct way of doing it.

    [x - (-1/2)]2 + [y-1/2]2 = (-1/2)2 + (1/2)2

    As for the radius, if I were to sum up the right hand side, the result would be (1/4) + (1/4) = 1/2

    From the general equation of the circle (x-a)2 + (y-b)2 = r2

    the value of r should be squared. By summing up the right hand side after squaring each fraction, it would not produce a squared value. Please correct me if I am wrong.

    Thank you for your help.
     
  7. Feb 4, 2009 #6
    Since when is the square root of 1/2 not a number? :)
     
  8. Feb 4, 2009 #7
    Oh.. Hahahha! Alright, I guess Im done with Question 1 with the help from you and Office_Shredder. Thank you both again.

    Would love some help with Question 2 though. :)
     
  9. Feb 4, 2009 #8

    nrqed

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    Watch out!
    [tex] e^{4x} [/tex] is not equal to [tex] e^4 e^x [/tex] !
     
  10. Feb 4, 2009 #9
    Thanks nrqed! Was looking for someone to ask whether if I could do that. Since I cant separate the 4 and the x, what do you reckon I should do for my next step? Or maybe even my first step was wrong?
     
  11. Feb 4, 2009 #10

    Office_Shredder

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    If you write [tex]e^{4x} = (e^{x})^4[/tex] etc., then you can make the substitution y=ex to get a quartic equation in y. You should be able to reduce that to something you can solve for y for, and then get x from y.
     
  12. Feb 5, 2009 #11
    Thank you Office_Shredder for your kind help. I tried to subsitute y=ex and got the quartic equation which unfortunately Im not very familiar with.

    9y4 - 9y-4 - 82y3 + 82y-3 + 9y2 - 9y-2 = 0

    I didnt know what to do with the negative powers so I mulitplied the equation with y4 to eliminate the negatives

    9y8 - 82y7 + 9y6 - 9y2 + 82y - 9 = 0

    which got me lost :(
     
    Last edited: Feb 5, 2009
  13. Feb 5, 2009 #12

    Office_Shredder

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    Look at how you can pair stuff up:

    9y8 - 9y2
    82y - 82y7
    9y6-9

    So if y6 = 1, then you get that each of those terms is zero, and hence you've found six (complex) solutions to that polynomial
     
  14. Feb 7, 2009 #13
    Thank you office_shredder for your kind reply again. So from y6 = 1. Which also means y =1?

    Therefore from y = ex

    1 = ex
    ln 1 = x ln e
    x = 0?
     
  15. Feb 7, 2009 #14

    Office_Shredder

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    No. Any polynomial of degree n has exactly n complex roots. So there are actually five complex roots of y6=1. Of course, these won't be relevant since you're only searching for real solutions to the original equation, but it guides us to realizing that there also are 8 roots of the degree eight polynomial. We find the other two by factoring: We now know y6-1 divides 9y8 - 82y'7 + 9y6 - 9y2 + 82y - 9 so we factor to get:

    9*(y6-1)*(y2 - 9y + 1) = 0

    So you should be able to find the other two roots from here
     
  16. Feb 7, 2009 #15
    Yes I got the other two roots from your guide office_shredder. Really appreciate all the help from you guys. This is a really educative and helpful forum. Thank you so much guys. :)
     
  17. Mar 4, 2009 #16

    HallsofIvy

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    How did you get this? If z= x+ iy then
    [tex]Re\frac{z-i}{z+ 1}= Re\frac{x+ i(y-1)}{(x+1)+ iy}[/tex]
    multiplying numerator and denominator by (x+1)- iy, we get
    [tex]\frac{x^2+ x- (y^2+y)}{(x+1)^2+ y^2}+ i (.... )[/tex]

    You just said that you were assuming it was a line. How did it become a circle in the next sentence?

     
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