FP1

Maths and statistics discussion, revision, exam and homework help.

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  1. 041087's Avatar
    • Full Member
    • Posts: 90
    FP1
    Hey, just started self-teaching fp1 yesterday and still on chapter one -complex numbers and this is the first exercise im having trouble with, and i haven't a clue how to do the following question

    Find the real numbers x and y , given that : 1/(x+iy) = 3-2i?

    This is my following steps:
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    i have no ideal what is the next step:mad:
    if you could help me by showing the working that would be much appreciated
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  2. notnek's Avatar
    • TSR Demigod
    • Location: Bangkok, Thailand
    Re: FP1
    Rationalising the algebraic part is not needed here and makes things more complicated than they need to be.


    The Comparing real and imaginary parts method would work better by doing this:

    \displaystyle \frac{1}{x+iy} = 3-2i \implies 1=3x+3iy-2ix+2y \implies 1+0i = (3x+2y)+(3y-2x)i

    You should find that comparing real and imaginary parts from here is much easier.


    Alternatively, you could change the equation to

    \displaystyle \frac{1}{3-2i}=x+iy

    and then rationalise the left hand side. This is probably the quickest method.
    Last edited by notnek; 30-06-2012 at 05:41.
  3. Micky76's Avatar
    • Full Member
    • Posts: 144
    Re: FP1
    (Original post by 041087)
    Hey, just started self-teaching fp1 yesterday and still on chapter one -complex numbers and this is the first exercise im having trouble with, and i haven't a clue how to do the following question

    Find the real numbers x and y , given that : 1/(x+iy) = 3-2i?

    This is my following steps:
    Click image for larger version. 

Name:	photo (1).jpg 
Views:	68 
Size:	350.6 KB 
ID:	160577
    Click image for larger version. 

Name:	photo.JPG 
Views:	44 
Size:	218.1 KB 
ID:	160578
    i have no ideal what is the next step:mad:
    if you could help me by showing the working that would be much appreciated
    Can I just ask what textbook do you study from?
  4. 041087's Avatar
    • Full Member
    • Posts: 90
    Re: FP1
    (Original post by Micky76)
    Can I just ask what textbook do you study from?
    Click image for larger version. 

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  5. 041087's Avatar
    • Full Member
    • Posts: 90
    Re: FP1
    (Original post by notnek)
    Rationalising the algebraic part is not needed here and makes things more complicated than they need to be.


    The Comparing real and imaginary parts method would work better by doing this:

    \displaystyle \frac{1}{x+iy} = 3-2i \implies 1=3x+3iy-2ix+2y \implies 1+0i = (3x+2y)+(3y-2x)i

    You should find that comparing real and imaginary parts from here is much easier.


    Alternatively, you could change the equation to

    \displaystyle \frac{1}{3-2i}=x+iy

    and then rationalise the left hand side. This is probably the quickest method.
    thanks
  6. aznkid66's Avatar
    • Exalted and Worshipped Member
    • Posts: 922
    Re: FP1
    The long+weird way:

    (I was bored. Also, don't hate on me detouring around all arithmetic that might include fractions. :P I hate fractions with a passion.)

    \dfrac{x}{x^2+y^2}=3

    \dfrac{y}{x^2+y^2}=2

    Note: From here, you can go straight to 2x=3y . However, note that

    \dfrac{y}{x^2+y^2}+1=3

    \dfrac{y}{x^2+y^2}+1=\dfrac{x}{x  ^2+y^2}

    y+x^2+y^2=x

    x^2+y^2-x+y=0

    Note: Quadratics are nice. ^-^

    Spoiler:
    Show
    \dfrac{2y}{x^2+y^2}=4=\dfrac{x}{  x^2+y^2}+1

    2y=x+x^2+y^2

    x^2+y^2+x-2y=0

    x^2+y^2+x-2y=x^2+y^2-x+y

    x-2y=-x+y


    2x=3y

    4x^2=9y^2

    x^2+y^2-x+y=0

    2x^2+2y^2-2x+2y=0

    \dfrac{4x^2}{2}+2y^2-2x+2y=0

    \dfrac{9y^2}{2}+2y^2-3y+2y=0

    9y^2+4y^2-6y+4y=0

    13y^2-2y=0

    y(13y-2)=0

    y=\dfrac{2}{13}

    x=\dfrac{3}{13}

    Note: y=x=0 is extraneous because of the denominator(s).
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