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Math Help - Complex D Name F D Number

  1. #1
    mathsjerk101
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    Complex D Name F D Number

    QUESTION 1)
    (1+i)(1+2i)(1+3i)........(1+ni)=A+iB
    show dat : tan^-1(1)+tan^-1(1/2)+tan^-1(1/3)..........tan^-1(1/n)=tan^-1(B/A)
    QUESTION 2)
    if x=Cos a +i Sin a;y=Cos b+i Sin b;z=Cos c +i Sin c,prove dat
    (1)1/x+1/y+1/z=0 whenevr x+y+z=0
    (2)xy+yz+zx=0 weneva x+y+z=0
    (3)xy/z+z/xy=2cos(a+b+c)
    ok basicaly al i noe is u shud use complex numbr properties..
    so plzzzz help
    Last edited by mathsjerk101; January 15th 2007 at 01:32 AM. Reason: just makin it more clear d question.......
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  2. #2
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    Quote Originally Posted by mathsjerk101 View Post
    fine here is d question:
    QUESTION 1)
    (1+i)(1+2i)(1+3i)........(1+ni)=A+iB
    When you have,
    1+ai
    You can write,
    \sqrt{1+a^2} \left( \frac{1}{\sqrt{1+a^2}}+\frac{a}{\sqrt{1+a^2}} i \right)
    Thus,
    sqrt{1+a^2} (\cos \theta_a + i \sin \theta_a )
    Where,
    \theta_a = \tan^{-1} a

    Thus, we have,
    \sqrt{1+1^2}(\cos \theta_1 +i\sin \theta_1)\cdot ... \cdot \sqrt{1+n^2}(\cos \theta_n+i\sin \theta_n )
    Thus,
    \sqrt{(1+1^2)\cdot ... \cdot (1+n^2)} (\cos (\theta_1 +... + \theta_n )+i\sin (\theta_1 + ... + \theta_n) )

    Thus,
    A=\sqrt{(1+1^2)...(1+n^2)}\cdot \cos (\tan^{-1} 1+...+\tan^{-1} n)

    B=\sqrt{(1+1^2)....(1+n^2)}\cdot \sin (\tan^{-1} 1+...+\tan^{-1} n)
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  3. #3
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    Quote Originally Posted by mathsjerk101 View Post
    QUESTION 2)
    if x=CiS a,y=CiS b,z=CiS c,prove dat
    (1)1/x+1/y+1/z=0 whenevr x+y+z=0
    (2)xy+yz+zx=0 weneva x+y+z=0
    (3)xy/z+z/xy=2cos(a+b+c)
    I do not understand the problem.

    Look at #2)
    x=1,y=2,z=-3
    It does not work.
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by ThePerfectHacker View Post
    When you have,
    1+ai
    You can write,
    \sqrt{1+a^2} \left( \frac{1}{\sqrt{1+a^2}}+\frac{a}{\sqrt{1+a^2}} i \right)
    Thus,
    sqrt{1+a^2} (\cos \theta_a + i \sin \theta_a )
    Where,
    \theta_a = \tan^{-1} a

    Thus, we have,
    \sqrt{1+1^2}(\cos \theta_1 +i\sin \theta_1)\cdot ... \cdot \sqrt{1+n^2}(\cos \theta_n+i\sin \theta_n )
    Thus,
    \sqrt{(1+1^2)\cdot ... \cdot (1+n^2)} (\cos (\theta_1 +... + \theta_n )+i\sin (\theta_1 + ... + \theta_n) )

    Thus,
    A=\sqrt{(1+1^2)...(1+n^2)}\cdot \cos (\tan^{-1} 1+...+\tan^{-1} n)

    B=\sqrt{(1+1^2)....(1+n^2)}\cdot \sin (\tan^{-1} 1+...+\tan^{-1} n)
    You do realise that this was not the question don't you?

    The question was to show that:

    atan(1)+atan(1/2)+atan(1/3)+...+atan(1/n)=atan(B/A)

    RonL
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