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Math Help - Complex Numbers

  1. #1
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    Complex Numbers

    I'm having difficulty working with complex numbers. In rectangular and polar forms.

    I have:

    1,94 - j0,43 = \frac{1}{\frac{1}{X} + \frac{1}{2-j2} + \frac{1}{10 +j0}}


    How can I calculate X in rectangular or polar?
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  2. #2
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    Quote Originally Posted by Apprentice123 View Post
    I'm having difficulty working with complex numbers. In rectangular and polar forms.

    I have:

    1,94 - j0,43 = \frac{1}{\frac{1}{X} + \frac{1}{2-j2} + \frac{1}{10 +j0}}


    How can I calculate X in rectangular or polar?
    First you could make some effort to render this unambiguous. What do you mean by j0, and what of j2?

    You can always as a first step resort to algebra to rearrange this into the form X=..


    CB
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  3. #3
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    Quote Originally Posted by Apprentice123 View Post
    I'm having difficulty working with complex numbers. In rectangular and polar forms.

    I have:

    1,94 - j0,43 = \frac{1}{\frac{1}{X} + \frac{1}{2-j2} + \frac{1}{10 +j0}}


    How can I calculate X in rectangular or polar?

    The basic algebraic manipulation required to solve this equation is no different to what you would need for an equation with real number.

    Clearly you will want to reciprocate something. so polar form of a complex number is handy because for z = R e^{i \theta} \ \  \ \frac{1}{z} = R^{-1} e^{-i \theta}

    Post again if you are still having troubles.

    Bobak
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  4. #4
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    Quote Originally Posted by CaptainBlack View Post
    First you could make some effort to render this unambiguous. What do you mean by j0, and what of j2?

    You can always as a first step resort to algebra to rearrange this into the form X=..


    CB
    In some countries, a comma is used for a decimal point. Also, in electrical engineering, the symbol j is used for \sqrt{-1} to avoid ambiguity ( i is used for something else).


    I assume this is actually

    1.94 + 0.43j = \frac{1}{\frac{1}{X} + \frac{1}{2 - 2j} + \frac{1}{10 + 0j}}.
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    Quote Originally Posted by Prove It View Post
    In some countries, a comma is used for a decimal point. Also, in electrical engineering, the symbol j is used for \sqrt{-1} to avoid ambiguity ( i is used for something else).
    Strange as it may seem but I know that backwards and forwards anagramatised.

    I assume this is actually

    1.94 + 0.43j = \frac{1}{\frac{1}{X} + \frac{1}{2 - 2j} + \frac{1}{10 + 0j}}.
    That only slightly helps, but it does not tell me what the OP intended but only your guess, well I can guess myself but refuse to.

    CB
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  6. #6
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    Quote Originally Posted by Prove It View Post
    In some countries, a comma is used for a decimal point. Also, in electrical engineering, the symbol j is used for \sqrt{-1} to avoid ambiguity ( i is used for something else).
    Strange as it may seem but I know that forwards and backwards anagrammatised.

    I assume this is actually

    1.94 + 0.43j = \frac{1}{\frac{1}{X} + \frac{1}{2 - 2j} + \frac{1}{10 + 0j}}.
    That only slightly helps since but it does not tell me what the OP intended but only your guess, well I can guess myself but refuse to.

    My request that the OP clarifies what they mean is for their benefit not mine. It is important that people learn to express themselves clearly without ambiguity (unless intended)

    CB
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  7. #7
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    Quote Originally Posted by Prove It View Post
    In some countries, a comma is used for a decimal point. Also, in electrical engineering, the symbol j is used for \sqrt{-1} to avoid ambiguity ( i is used for something else).


    I assume this is actually

    1.94 + 0.43j = \frac{1}{\frac{1}{X} + \frac{1}{2 - 2j} + \frac{1}{10 + 0j}}.
    You're right, j is the imaginary part
    I know not how I calculate the division of complex numbers:

    \frac{1}{2 - 2j} and \frac{1}{10 + 0j}

    0j Just to indicate that the imaginary part is 0
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  8. #8
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    {\frac{1}{z} = \frac{{\overline z }}{{\left| z \right|^2 }}}

    So \frac{1}{2-2i}=\frac{2+2i}{8}.
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  9. #9
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    Quote Originally Posted by Plato View Post
    {\frac{1}{z} = \frac{{\overline z }}{{\left| z \right|^2 }}}

    So \frac{1}{2-2i}=\frac{2+2i}{8}.
    I find:

    1,94 + 0,43i = \frac{1}{\frac{1}{x} + 0,35 + 0,25i}

    x = (1,94 + 0,43i)(0,35 + 0,25i) = 0,5715+ 0,6355i

    Is correct ?
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