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Dividing complex numbers

  1. Jun 6, 2014 #1
    The problem


    [itex]H(e^{j0.2\pi}) = {\frac{1 - 1.25e^{-j0.2\pi}}{1 - 0.8e^{-j0.2\pi}}}[/itex]

    Solves to [itex]H(e^{j0.2\pi}) = 1.25e^{j0.210\pi}[/itex]

    Attempts

    I'm really not sure how to get that answer, but I've tried a number of different approaches
    • Multiplying by complex conjugate
    • Multiplying by real conjugate
    • Multiplying by complex & real conjugates
    • Algebraic division
    • Converting to polar form

    None of which give me anything close to the answer. I feel like there must be an easier way to solve this. Is there an approach that I don't know of? Thanks
     
  2. jcsd
  3. Jun 6, 2014 #2

    jbunniii

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    After some fiddling, I was able to show that the magnitude is indeed equal to 1.25. Here is a suggestion that should put you on the right path.

    If we put ##a = 1.25##, ##b = 0.8##, ##c = \cos(0.2\pi)##, and ##s = \sin(0.2\pi)##, you can express ##|H(e^{j0.2\pi})|^2## as
    $$\frac{(1 - ac)^2 + (as)^2}{(1 - bc)^2 + (bs)^2}$$
    Simplify as much as you can, using the facts that ##c^2 + s^2 = 1## (always true), and (useful observation for this problem) ##ab = 1##. After a fair bit of work, you should be able to reduce the answer to ##a^2##, which is what you want.

    After you get the magnitude, the angle may be more straightforward. (I haven't tried it yet.)
     
  4. Jun 6, 2014 #3

    SteamKing

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    1. Use Euler's formula to express the exponentials in the numerator and denominator in rectangular form.
    2. In rectangular form, carry out the indicated subtractions.
    3. Convert the numerator and denominator back into polar form.
    4. Carry out the indicated division operation.

    Manipulating complex numbers is a pain because addition/subtraction work best with rectangular form and
    multiplication/division work best in polar form.

    If you've done calculations already, please post them. You may have made an arithmetic mistake.
     
  5. Jun 6, 2014 #4

    jbunniii

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    Ha, after doing all that algebra for my first post, I just saw a shortcut which allows you to do it in just a couple of lines.

    Hint: pull a factor out of the numerator which will result in a fraction of a very nice form. You have to exploit the fact that ##1.25 \times 0.8 = 1##, so this isn't a trick which will work in general.
     
  6. Jun 6, 2014 #5

    LCKurtz

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    I'm guessing that angle bit is going to be easier said than done. I've been tinkering around recently with trying to really understand impedance (just for something to do, my only AC circuits course was in 1961 and I'm not an EE). This looks a lot like the calculations necessary to show the phasor transformation$$
    H:~\cos(\omega t + \phi)\rightarrow e^{i\phi}$$is additive. In fact, it makes me wonder if this problem didn't come from a circuit impedance problem and could be worked indirectly another way. Understand, I'm just musing out loud here. I could be barking up the wrong tree entirely and I'm likely not being very helpful. :frown:
     
  7. Jun 6, 2014 #6

    SteamKing

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    'j' is favored by EEs as the imaginary unit so that it doesn't get confused with 'i' for current, so your supposition is probably correct.
     
  8. Jun 6, 2014 #7

    AlephZero

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    The usual way to simplify this sort of fraction is make the denominator real, by multiplying top and bottom by the complex conjugate of the denominator, ##1 + 0.8e^{-j0.2\pi}##.

    If that was what you meant by "Multiplying by complex conjugate", you probably make a mistake with the arithmetic.

    But I don't know what you mean by "Multiplying by real conjugate" or "Multiplying by complex & real conjugates", so maybe tried to do something different (and wrong).
     
  9. Jun 7, 2014 #8

    LCKurtz

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    In case anybody cares, the exact value of the angle is$$
    \phi = \arctan\left(\frac {(1071+369\sqrt 5)\sqrt{10-2\sqrt 5}} {5736}
    \right)=.2097653805\pi$$
     
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