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Complex Multiplication

  1. Jul 23, 2014 #1
    Is this statement true?

    [tex]

    \bf{Re}\{z_1 \times z_2\} = 0 \,\, and \,\, z_2 \ne 0 \Rightarrow z_1 = 0 \,\,\, where \, z_1; z_2 \in \rm{C}

    [/tex]

    Thanks.
     
  2. jcsd
  3. Jul 23, 2014 #2

    Mentallic

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    No. Can you find an example of where it won't work?

    Let [itex]z_1=a+bi; z_2=x+yi;[/itex] and choose x and y such that they're not both zero, and a,b are not both zero either.
     
  4. Jul 23, 2014 #3

    George Jones

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    What happens if the polar form of complex numbers is used instead of the rectangular form?
     
  5. Jul 23, 2014 #4
    That's what I'm getting at actually. In one text book it's presented in polar form and it says it's true in an attempt to apply Voltage Kirchhoff's law to sum of phasors.
     
  6. Jul 23, 2014 #5

    George Jones

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    Write ##z_1 = r_1 e^{i\theta_1}## and ##z_2 = r_2 e^{i\theta_2}##. Set ##0 = z_1 z_1##. What do you get?
     
  7. Jul 23, 2014 #6
    Oh man you are smart!!! Thanks.
     
  8. Jul 23, 2014 #7
    Sorry I forget something. The Real part of the multiplication is 0, which still does not imply one of the complex numbers is 0.
     
  9. Jul 23, 2014 #8

    George Jones

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    If ##z = re^{i\theta} = a + ib##, then ##z = 0## if and only if ##0 = r = \sqrt{a^2 + b^2}## if and only if ##0 = a = b##.
     
  10. Jul 23, 2014 #9

    Mentallic

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    I'm confused. Are we looking at the same thing here?

    Let [itex]z_1=z_2=1+i[/itex], then [itex]z_1z_2 = (1+i)^2 = 1+2i-1 = 2i[/itex], hence [itex]Re(z_1z_2)=0[/itex] which disproves the statement above because [itex]z_1,z_2\neq 0[/itex].
     
  11. Jul 23, 2014 #10

    George Jones

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    We are looking at the same thing, but one of us (me) isn't able to read. No matter how many times I read it, I couldn't see the ##\bf{Re}## in

    Even
    still didn't clue me in.

    Apparently not. Sorry seminum and Mentallic.

    I think polar form still helps (to show that the statement is incorrect).

    $$z_1 z_2 = r_1 r_2 e^{i \left( \theta_1 + \theta_2 \right)}$$

    This is purely imaginary for non-zero ##z_1## and ##z_1## if ##\theta_1 + \theta_2 = \pi/2 +n\pi##. Mentallic's example has ##\theta_1 = \theta_2 = \pi/4 ##. Another simple example is ##z_1 = 1## and ##z_2 = i##.
     
  12. Jul 24, 2014 #11
    Yes I could disprove the statement using rectangular form. However based on my understanding the following text says the opposite:

    Electric Circuits 9th Ed by Nilsson and Riedel
    Page 321, Kirchhoff's Law in the Frequency Domain
     
  13. Jul 24, 2014 #12

    George Jones

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    What if the line after (9.40),
    is changed to
     
  14. Jul 24, 2014 #13

    Curious3141

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    If you do the (simple) algebra in rectangular form, you'll quickly find that:

    $$\bf{Re}(z_1 \times z_2) = 0 \iff \bf{Re}(z_1)\bf{Re}(z_2) = \bf{Im}(z_1)\bf{Im}(z_2)$$


    and that's the only condition that needs to be satisfied. Mentallic's example is merely the special case with ##\bf{Re}(z_1) = \bf{Re}(z_2) = \bf{Im}(z_1) = \bf{Im}(z_2) = 1##.
     
  15. Jul 24, 2014 #14

    George Jones

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    Curious3141, the text has given the right result for the wrong reason. In more detail: the text has written (for ##z## independent of ##t##) ##0 = \bf{Re} \left( z e^{j\omega t} \right\}## implies that ##z=0.##

    $$0 = 2\bf{Re}\left\{ z e^{j\omega t} \right\} = z e^{j\omega t} + \left( z e^{j\omega t} \right)^* = z e^{j\omega t} + z^* e^{-j\omega t}$$

    What does ##t = 0 ## give?

    What does ##t = \pi/\left(2\omega\right) ## give?
     
  16. Jul 24, 2014 #15

    Curious3141

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    Hi George, I don't have the text, but going by what you've written, if we let ##z = re^{j\theta}##, then the condition implies that ##\bf{Re}(re^{j(\omega t + \theta)}) = 0 \implies r\cos({\omega t + \theta}) = 0 \implies r = 0 \space \rm{or} \space \omega t + \theta = \frac{\pi}{2} + 2k\pi##.

    Since the latter is excluded by independence between ##z \space \rm{and}\space t##, the only possibility is ##r=0 \implies z = 0##.
     
  17. Jul 24, 2014 #16

    George Jones

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    Yes.

    More explicitly, ##t=0## gives ##0=\bf{Re} \left\{ z\right\}##, and ##t = \pi/\left(2\omega\right)## gives ##0=\bf{Im} \left\{ z\right\}##.
     
  18. Jul 24, 2014 #17
    I was just looking around on pf and I came across this thinking I would understand it. I got lost on the first post :(
     
  19. Jul 24, 2014 #18
    Gotcha! Thanks.
     
  20. Jul 24, 2014 #19

    Mentallic

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    If you have two complex numbers,

    [tex]z_1=a+ib[/tex]

    [tex]z_2=x+iy[/tex]

    and multiply them, of course you get

    [tex]z_1z_2=(a+ib)(x+iy)=ax-by+i(ay+bx)[/tex]

    and [itex]Re(z_1z_2) = ax-by[/itex] is the real part of that result (just as [itex]Re(z_1)=a, Re(z_2)=x[/itex]).

    So the question is asking, if [itex]z_2\neq 0[/itex] which means that x and y aren't both zero (one of them can be zero) then are we forced to choose [itex]z_1=0[/itex] so that we can end up with [itex]Re(z_1z_2)=0[/itex]?

    The answer is no because as we can see, [itex]Re(z_1z_2)=ax-by=0[/itex] and for simplicity, if we choose x=y where x and y aren't zero, then we get [itex]ax-by=ax-bx=x(a-b)=0[/itex] which means we end up with a-b=0, a=b. So we don't need to choose [itex]z_1=0[/itex] to satisfy the result, we can choose [itex]z_1=a+ia[/itex] for any a that's non-zero. Also, if we didn't restrict x=y then we'd find other combinations that satisfy the result as well.

    Hence my example earlier with choosing [itex]z_1=z_2=1+i[/itex].
     
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