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Homework Help: Product of two complex numbers = 0

  1. Aug 28, 2012 #1
    1. The problem statement, all variables and given/known data
    Let z and w be two complex numbers such that zw = 0. Show either z = 0 or w = 0.
    Hint: try working in exponential polar form

    2. Relevant equations
    z = re

    3. The attempt at a solution
    I have absolutely no clue.
  2. jcsd
  3. Aug 28, 2012 #2
    What happens if you write two complex numbers in polar form, then multiply them together and require the result is equal to zero? So [itex] z=re^{i\theta}, w = qe^{i\phi} [/itex],zw = ?
  4. Aug 28, 2012 #3


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    Have you tried that yet? What does the question look like in exponential polar form?
  5. Aug 28, 2012 #4
    so [itex]zw = re^{i\theta} \times qe^{i\phi}[/itex]

    and since neither [itex]re^{i\theta}[/itex] or [itex]qe^{i\phi}[/itex] can be negative, either [itex]z = 0[/itex] or [itex]w = 0[/itex]. Is this correct?
  6. Aug 28, 2012 #5
    Not in the slightest. Does negative mean anything when it comes to complex numbers? Also, when imaginary exponents are involved, [itex]e^{a+b\cdot i}[/itex] can be anything. Basic identities about exponents, one of which should be screaming at you, still apply.
  7. Aug 28, 2012 #6


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    Correct, except what does them being negative have to do with anything?

    Edit: what I mean is, correct answer, totally wrong reasoning.
  8. Aug 28, 2012 #7
    Right, whoops. So either z = 0 or w = 0 because of the zero-product property? i.e. "if ab = 0, then either a = 0 or b = 0"? Why is working in exponential polar form at all necessary or even helpful then? :S

    [itex]rqe^{i(\theta + \phi)}[/itex] ?

    and then since [itex]e^{i(\theta + \phi)} > 0[/itex] either [itex]r=0[/itex] or [itex]q=0[/itex], therefore either [itex]z = 0[/itex] or [itex]w = 0[/itex] ?
  9. Aug 28, 2012 #8
    It certainly can be smaller than 0, but can it be exactly 0?
  10. Aug 28, 2012 #9
    Oh yeah, so it should be [itex]e^{i(\theta + \phi)} \not= 0[/itex] then?

    I think it's all correct other than that, right?
  11. Aug 28, 2012 #10
    The point is that we already know it works for reals. Let's assume it works for reals. Now prove it works for complex numbers. Relying on what you're trying to prove won't get you anywhere. If you really do need a hint,

    Ever heard of the absolute value function and how it works on complex numbers?
  12. Aug 28, 2012 #11

    Ray Vickson

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    You can also do it directly: if z = x + i*y and w = u + i*v with x,y,u,v real, then
    zw = (xu - vy) + i*(xv + yu), so we need xu = vy and xv = -uy. You can consider several (easy) cases such as u =/= 0 vs, u = 0, and so conclude that in the first case you *must have* x = y = 0, while in the second case you have either v = 0 (so you are done) or else you have x = y = 0.

    Last edited: Aug 28, 2012
  13. Aug 28, 2012 #12
    I'm a little confused, could you explain how this is incorrect?

    [itex]zw = 0[/itex]
    [itex]zw = re^{i\theta} \times qe^{i\phi}[/itex]
    [itex]zw = rqe^{i(\theta + \phi)}[/itex]
    [itex]e^{i(\theta + \phi)} \not= 0[/itex]
    [itex]\therefore r=0[/itex] or [itex]q=0[/itex]
    [itex]\therefore z = 0[/itex] or [itex]w = 0[/itex]

    I guess you could also do it this way to eliminate [itex]i[/itex] altogether so that you're only working with reals? This doesn't involve exponential polar form though so I'm not sure if this is how we're expected to solve it.

    [itex]z = a + ib[/itex]
    [itex]w = c + id[/itex]
    [itex]|zw| = |z||w| = 0[/itex]
    [itex]\sqrt{a^2 + b^2} \times \sqrt{c^2 + d^2} = 0[/itex]
    [itex]\therefore a[/itex] and [itex]b = 0[/itex]
    or [itex]c[/itex] and [itex]d = 0[/itex]
    [itex]\therefore z = 0[/itex] or [itex]w = 0[/itex]
    Last edited: Aug 29, 2012
  14. Aug 28, 2012 #13
    If [itex]z = 0[/itex], you are done. Suppose [itex]z \neq 0[/itex]. Then, the multiplicative inverse of z exists. But, then:
    w = \frac{1}{z} \cdot 0 = ?
  15. Aug 29, 2012 #14
    This has confused me even more. :S

    also I cringe every time I re-read this:
    I was very very tired. :frown:
  16. Aug 29, 2012 #15
    It's this step. To justify it, we need to rely on the very identity we are trying to prove, resulting in circular reasoning. Hint: find some way to introduce an absolute value into this equation.
  17. Aug 29, 2012 #16


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    Because this is what you are being asked to prove! That is, you are asked to prove that the "zero product property" holds for complex numbers as well as real numbers.

    NO! This not true. For example, [itex]e^{i\pi}= -1[/itex]. But, as you were told before, there is no "less than" or "larger than" relation on the complex numbers.

    It is, however, true that [itex]e^{ix}[/itex], for x real, is never 0.
  18. Aug 29, 2012 #17
    Uh... I understand what you're saying about what's wrong with my proof but I really don't know what to do even with the hint. :confused:

    I mean I know [itex]r = |z|[/itex] and [itex]q = |w|[/itex] so [itex]rq=|zw|[/itex] but I don't know where that gets me.
  19. Aug 29, 2012 #18
    Try taking the absolute value of both sides.
  20. Aug 29, 2012 #19

    Ray Vickson

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    I gave you an alternative suggestion in my previous posting. Have you looked at it? Have you tried to complete the argument I gave?

  21. Aug 29, 2012 #20


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    It's not circular reasoning. It's safe to assume here that for real numbers x and y, that xy=0 implies that x=0 or y=0, and this is what the proof is relying on as r and q are both real numbers. The OP is being asked to show the same property holds for complex numbers.
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