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Norm of a Complex Number

  1. Jan 26, 2013 #1
    This question might be elementary:
    If the norm of two complex numbers is equal, can we deduce that the two complex numbers are equal.
    I know in ℝ we can just look at this as an absolute value, but what about ℂ?

    So mainly:

    let |z| = |w|*|r| can we say → z = w*r ?

  2. jcsd
  3. Jan 26, 2013 #2


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    It can be violated with real numbers, and every real number is a complex number.

    The other direction is true, of course.
  4. Jan 26, 2013 #3
    Yes because square roots are multiplicative, but your statement above is false on the following grounds.

    assuming w and r are complex you would have to say that:
    |z| = |w|*|r| we can say → -z = w*r
    Last edited: Jan 27, 2013
  5. Jan 26, 2013 #4


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    That's a strange question. You appear to know that if |a|= |b| for real numbers, then it does NOT follow that a= b but are you thinking that with complex numbers we might not have that ambiguity? In fact, for complex numbers the situation is much worse!

    In the real numbers, if |a|= 1 then a can be either 1 or -1. In the complex numbers there are an infinite number of possible values for a. There exist an infinite number of complex numbers, a, such that |a|= 1.
    Last edited by a moderator: Jan 27, 2013
  6. Jan 26, 2013 #5
    Complex numbers, much like vectors, are quantities defined by both a modulus (norm) AND an argument (direction). An infinity of complex numbers share the same norm, but have different arguments. The other way around is also true.

    Therefore, |z| = |w| does not imply z = w.
  7. Jan 26, 2013 #6


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    Draw a circle in the complex plane, centered at 0, with radius R. Every point on that circle has norm equal to R. Thus, except for the R = 0 case, there are infinitely many points with the same norm.
  8. Jan 27, 2013 #7
    lol...indeed it is a strange question. I just needed to brush up on my linear algebra a little and apply the vector normalization to get an equality. I don't know why this idea crossed my mind.

    Thank you all for the great clarifications.
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