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Thread: Real Number

  1. #1
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    Post Real Number

    Good evening all of our class is having a online debate with the following question, any feedback would be appreciated:



    In the Real Number realm,
    ab = 0 a = 0 or b = 0
    Is the same theorem true in the Complex Number realm? (Why or why not?)
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  2. #2
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    Quote Originally Posted by AlgebraicallyChallenged View Post
    Good evening all of our class is having a online debate with the following question, any feedback would be appreciated:



    In the Real Number realm,
    ab = 0 a = 0 or b = 0
    Is the same theorem true in the Complex Number realm? (Why or why not?)
    In the complex number 'realm', for a complex number to be considered equal to $\displaystyle 0$, then both it's real and imaginary parts must be equal to zero!

    $\displaystyle z_1 = a+ib $
    $\displaystyle z_2 = c+id $

    $\displaystyle z_1 \times z_2 = ac+adi+cbi-bd = (ac-bd) + i(ad+cb) = 0 + 0i $

    Hence $\displaystyle ac-bd = 0 $ and $\displaystyle ad+cb=0 $ for $\displaystyle z_1 \times z_2=0$

    So let's look at the case where $\displaystyle z_1 = a+ib =0+0i$. For this $\displaystyle a=0$ and $\displaystyle b = 0$, which means that $\displaystyle ac-bd = 0(c)-(0)d = 0$. So the first equation is satisfied. $\displaystyle (0)d+(0)c = 0$. So the 2nd is satisfied.

    Now look at the case where $\displaystyle z_2 = c+id =0+0i$. For this c=0 and d = 0, which means that $\displaystyle ac-bd = 0(a)-(0)b = 0$. So the first equation is satisfied. $\displaystyle (0)b+(0)b = 0$. So the 2nd is satisfied.
    Last edited by Mush; Jan 6th 2009 at 08:46 PM.
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  3. #3
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    Ahhh I see what you were asking. But the question remains, are there any combinations of $\displaystyle z_1$ and $\displaystyle z_2$ for which their product is zero, but for which neither of them are 0?!

    Well. Let's try this then. For a complex number to be zero, then it's modulus must be zero, yes? So let's find the modulus of our product.

    $\displaystyle |z_1.z_2| = |(ac-bd) + i(ad+bc)|$

    $\displaystyle = \sqrt{(ac-bd)^2 + (ad+bc)^2}$

    $\displaystyle = \sqrt{(ac)^2-2abcd+(bd)^2 + (ad)^2+2abcd+(bc)^2}$

    $\displaystyle = \sqrt{(ac)^2+(bd)^2 + (ad)^2+(bc)^2} = 0$

    Clearly, if this is zero, then the expression inside the square root is zero!

    $\displaystyle = (ac)^2+(bd)^2 + (ad)^2+(bc)^2 = 0$

    $\displaystyle = a^2(c^2+d^2)+b^2(c^2+d^2) = 0$

    $\displaystyle = (a^2+b^2)(c^2+d^2) = 0$

    Hence, by the logic of REAL numbers (a, b, c and d must all be real, remember!) either:

    $\displaystyle a^2+b^2 = 0$

    OR

    $\displaystyle c^2+d^2 = 0$

    For these to be true:

    $\displaystyle a = \pm \sqrt{-b^2} $

    or

    $\displaystyle c = \pm \sqrt{-d^2} $

    $\displaystyle b^2$ and $\displaystyle d^2$ are always positive numbers, which means that the solutions to these 2 equations for all values of b and d are not real solutions, but purely imaginary solutions. And by the definition of the complex numbers $\displaystyle z_1$ and $\displaystyle z_2$, a, b, c and d must be REAL numbers. Hence there are no two non-zero complex numbers whose product is zero, and hence if $\displaystyle z_1.z_2 = 0 +0i$ then $\displaystyle z_1 = 0+0i $ or $\displaystyle z_2 = 0+0i $
    Last edited by Mush; Jan 7th 2009 at 06:54 AM.
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