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Math Help - 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 0, then both it's real and imaginary parts must be equal to zero!

     z_1 = a+ib
     z_2 = c+id

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

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

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

    Now look at the case where  z_2 = c+id =0+0i. For this c=0 and d = 0, which means that ac-bd = 0(a)-(0)b = 0. So the first equation is satisfied. (0)b+(0)b = 0. So the 2nd is satisfied.
    Last edited by Mush; January 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  z_1 and  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.

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

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

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

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

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

     = (ac)^2+(bd)^2 + (ad)^2+(bc)^2 = 0

     = a^2(c^2+d^2)+b^2(c^2+d^2) = 0

     = (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:

     a^2+b^2 = 0

    OR

     c^2+d^2 = 0

    For these to be true:

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

    or

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

    b^2 and 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  z_1 and  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  z_1.z_2 = 0 +0i then z_1 = 0+0i or z_2 = 0+0i
    Last edited by Mush; January 7th 2009 at 06:54 AM.
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