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 è a = 0 or b = 0
Is the same theorem true in the Complex Number realm? (Why or why not?)
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 è 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.
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 $