My question is very simple that is: Is Complex number closed under division? Can we consider this 0+0i as complex numbers? or it is not a complex number.
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$\begingroup$For any nonzero $z=a+bi$, we have a multiplicative inverse $z^{-1}=\frac{a-bi}{a^2+b^2}$. This being the case, you can divide by $z$ by multiplying with $z^{-1}=\frac{a-bi}{a^2+b^2}$.
But the additive identity zero (given by $0+i0$ in this case) cannot have a multiplicative inverse, so you cannot divide by it in this ring, nor in any field.
In this sense, fields (actually even division rings) are as closed as possible under division as you can get.
Added after your question persisted in the comments:
It is not sensible to ask if a set is closed under an operation that isn't even defined on all the set. If there are elements that can't participate in the division, then it is hardly fair to say that the set "isn't closed" under an operation.
The best one can say is that division (where defined) always results in another element of $\Bbb C$. Said another way, the result of a division in $\Bbb C$ will not fall outside of $\Bbb C$. But this is not exactly saying that "$\Bbb C$ is closed under division."
$\endgroup$ 3 $\begingroup$$0+0i$ is the same as $0$. $0$ is a complex number. It is the zero element of the field $\mathbb{C}$. In a field the zero element cannot have a multiplicative inverse because the equation $$ 0 \cdot x = 1 $$ cannot not have a solution. This means one cannot divide by $0$.
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