# Division of a complex number by zero

• I
Hi
I know that division of a real number by zero is not defined. I just came across the following in a textbook on Complex Analysis by Priestley , " we are allowed to divide a complex number by zero as long as the complex number ≠ 0 "
Is this correct ? What happens if the complex number is purely real ?
Thanks

fresh_42
Mentor
This is nonsense. Complex numbers are a field. A field consists of two groups, one for addition and one for multiplication. They are related by the distributive law. Zero is no element of the multiplicative group, so the question doesn't even arise.

More context would be helpful here. E.g. if we consider complex functions such as ##f(z)=\dfrac{1}{(z-a)^n}## then we can investigate the kind of the singularity.

If it is a book about non standard analysis, then certain constructions with infinities are possible, but I don't know the details.

In any ordinary sense this statement is false. We are allowed to divide zero by a complex number which isn't zero. This is because of the distributive law:
$$\dfrac{0}{z} = (1-1)\cdot z^{-1} = 1\cdot z^{-1} - 1\cdot z^{-1} = z^{-1} -z^{-1} =0$$

• FactChecker
FactChecker
Gold Member
Division by zero is not allowed in complex numbers. If the behavior is good enough, it may be possible to extend the definition of a function in a smooth way where the denominator is 0, but that is not the same as dividing by zero. For instance, we can smoothly extend ##f(z) = (z-1)(z-2)/(z-1)## to equal -1 at ##z=1##. That is actually defining a new function and is not the same as doing an actual division by zero.

Thanks for your replies. They totally make sense. But I found it another book , Complex Analysis by Howie where it states , the convention is c/0 = ∞ where c is a finite complex number.
In both books I found the information in a section on Mobius Transformations and the extended complex plane

fresh_42
Mentor
This is rather sloppy and should not be used. It is un-mathematical. The same can be done with the reals, but it is not precise.

The Möbius transformation has a singularity, and the values tend to infinity the closer we get to that singularity. However, in complex calculus such singularities are classified and they have different consequences if we e.g. integrate around them. To throw all of them in a container labeled infinity is a bit too easy and creates more problems than it solves.

FactChecker
Gold Member
In both books I found the information in a section on Mobius Transformations and the extended complex plane
We should be careful here. The extended complex plane (aka Riemann sphere ) is not really the same. It has a point for ##\infty##, but that point does not have the usual algebraic properties of complex numbers. For instance, ##0 * \infty## is undefined.

That being said, the Riemann sphere can be useful. Riemann was not a dummy.

fresh_42
Mentor
That being said, the Riemann sphere can be useful. Riemann was not a dummy.
Yes, but we need the projective complex line not the complex plane for it! This makes a big difference.

• FactChecker
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