Difference of two functions sharing the same poles pole-free?

[Grothard]

If f(z) and g(z) share all the same poles, is f(z)-g(z) pole-free? I feel like this would be true, but I can't really come up with a proof for it.

 

[micromass]

No, consider 1/x and 1/(2x). These both have poles at zero.

But 1/x - 1/(2x)=1/(2x) which also has a pole at zero.

Or did I misunderstood the question?

 

[marcusl]

Certainly not. Here's the simplest counter-example: f = a*g where a is a constant, so f-g has the same poles as f. More complicated examples exist as well.