Taylor series of complex function confusion with big O notation

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Suppose $u(x,t)$ is a function of two real numbers that outputs a complex number.

Usually I would have $u(x,t+k) = u(x,t) + k\frac{\partial u}{\partial t}(x,t) + \frac{1}{2}k^2 \frac{\partial ^2 u}{\partial t^2}(x,t) + O(k^3)$

But now that doesn't seem entirely accurate, since $O(k^3)$ is always real (right?). So what would be the correct notation in this case?

More specifically, I'm interested in evaluating the expressions:

$\frac{u(x, t+k) - u(x,t)}{k}$, and $\frac{u(x-h, t) -2u(x,t) + u(x+h, t)}{h^2}$

In the real case, I'd say that

$\frac{u(x, t+k) - u(x,t)}{k} = \frac{\partial u}{\partial t}(x,t) + O(k)$ and $\frac{u(x-h, t) -2u(x,t) + u(x+h, t)}{h^2} = \frac{\partial ^2u}{\partial x^2}(x,t) + O(h^2)$

What is the complex equivalent? Is this notation also correct in the complex sense?

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Let's use the following definition:

for a function $f$ defined in a neighbourhood of $0$ and a non-negative function $g$ defined at least on the same neighbourhood, we say that $f(z) = O(g(z))$ if there exist $r>0$ and $M>0$ such that for all $z$ with $|z|<r$ it holds that $|f(z)|\le M\cdot g(z)$.

Theis definition works whether we are talking about real or complex functions. You just have to make sure that the functions you are talking about is Lipschitz continuously differentiable enough times in the variable that you care.