# Question about derivatives of complex-valued functions

For $z \in \mathbb{C}, t \in \mathbb{R}, \\f : \mathbb{C} \times \mathbb{R} \to \mathbb{C}, \\a : \mathbb{C} \times \mathbb{R} \to \mathbb{R}, \\b : \mathbb{C} \times \mathbb{R} \to \mathbb{R}$

And given that $\frac{\partial f(z,t)}{\partial t} = a(z,t) +ib(z,t)$, under what conditions will the following equations be true?

$\frac{\partial f^R(z,t)}{\partial t} = a(z,t) \\ \frac{\partial f^I(z,t)}{\partial t} = b(z,t)$

where $f^R : \mathbb{C} \times \mathbb{R} \to \mathbb{R}$ and $f^I : \mathbb{C} \times \mathbb{R} \to \mathbb{R}$ such that $f(z,t) = f^R(z,t)+if^I(z,t)$.

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Under any condition. Derivation is linear, so if $f=f^R+i\,f^I$ then $$\frac{\partial f}{\partial t}=\frac{\partial f^R}{\partial t}+i\,\frac{\partial f^I}{\partial t}.$$ The real and imaginary parts of a complex number are unique, so you get your equality.
Godd point.$\$ – Martin Argerami Nov 28 '12 at 12:42
@ChristianClason Couldn't it happen that $\frac{\partial f^R}{\partial t}$ or $\frac{\partial f^I}{\partial t}$ are functions with complex range(rather than real)? If so, then equations which i wrote may not hold. – Sunny88 Nov 28 '12 at 12:55