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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.

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    Note that you were asking about the derivative with respect to $t$; a complex function $g(z)$ with real range (like the real or imaginary part of a complex function) cannot be differentiable with respect to $z$.2012-11-28