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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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1 Answer 1

up vote 3 down vote accepted

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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Sorry for being pedantic, but that should read 'any condition' (otherwise it would mean that the statement is false, and confusion follows). –  Christian Clason Nov 28 '12 at 12:37
    
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
    
No, because the real part and the imaginary part have real range by definition. –  Christian Clason Nov 28 '12 at 13:01
    
@ChristianClason I see, I was a bit unsure whether function with real range can have derivative with complex range, but I followed the definition of the derivative now, and it seems that you are right, since derivative of a function with real range is just a limit of difference between two real numbers, and it can not be complex. –  Sunny88 Nov 28 '12 at 13:08

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