# Simple question about derivatives of complex functions

Is it true that for any complex function $f(z, t)$, the following equation is correct? $\frac{\partial f(z,t)}{\partial t} = \frac{\partial f^R(z,t)}{\partial t} + i \frac{\partial f^I(z,t)}{\partial t}$,

where $f(z,t) = f^R(z,t) + i f^I(z,t)$, z is complex number, t is real number(time), and $f(z,t)$ is complex number for all z and t.

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I take it you are assuming all those derivatives exist. – Gerry Myerson Nov 27 '12 at 11:47
@GerryMyerson yes, I assume that they exist. – Sunny88 Nov 27 '12 at 12:02
@GerryMyerson Doesn't the existence of $\frac{\partial f}{\partial t}(z,t)$ imply that both $\frac{\partial f^R}{\partial t}(z,t)$ and $\frac{\partial f^I}{\partial t}(z,t)$ exist? – Sunny88 Nov 28 '12 at 14:50

Yes, it is true. You can check this with the limit definition of the derivative: $$\frac{\partial f}{\partial t}(z,t) := \lim_{h\to 0} \frac{f(z,t+h) - f(z,t)}{h}.$$ If you split the quotient on the right hand side into real and imaginary parts, you get $$\frac{f(z,t+h) - f(z,t)}{h} = \frac{f^R(z,t+h) - f^R(t,h)}{h} + i\frac{f^I(z,t+h) - f^I(z,t)}{h}.$$ Taking the limit at $h\to 0$ (assuming all the limits exist) you are left with $$\frac{\partial f}{\partial t}(z,t) = \frac{\partial f^R}{\partial t}(z,t) + i\frac{\partial f^I}{\partial t}(z,t).$$
Doesn't the existence of $\frac{\partial f}{\partial t}(z,t)$ imply that both $\frac{\partial f^R}{\partial t}(z,t)$ and $\frac{\partial f^I}{\partial t}(z,t)$ exist? – Sunny88 Nov 28 '12 at 14:49
Yes, because the limit of a complex function $g(h)$ as $h\to 0$ exists if and only if the limits of its real and imaginary parts $g^R(h)$ and $g^I(h)$ exist as $h\to 0$. – froggie Nov 28 '12 at 15:54