Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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
add comment

1 Answer

up vote 2 down vote accepted

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 $$\frac{\partial f}{\partial t}(z,t) = \frac{\partial f^R}{\partial t}(z,t) + i\frac{\partial f^I}{\partial t}(z,t).$$

share|improve this answer
    
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
1  
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.