Question

quick question
Are all real differentiable functions when pushed to complex plane are still differentiable ?

For example, f(x) = x^2 is differentiable when x is real. Is it still differentiable when x is complex ?

Answer 1

I guess so, maybe this is what you're looking for http://mathworld.wolfram.com/Cauchy-RiemannEquations.html

Answer 2

\(f(z) = z^2\)
\[f'(z_0)=\lim\limits_{\Delta z\to 0}~\dfrac{f(z_0 + \Delta z)-f(z_0)}{\Delta z}\]

If i mimic the "real limits" argument, I do get \(f'(z_0) = 2z_0\)

was wondering if this is true for all real differentiable functions

Answer 3

Woah, that's interesting

Answer 4

hey

Answer 5

hey doggy knw how to help me

Answer 6

empty can you help me in trignometry

Answer 7

nope, he is mine for next 30 mnts or so

Answer 8

Sorry I am not good at trig.

Quick answer ganeshie: yeah you can haha.

Answer 9

ok empty
sorry

Answer 10

:'(

Answer 11

for now, im convincing myself saying that if it is differentiable in 1D then it is definitely differentiable in 2D etc etc...

Answer 12

ok empty
sorry

Answer 13

because when we're moving to complex plane from real, we're just literally incrementing the dimension of both domain and range of a function, "i think"

Answer 14

Yeah, well I mean if we can construct a function out of a power series then we really turn the problem into like understanding:

\[f(z)=z^n\] and from there the rest should follow.

Answer 15

Ahh are you saying power series doesn't bother whether z is real or ccomplex ?

Answer 16

If that is the case, this must be true
\[f'(x) = g(x) \implies f'(z) = g(z)\]

where \(x\) is real and \(z\) is complex

Answer 17

So a quick way to show this:

\[f'(z)=\lim_{\Delta z \to 0} \frac{f(z+\Delta z) -f(z)}{\Delta z}\]
\[f'(z)=\lim_{\Delta z \to 0} \frac{[z^n+(n-1)z^{n-1}\Delta z +S] -z^n}{\Delta z}\]
So S is the sum from the binomial expansion with all the \(\Delta z\) terms with exponents higher than 1. Now we subtract the \(z^n\) terms and divide the \(\Delta z\) to get:

\[f'(z)=\lim_{\Delta z \to 0} \left[(n-1)z^{n-1}+\frac{S}{\Delta z} \right]\] Since S has only \(\Delta z\) terms with positive exponents greater than 2, dividing out 1 of them means the limit will take all these guys to zero just leaving us with this other term withohut anything.

And then yeah, we can use the power series to extend our functions from real to complex values, I think they call this "analytic contnuation" but I might be wrong in the terminology but that's how they do it. That's why we just can look at the power series to compare for this:

\[e^{ix} = \cos x+ i \sin x\] for instance :P

Answer 18

That looks neat! so proving \(z^n\) is differentiable proves all real functions that can be represented as power series are still differentiable in complex plane xD

Answer 19

Yeah, exactly, so the weird part is that this seems to be more about differentiation in a direction rather than before, and there's a bunch of weird and cool things that complex differentiation allows us to do that we didn't have before such as if you can differentiate it once, it's infinitely differentiable. I'm probably not the person to describe the rigorous side of this, but I have more or less an intuitive understanding of this.

But the main thing is that in the past you would say the left hand and right hand limit have to be equal, in this case we are saying that there are an infinite number of ways to approach the point, and all those give us the same derivative at that value, so in a sense complex differentiability is much more intense.

I can describe some intuitive things like the Cauchy Riemann equations (which really are just the same as talking about curl and divergence).

Answer 20

That sounds amazing, I think this is also useful, it's not very long either https://en.wikipedia.org/wiki/Holomorphic_function

Answer 21

I think I'm getting it slowly, both domain and range are 2D, so clearly the left and right limits are not sufficient anymore. for a complex function to be differentiable, the limit of difference quotient must exist and equal in all directions.. this looks more or less like the limit of a function of two varibales