Imaginary Calculus

[NorthWind]

First off, mornin' ladies and gentlemen of Science Forums. I'm currently an AP Calculus student, and I've noticed that the College Board curriculum for calculus is kept strictly to real numbers.

So that got me wondering: can you have imaginary calculus? I would guess that the answer is yes; if so, are the rules the same as real calculus?

 

I did some messing around with f(z) = (z + zi). I haven't done anything with imaginary numbers for a while, but I think the graph of this would be a vector with a constant slope i? (Can you take slopes on the imaginary plane like a linear function?) I tried to find a derivative of z+zi that equaled i and found that if I took the derivative of zi over the derivative of z (zi's derivative being i; z's derivative being 1) I get i. So I did this like a parametric.

 

Is any of this correct?

 

Thanks,

N

[BigMoosie]

I believe the derivative would be 1+i:

 

f(z) = (1+i)z

f'(z) = 1+i

[Mr Skeptic]

If you consider i to be a constant, and you can do calculus with constants, then you should have no trouble doing calculus with imaginary numbers.

[Royston]

Plus (although I know this is slightly different to what the OP is asking) they're sometimes unavoidable when it comes to solving differential equations.

[WhataBohr]

for a complex function to be differentiable on an open set i believe to has to satisfy the Cauchy Riemann equations. du/dy= -dv/dw (partial dirivatives)

[ydoaPs]

If you consider i to be a constant, and you can do calculus with constants, then you should have no trouble doing calculus with imaginary numbers.

 

That's what I was thinking. I would treat i as a constant(after all, it is ).

[Bignose]

There is a whole extension -- Find a good "Complex Analysis" book and you'll see that there is quite a lot in addition to just the calculus of real variables.

[WhataBohr]

Complex Analysis is not that simple... deals withs residues and poles etc and criteria for functions that are analytical... if u treat z as a real constant u are missing a whole lot. Good luck...

[Royston]

Is u a variable, or does u = you ? I know it's the latter, but it's not a good idea to use SMS on here.

[Bignose]

Complex Analysis is not that simple... deals withs residues and poles etc and criteria for functions that are analytical... if u treat z as a real constant u are missing a whole lot. Good luck...

 

I'm kind of confused by this response because who said complex analysis was simple and who said to treat z as a constant? i is a constant, but obviously not z = x + yi.

 

Furthermore, I agree completely with what Snail wrote... especially in light of your last post where you actually used u as a variable! Though you didn't define what u, v, y, or w was...

[WhataBohr]

my point was that complex analysis deals with many different topics that the poster ignoring and is worth learning more about. ie residues , poles etc

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f(z) = u(x,y) + iv(x,y) for those math guys so apply Cauchy riemann equations to that....

[D H]

Complex analysis is in a way a lot easier than real analysis. Many colleges teach complex analysis before real analysis for this very reason.

[Bignose]

my point was that complex analysis deals with many different topics that the poster ignoring and is worth learning more about. ie residues , poles etc

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f(z) = u(x,y) + iv(x,y) for those math guys so apply Cauchy riemann equations to that....

 

OK, that's fine, but again no one wrote that it was easy or to treat z as a constant, so it was kind of a weird response to the thread.

 

Also, you still didn't actually define what w was. I'm sure it was a typo (since it doesn't really make any sense), but it is still bad form to just leave any variables undefined -- especially as the OP is an obvious beginner. At least link to a source with more complete information, like linking to: http://mathworld.wolfram.com/Cauchy-RiemannEquations.html

Might be nice to say a word or two about them (like, C-R equations come from differentiating a complex function f(z) with respect to z = x + yi i.e. with respect to dz = dx + dyi.

 

Finally, this forum does have a LaTeX capability that will make the math easier to write and read: a la

 

[math]\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}[/math] and

[math]\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}[/math]

 

much nicer looking than trying to write out "partial derivatives" where necessary and write the fractions using only ()'s and the / key.

[WhataBohr]

thanks for the info about latex capability for i am new to this site. I will try to be more complete in my discussions....