Question

partial differentiation question with complex numbers

(i have very little experience, i have a few relevant books but have not started yet and im curious)

Answer 1

does this form of chain rule work:

\[\frac{dh( f(t) + g(t)i) }{dt} = h'( f(t) + g(t)i ).(f'(t) + g'(t) )\]

i realise now its not really partial differentiation ( i dont think it is anyway)

Answer 2

it works for some things i've tried it with, like
\[ x^i\]

Answer 3

looks like I am also illiterate on this matter.

Answer 4

it could be a chain rule ... f(t) and g(t) being domain of h

Answer 5

sorry, i meant :

(x+i)^i

Answer 6

http://www.physicsforums.com/showthread.php?t=53374
says that partial derivative for complex numbers work exactly like for real numbers.

Answer 7

im pretty sure it works

Answer 8

aw sweeet thats just what im looking for

Answer 9

a complex number with complex power?? i haven't seen that.

Answer 10

http://www.wolframalpha.com/input/?i=%281%2Bi%29%5Ei
there must be some way outta here

Answer 11

let's change x+i into e's

Answer 12

i guess that might solve our problem.

Answer 13

aw they're pretty cool, when we did complex numbers for the first time last year i just spent lessons seeing how much i could figure out, so im not 100% on whether im allowed to do all this:
\[a+bi = R(\cos\theta + isin\theta) = e^{\ln{R} + \theta i }\]
\[\ln(a+bi) = \ln|a+bi| + Arg(a+bi)\]
\[= \ln(\sqrt{a^2 + b^2}) + \arctan(b/a)\]
\[(a + bi)^{c+di} = e^{(c+di)\ln({a+bi})}\]
\[e^{(c+di)\ln(a+bi)} = e^{(c+di)(\ln(a^2 + b^2) + \arctan(b/a))}\]
etc etc, it gets complicated quite quickly, so a chain rule would be very useful

Answer 14

yeah thats the approach, using e to split it into f(x) + g(x)i

Answer 15

i guess it does.

Answer 16

@experimentX a complex number with complex power?? i haven't seen that.

Then you might be interested in \( i^i \approx 0.2079\)

http://www.wolframalpha.com/input/?i=I%5EI

Answer 17

@phi truly interesting