Question

can somone help me with this derivative?

\[\frac{d}{dz} (1+i)^{z^2}\] assume an appropriate branch cut when necessary.

Answer 1

Ughhh I've got a test on this stuff next week :(
I Better get studying!!

Answer 2

mine is tomorrow!

Answer 3

you can try this with a u substitution if you know how? does that sound a little familiar?

Answer 4

So we would use the differentiation rules for exponential functions yes?
\[\Large\bf\sf =\quad (1+\mathcal i)^{z^2}\cdot \ln(1+\mathcal i)\frac{d}{dz}z^2\]This approach maybe? :O

Answer 5

Woops I should call that \(\Large\bf\sf log(1+\mathcal i)\)
And then we expand out the log after that.

Mmmmm am I doing this right? +_+
Hopefully you have some idea...

Answer 6

I don't understand how you are getting that log?

Answer 7

\[\Large\bf\sf (a^x)'\quad=\quad a^x \ln a\]

Answer 8

If y = x^x then ln y = ln (x^x)
so...
lny = xlnx

Answer 9

Remember back to taking the derivative of exponential functions? :)
The log didn't often show up because we usually dealt with e as a base.
And ln(e)=1 so it was ignored.

Answer 10

oh right. haha. now I feel dumb.

Answer 11

And then you'd have to also expand out that log,\[\Large\bf\sf \log(1+\mathcal i)\quad=\quad \ln|1+\mathcal i|+\arg(1+\mathcal i)\]

Answer 12

yeah I could. but the answer you did is correct.

Answer 13

And ummmm define a branch cut somewhere that works for the differentation or something?
Blah I haven't looked at my notes enough yet..

Answer 14

Oh ok :) heh

Answer 15

the question says assume appropriate branch cuts where neccesary, so I can just assume this is analytic. I just forgot my first year calc haha

Answer 16

Ohhh interesting! :)

Answer 17

when the value of log(z) is specified, \(c^z\) is an entire function and \((c^z)' = c^z \log(z)\). It was sitting right there in my textbook! thanks a lot!

Answer 18

Oh how bout that!
np \c:/