Hi,
I'm doing revision for an exam in complex analysis at the mo. I've been going through examples and I've found a couple that I've never seen before and have no idea how to do-any help would be much appreciated!
a)
Find all values of the following:
and log (i+1)
(Can't figure it out in Latex, I'm only learning!)
b) Let D= C\(-oo,0], and suppose log 1=0
Find (1-i)^4i and [e(-1-ROOT(3)i)/2]^(3PIi)
(Again if someone could show me this in latex it'd be appreciated, I kept getting 1-i^4 when I tried)
c)Given D= C\(-oo,0] and =exp(PI/2), its principle value, show that
(-1+(ROOT(3))i)^(3/2)= -2ROOT2
Please help, I'm so confused. I'm thinking it has to do with the section we did on simply connected domains and isolated singularities of an analytic domain
but I can't find any examples to follow and I'm lost!
Thanks
PS I'm trying with the latex but its slow so sorry about that!
A. For all values of , what we want is, from the definition of the logarithm, all z such that .
This is, of course, , where n is any integer.
Similarly, note that and , and use these to find all z such that .
B. Our branch cut (and log(1)=0) gives us for any z not on the cut. Thus we use the fact that for any complex z to get
So you can use this (along with basic exponentiation rules like and ) for and for
C. Are you sure that it's and not ? I dont think the former is possible:
, so and .
Thus gives , but .
I hope this helps point you in the right direction.
--Kevin C.
Ok, just working on the first time (I'm slow at these)
Do I need the perfect hackers comment?
log i= log|i|+iarg(i)
=log 1 +i(PI/2 +2nPI)
=0 + i(PI/2 +2nPI)
Can you just use the fact that z=|z|e^(iargz) and state answer as z=e^(i)(PI/2+2nPI) for example in the first case?
For the second does the final answer become z=[ROOT(2)][e^(i)(7PI/4 + 2nPI)] ?
In part b)
Does the answer come out something like (1-i)^4i= e^(-4logROOT(2)-4i(7PI/4 +2nPI)) ?
and the second comes out as e^(3PI(-1/ROOT3 +2nPI +ilog e^2)?
In part c) you are right but it is supposed to be (-i)^i=exp(PI/2)
Thanks so much for the help!