find the principle value of
Name:
Anonymous
2009-05-16 22:43
the modulus of (-i)^2+i
Name:
Anonymous
2009-05-17 17:13
anyone?
Name:
4tran
2009-05-17 18:20
(-i)^(2+i) or i + (-i)^2?
latter is i - 1
former is e^((2+i)(3i pi/2)) = - e^(-3pi/2)
I think that's what principle value means... what do you mean by modulus?
Name:
Anonymous
2009-05-18 19:55
similar problem, im told to evalute this using the principle value for the power
\mid(1-i)^{2i-1}\mid
Name:
Anonymous
2009-05-19 7:52
Principle value of the modulus?
You mean the principle value of the argument?
Name:
Anonymous
2009-05-19 12:19
You mean the principle value of the argument?
Yes.
>>1 was stated badly, ignore it. I need to solve
>>4
I've almost got it
(1-i) = \sqrt2e^{-4/\pi}
Principle value is
ln(\sqrt2) {-i\pi/4}
So
|(1-i)^{2i-1}|=|e^{(ln(\sqrt2)-i\pi/4)(2i-1)} |
|(1-i)^{2i-1}|=|e^{(2iln\sqrt2 +\pi/2-ln\sqrt2+i\pi/4} |
Anyone know how to tidy up the last term so it's not in exponential form?
Name:
Anonymous
2009-05-20 13:21
>>6
Im just guessing because i haven't taken a complex analysis class yet. But perhaps Euler's Formula is of some use?
Name:
Anonymous
2009-05-20 19:13
>>6
well you get you can just change the addition into multiplication and use the fact aln(b) = ln(b^a), but you'll still get some exponentiation in there
Name:
Anonymous
2009-06-01 14:10
(-i)^2+i = i-1
modulus is root 2
principle argument is 3Pi/4
am I missing something?