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Callisto
Complex number. For the exponential form of a complex number, \(a + bi = e^{\log \sqrt{a^2 + b^2} + i (tan^{-1}\frac{b}{a})}\), why is it using common log, but not natural log or log with other base?
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sometimes ... we need r and e^itheta as one ... |dw:1345042767656:dw|
|dw:1345042823054:dw| yep ... my bad :))
ln .. since in usually draw ... ln seems like in .. and one user mistook for lnt for int
using this ... try to find the value of i^i
Your show time again :|
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Not quite understand..
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\[a + bi = e^{\ln \sqrt{a^2 + b^2} + i (tan^{-1}\frac{b}{a})}\]Put a = 0 and b = 1 into the formula. \[i = 0 + 1i \]\[= e^{\ln \sqrt{0^2 + 1^2} + i (tan^{-1}\frac{1}{0})}\]\[=e^{\ln (1) + i (tan^{-1}(undefined))}\]\[=e^{i (tan^{-1}(undefined))}\]It doesn't seem right at all!
I know... but that would be \[=e^{\frac{i\pi}{2}}\]Something goes wrong here?!
it's all right ... now complete the problem.
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Oh! \[i^i = (e^{\frac{i\pi}{2}})^{i}=(e^{\frac{i^2\pi}{2}})=e^{\frac{-\pi}{2}}\] The problem was that I didn't know what I was doing!
http://www.wolframalpha.com/input/?i=i^i+%3D+e^%28-pi%2F2%29 you are right!!
yw!! you can simplify things of the form log(a+ib)
no not really ... http://www.wolframalpha.com/input/?i=log%281+%2B+isqrt%283%29%29
*lost* If \(a + bi = e^{\log \sqrt{a^2 + b^2} + i (tan^{-1}\frac{b}{a})}\) Then, \(\ln (a + bi) = \ln (e^{\log \sqrt{a^2 + b^2} + i (tan^{-1}\frac{b}{a})})\) \(\ln (a + bi) = \log \sqrt{a^2 + b^2} + i (tan^{-1}\frac{b}{a})\) ?! Looks strange enough to me :|
yep ... that's right!!
http://www.wolframalpha.com/input/?i=log%281+%2B+2i%29+%3D+log%28sqrt%285%29%29+%2B+i+arctan%282%29
np :) you might find this interesting http://www.wolframalpha.com/input/?i=sin%283+%2B+i4%29
basically ... |dw:1345045135680:dw| using this identity ... you can prove any trigonometric identity you did back in school. though redoing them again ... is crazy.
Never mind... I guess I'll spend some time re-doing them again.. To see if I really understand! Thanks!
yw!! best of luck!!
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