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Callisto
 4 years ago
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?
Callisto
 4 years ago
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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experimentX
 4 years ago
Best ResponseYou've already chosen the best response.3dw:1345042462489:dw

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.3dw:1345042511668:dw

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.3dw:1345042599891:dw

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.3dw:1345042661317:dw

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.3sometimes ... we need r and e^itheta as one ... dw:1345042767656:dw

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.3dw:1345042823054:dw yep ... my bad :))

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.3ln .. since in usually draw ... ln seems like in .. and one user mistook for lnt for int

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.3using this ... try to find the value of i^i

Callisto
 4 years ago
Best ResponseYou've already chosen the best response.0Your show time again :

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.3dw:1345043073075:dw

Callisto
 4 years ago
Best ResponseYou've already chosen the best response.0Not quite understand..

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.3dw:1345043197633:dw

Callisto
 4 years ago
Best ResponseYou've already chosen the best response.0\[a + bi = e^{\ln \sqrt{a^2 + b^2} + i (tan^{1}\frac{b}{a})}\]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!

Callisto
 4 years ago
Best ResponseYou've already chosen the best response.0I know... but that would be \[=e^{\frac{i\pi}{2}}\]Something goes wrong here?!

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.3it's all right ... now complete the problem.

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.3dw:1345043811417:dw

Callisto
 4 years ago
Best ResponseYou've already chosen the best response.0Oh! \[i^i = (e^{\frac{i\pi}{2}})^{i}=(e^{\frac{i^2\pi}{2}})=e^{\frac{\pi}{2}}\] The problem was that I didn't know what I was doing!

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.3http://www.wolframalpha.com/input/?i=i^i+%3D+e^%28pi%2F2%29 you are right!!

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.3yw!! you can simplify things of the form log(a+ib)

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.3no not really ... http://www.wolframalpha.com/input/?i=log%281+%2B+isqrt%283%29%29

Callisto
 4 years ago
Best ResponseYou've already chosen the best response.0*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 :

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.3yep ... that's right!!

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.3http://www.wolframalpha.com/input/?i=log%281+%2B+2i%29+%3D+log%28sqrt%285%29%29+%2B+i+arctan%282%29

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.3np :) you might find this interesting http://www.wolframalpha.com/input/?i=sin%283+%2B+i4%29

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.3basically ... dw:1345045135680:dw using this identity ... you can prove any trigonometric identity you did back in school. though redoing them again ... is crazy.

Callisto
 4 years ago
Best ResponseYou've already chosen the best response.0Never mind... I guess I'll spend some time redoing them again.. To see if I really understand! Thanks!

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.3yw!! best of luck!!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1348318962769:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1348319040083:dw
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