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Callisto
Group Title
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?
 one year ago
 one year ago
Callisto Group Title
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?
 one year ago
 one year ago

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experimentX Group TitleBest ResponseYou've already chosen the best response.3
no quite ... !!
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.0
...... Yes?!
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.3
dw:1345042462489:dw
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.3
dw:1345042511668:dw
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.3
dw:1345042599891:dw
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.3
dw:1345042661317:dw
 one year ago

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

Callisto Group TitleBest ResponseYou've already chosen the best response.0
dw:1345042836751:dw
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.3
dw:1345042823054:dw yep ... my bad :))
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.0
So, is it ln or log?
 one year ago

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

experimentX Group TitleBest ResponseYou've already chosen the best response.3
using this ... try to find the value of i^i
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.0
Your show time again :
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.3
dw:1345043073075:dw
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.0
Not quite understand..
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.3
dw:1345043197633:dw
 one year ago

Callisto Group TitleBest 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!
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.3
that would be pi/2
 one year ago

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

experimentX Group TitleBest ResponseYou've already chosen the best response.3
it's all right ... now complete the problem.
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.0
May I know.. how?
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.3
dw:1345043811417:dw
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.0
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!
 one year ago

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

Callisto Group TitleBest ResponseYou've already chosen the best response.0
Thanks a ton!!! :)
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.3
yw!! you can simplify things of the form log(a+ib)
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.0
That's another thing?
 one year ago

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

Callisto Group TitleBest 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 :
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.3
yep ... that's right!!
 one year ago

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

Callisto Group TitleBest ResponseYou've already chosen the best response.0
Thanks again!
 one year ago

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

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

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

experimentX Group TitleBest ResponseYou've already chosen the best response.3
yw!! best of luck!!
 one year ago

mahmit2012 Group TitleBest ResponseYou've already chosen the best response.1
dw:1348318962769:dw
 one year ago

mahmit2012 Group TitleBest ResponseYou've already chosen the best response.1
dw:1348319040083:dw
 one year ago
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