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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?
 2 years ago
 2 years 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?
 2 years ago
 2 years ago

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

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

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

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

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

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

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

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

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

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

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

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

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

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

experimentX Group TitleBest ResponseYou've already chosen the best response.3
dw:1345043197633:dw
 2 years 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!
 2 years ago

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

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

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

experimentX Group TitleBest ResponseYou've already chosen the best response.3
dw:1345043811417:dw
 2 years 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!
 2 years 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!!
 2 years ago

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

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

Callisto Group TitleBest ResponseYou've already chosen the best response.0
That's another thing?
 2 years 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
 2 years 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 :
 2 years ago

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

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

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

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

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