## Callisto 3 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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no quite ... !!

2. Callisto

...... Yes?!

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sometimes ... we need r and e^itheta as one ... |dw:1345042767656:dw|

8. Callisto

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|dw:1345042823054:dw| yep ... my bad :))

10. Callisto

So, is it ln or log?

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ln .. since in usually draw ... ln seems like in .. and one user mistook for lnt for int

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using this ... try to find the value of i^i

13. Callisto

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15. Callisto

Not quite understand..

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17. Callisto

$a + bi = e^{\ln \sqrt{a^2 + b^2} + i (tan^{-1}\frac{b}{a})}$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!

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that would be pi/2

19. Callisto

I know... but that would be $=e^{\frac{i\pi}{2}}$Something goes wrong here?!

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it's all right ... now complete the problem.

21. Callisto

May I know.. how?

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23. Callisto

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!

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http://www.wolframalpha.com/input/?i=i^i+%3D+e^%28-pi%2F2%29 you are right!!

25. Callisto

Thanks a ton!!! :)

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yw!! you can simplify things of the form log(a+ib)

27. Callisto

That's another thing?

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no not really ... http://www.wolframalpha.com/input/?i=log%281+%2B+isqrt%283%29%29

29. Callisto

*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 :|

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yep ... that's right!!

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32. Callisto

Thanks again!

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np :) you might find this interesting http://www.wolframalpha.com/input/?i=sin%283+%2B+i4%29

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basically ... |dw:1345045135680:dw| using this identity ... you can prove any trigonometric identity you did back in school. though redoing them again ... is crazy.

35. Callisto

Never mind... I guess I'll spend some time re-doing them again.. To see if I really understand! Thanks!

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yw!! best of luck!!

37. mahmit2012

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38. mahmit2012

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