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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

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1. experimentX

no quite ... !!

2. Callisto

...... Yes?!

3. experimentX

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4. experimentX

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5. experimentX

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6. experimentX

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7. experimentX

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

8. Callisto

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9. experimentX

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

10. Callisto

So, is it ln or log?

11. experimentX

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

12. experimentX

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

13. Callisto

Your show time again :|

14. experimentX

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15. Callisto

Not quite understand..

16. experimentX

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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!

18. experimentX

that would be pi/2

19. Callisto

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

20. experimentX

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

21. Callisto

May I know.. how?

22. experimentX

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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!

24. experimentX

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

25. Callisto

Thanks a ton!!! :)

26. experimentX

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

27. Callisto

That's another thing?

28. experimentX

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 :|

30. experimentX

yep ... that's right!!

31. experimentX
32. Callisto

Thanks again!

33. experimentX

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

34. experimentX

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!

36. experimentX

yw!! best of luck!!

37. mahmit2012

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38. mahmit2012

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