Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

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?

I got my questions answered at in under 10 minutes. Go to now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly


Get your free account and access expert answers to this and thousands of other questions

no quite ... !!
...... Yes?!

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

sometimes ... we need r and e^itheta as one ... |dw:1345042767656:dw|
|dw:1345042823054:dw| yep ... my bad :))
So, is it ln or log?
ln .. since in usually draw ... ln seems like in .. and one user mistook for lnt for int
using this ... try to find the value of i^i
Your show time again :|
Not quite understand..
\[a + bi = e^{\ln \sqrt{a^2 + b^2} + i (tan^{-1}\frac{b}{a})}\]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!
that would be pi/2
I know... but that would be \[=e^{\frac{i\pi}{2}}\]Something goes wrong here?!
it's all right ... now complete the problem.
May I know.. how?
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!^i+%3D+e^%28-pi%2F2%29 you are right!!
Thanks a ton!!! :)
yw!! you can simplify things of the form log(a+ib)
That's another thing?
no not really ...
*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 :|
yep ... that's right!!
Thanks again!
np :) you might find this interesting
basically ... |dw:1345045135680:dw| using this identity ... you can prove any trigonometric identity you did back in school. though redoing them again ... is crazy.
Never mind... I guess I'll spend some time re-doing them again.. To see if I really understand! Thanks!
yw!! best of luck!!

Not the answer you are looking for?

Search for more explanations.

Ask your own question