Thread: Natural Logarithm of a Complex Number

1. Natural Logarithm of a Complex Number

I'm not sure how to go about solving this little question...

By expressing z = i*r*e^(i*theta) in terms of 'z'*e^(i*theta) ('z' is the modulus of z), find the natural logarithm of z.

I have thought of a couple of different approaches but the answers end up looking quite inelegant, so I think they may be wrong... and there's no answer in the back of the book...

Thank you very much in advance!

Jessica.

But the usual way of doing this is: $\displaystyle \log (z) = \ln \left( {\left| z \right|} \right) + i\arg (z)$.

Is that close to what ou mean?

3. Yes, I know that's how the general result looks, but I was wondering what i*r*exp(i*theta) looks like in this form. I have a result, and it looks like this:

log z = ln r + i(theta + pi/4 + 2pi*n)

However I don't think this is right.

4. Because $\displaystyle i = e^{i\left( {\frac{\pi }{2}} \right)}$ then $\displaystyle ire^\Theta = re^{i\left( {\Theta + \frac{\pi }{2}} \right)}$.

5. Hello,
Originally Posted by j_clough
Yes, I know that's how the general result looks, but I was wondering what i*r*exp(i*theta) looks like in this form. I have a result, and it looks like this:

log z = ln r + i(theta + pi/4 + 2pi*n)

However I don't think this is right.
The incorrect part is pi/4.

Note that $\displaystyle i=0+1 \times i=\cos \tfrac \pi 2+i \sin \tfrac \pi 2=e^{i \frac \pi 2}$

6. Ahh, I didn't see that! Thank you very much, Moo.