# Thread: Horribly Confused With Complex Logarithms

1. I'm getting myself all confused with complex logarithms. I'll try to explain why. One identity with complex logarithms is ln(z^c)=cln(z)+2πik, with k an integer. This is, of course, a more general case of ln(e^c)=c+2πik, but it doesn't always work the same! Lets say we are evaluating ln(e^i). Using the latter identity, it is i+2πik, which is, logically, the correct answer, but using the first identity, you get iln(e)+2πik, which is i(1+2πin)+2πik=i+2πn+2πik...! What! Obviously e^(i+2π) doesn't equal e^i. Another example, ln(1)=ln(e^2πi)=2πi(1+2πin)+2πik=2πi+4π^2n+2 πik

And, I have another problem. I have this when I try to solve an equation 10^z=e^πi, so I take ln of both sides zln(10)=πi+2πik and then z=(πi+2πik)/ln(10), where ln(10) in the denominator is infinite answered and will give solutions that don't work! I'm clearly doing something wrong, so someone please help me!  2.

3. Originally Posted by Muon321 I'm getting myself all confused with complex logarithms. I'll try to explain why. One identity with complex logarithms is ln(z^c)=cln(z)+2πik, with k an integer. This is, of course, a more general case of ln(e^c)=c+2πik, but it doesn't always work the same! Lets say we are evaluating ln(e^i). Using the latter identity, it is i+2πik, which is, logically, the correct answer, but using the first identity, you get iln(e)+2πik, which is i(1+2πin)+2πik=i+2πn+2πik...! What! Obviously e^(i+2π) doesn't equal e^i. Another example, ln(1)=ln(e^2πi)=2πi(1+2πin)+2πik=2πi+4π^2n+2 πik
You're misapplying the second identity to a real argument. Just consider what happens if you try to evaluate ln(e) your way. So don't do that. And, I have another problem. I have this when I try to solve an equation 10^z=e^πi, so I take ln of both sides zln(10)=πi+2πik and then z=(πi+2πik)/ln(10), where ln(10) in the denominator is infinite answered and will give solutions that don't work! I'm clearly doing something wrong, so someone please help me!
What do you mean by "infinite answered?" Do you mean that you are bothered by the decimal representation of ln(10) requiring an infinite number of digits? Or do you mean that, when you misapply your second identity to ln(10), you re-encounter your earlier difficulty?  Bookmarks
 Posting Permissions
 You may not post new threads You may not post replies You may not post attachments You may not edit your posts   BB code is On Smilies are On [IMG] code is On [VIDEO] code is On HTML code is Off Trackbacks are Off Pingbacks are Off Refbacks are On Terms of Use Agreement