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May 6th, 2014, 12:58 PM  #1 
Newbie Joined: May 2014 From: Estonia Posts: 5 Thanks: 0  Horribly Confused With Complex Logarithms
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! 
May 6th, 2014, 01:20 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,556 Thanks: 600 
Don't add 2mπi in the middle.

May 6th, 2014, 01:29 PM  #3 
Newbie Joined: May 2014 From: Estonia Posts: 5 Thanks: 0 
But why wouldn't you? The identity is $\displaystyle ln(z^c)=cln(z)+2πik$, not $\displaystyle ln(z^c)=cLn(z)+2πik$, so ln(z) would be infinite answered, wouldn't it?
Last edited by Muon321; May 6th, 2014 at 02:25 PM. 
May 7th, 2014, 11:58 AM  #4 
Math Team Joined: Aug 2012 From: Sana'a , Yemen Posts: 1,177 Thanks: 44 Math Focus: Theory of analytic functions 
You are claiming that $$\log(z^c) = c \log(z)+ 2k\pi i$$ How to prove this ? 

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