Calculating Complex Logarithms

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Homework Help Overview

The discussion revolves around calculating the complex logarithm of the expression ##\log (e^{1+2i})##. Participants are exploring the definitions and properties of logarithms in the context of complex numbers.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to apply the definition of the logarithm for complex numbers, specifically questioning the interpretation of the components of the expression, such as the modulus and argument. There is also discussion about the use of the natural logarithm in complex analysis.

Discussion Status

Some participants have provided insights into the calculation process and the implications of the definitions used. There is recognition of different interpretations and approaches to the problem, but no explicit consensus has been reached.

Contextual Notes

Participants are navigating the complexities of logarithmic functions in the realm of complex numbers, with some uncertainty about the application of the natural logarithm and its implications for the problem at hand.

Bashyboy
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Hello everyone,

I am asked to calculate ##\log (e^{1+2i})##, and I would appreciate it if someone could verify my calculation..

My textbook defines ##\log z## as ##\log z = \ln |z| + i \arg z##.

##\log (e^{1+2i}) = \ln |e^{1+2i}| + i \arg(e^{1+2i}) \iff##

##\log (e^{1 + 2i}) = \ln|e e^{2i}| + i \arg (e e^{2i}) \iff##

##\\log (e^{1+2i}) = \ln e + i \arg (e e^{2i}) \iff##

##\log (e^{1+2i}) = 1 + i \arg (e e^{2i})##

Here is the step that I am not entirely certain about. I know that ##e^{2i}## is the exponential representation of the complex point whose angle is ##2##. But I am wondering, would the coefficient ##e## just represent the distance of the complex point from the origin? More generally, the complex exponential function ##e^z## is defined as ##e^z = e^x e^{iy}##, and so ##e^x## would represent the distance of the complex point ##e^{iy}##?

If this is so, then the answer would seem to be

##\log (e^{1+2i}) = 1 + i(2 + 2 n \pi)##

Is this correct?
 
Last edited:
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Bashyboy said:
Hello everyone,

I am asked to calculate ##\log (e^{1+2i})##, and I would appreciate it if someone could verify my calculation..

My textbook defines ##\log z## as ##\log z = \ln |z| + i \arg z##.

##\log (e^{1+2i}) = \ln |e^{1+2i}| + i \arg(e^{1+2i}) \iff##

##\log (e^{1 + 2i}) = \ln|e e^{2i}| + i \arg (e e^{2i}) \iff##

##\\log (e^{1+2i}) = \ln e + i \arg (e e^{2i}) \iff##

##\log (e^{1+2i}) = 1 + i \arg (e e^{2i})##

Here is the step that I am not entirely certain about. I know that ##e^{2i}## is the exponential representation of the complex point whose angle is ##2##. But I am wondering, would the coefficient ##e## just represent the distance of the complex point from the origin? More generally, the complex exponential function ##e^z## is defined as ##e^z = e^x e^{iy}##, and so ##e^x## would represent the distance of the complex point ##e^{iy}##?

If this is so, then the answer would seem to be

##\log (e^{1+2i}) = 1 + i(2 + 2 n \pi)##

Is this correct?
Yes, but it is rather a long way about. You can straight away write down that one value of ln(ez) is z. The only question is what other values are there. If w is another value then w = ln(ez). Exponentiate both sides and see what develops.
 
I am not certain I see where the quantity ##\ln e^z## arises. If recall correctly, my professor said that we do not use the natural logarithm when dealing with complex numbers.
 
Bashyboy said:
I am not certain I see where the quantity ##\ln e^z## arises. If recall correctly, my professor said that we do not use the natural logarithm when dealing with complex numbers.
OK, I see - you have to start with the definition you quoted, and your method is fine.
 

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