- #1
Bashyboy
- 1,421
- 5
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?
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?
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