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2bornot2b
 4 years ago
What is the difference between exp(z) and \(e^z\). z is complex.
2bornot2b
 4 years ago
What is the difference between exp(z) and \(e^z\). z is complex.

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I didn't know there was any

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0notation i think unless i am sadly mistaken they are the same

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.0no difference  they are synonyms

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1since exponents can get pretty messy; the exp(...) notation is for clarity

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1the of exp(..) as a similar notation to log(...)

2bornot2b
 4 years ago
Best ResponseYou've already chosen the best response.0I have a book which defines exp(z) as \(e^x(cos y +isiny) \) where z=x+iy and it defines \(e^z~as~exp(z~Log~e)\) where Log is used to denote the multivalued Logarithmic function

2bornot2b
 4 years ago
Best ResponseYou've already chosen the best response.0\(Logz=e^{logr}+i\theta +2ni\pi\) where \(z=r(cos\theta+isin\theta)\)

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1http://math.furman.edu/~dcs/courses/math39/lectures/lecture17.pdf this might be useful

2bornot2b
 4 years ago
Best ResponseYou've already chosen the best response.0So Log e is a multivalued function. It has an infinite number of values and 1 is one of the values it takes.

2bornot2b
 4 years ago
Best ResponseYou've already chosen the best response.0Can you all guide me somewhere, like a book or may be an user of OS, who can help on this

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0actually as i recall the notation Log is single values whereas log is multivalued, but i could be wrong

2bornot2b
 4 years ago
Best ResponseYou've already chosen the best response.0Yes, right, that is why I mentioned it

2bornot2b
 4 years ago
Best ResponseYou've already chosen the best response.0So as to avoid any confusion. @satellite73 you are right, but some books write it the other way round.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\log(z)=\ln(z)+i\theta\] for any \(\theta\) whereas \[Log(z)=\ln(x)+i\theta\] for \[\pi\leq\theta\leq\pi\]

2bornot2b
 4 years ago
Best ResponseYou've already chosen the best response.0@Zarkon can you help please?

2bornot2b
 4 years ago
Best ResponseYou've already chosen the best response.0@amistre64 You have provided a great link. So I choose your answer as the best answer.
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