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What is the difference between exp(z) and \(e^z\). z is complex.
 one year ago
 one year ago
What is the difference between exp(z) and \(e^z\). z is complex.
 one year ago
 one year ago

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imranmeah91Best ResponseYou've already chosen the best response.0
I didn't know there was any
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
notation i think unless i am sadly mistaken they are the same
 one year ago

asnaseerBest ResponseYou've already chosen the best response.0
no difference  they are synonyms
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
since exponents can get pretty messy; the exp(...) notation is for clarity
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
the of exp(..) as a similar notation to log(...)
 one year ago

2bornot2bBest ResponseYou've already chosen the best response.0
I 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
 one year ago

2bornot2bBest ResponseYou've already chosen the best response.0
\(Logz=e^{logr}+i\theta +2ni\pi\) where \(z=r(cos\theta+isin\theta)\)
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
http://math.furman.edu/~dcs/courses/math39/lectures/lecture17.pdf this might be useful
 one year ago

2bornot2bBest ResponseYou've already chosen the best response.0
So Log e is a multivalued function. It has an infinite number of values and 1 is one of the values it takes.
 one year ago

2bornot2bBest ResponseYou've already chosen the best response.0
Can you all guide me somewhere, like a book or may be an user of OS, who can help on this
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
actually as i recall the notation Log is single values whereas log is multivalued, but i could be wrong
 one year ago

2bornot2bBest ResponseYou've already chosen the best response.0
Yes, right, that is why I mentioned it
 one year ago

2bornot2bBest ResponseYou've already chosen the best response.0
So as to avoid any confusion. @satellite73 you are right, but some books write it the other way round.
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
\[\log(z)=\ln(z)+i\theta\] for any \(\theta\) whereas \[Log(z)=\ln(x)+i\theta\] for \[\pi\leq\theta\leq\pi\]
 one year ago

2bornot2bBest ResponseYou've already chosen the best response.0
@Zarkon can you help please?
 one year ago

2bornot2bBest ResponseYou've already chosen the best response.0
@amistre64 You have provided a great link. So I choose your answer as the best answer.
 one year ago
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