Mixing
Exponential Form of Complex Numbers - Why e? [duplicate]
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This question already has an answer here:
How to prove Euler's formula: $e^{ivarphi}=cos(varphi) +isin(varphi)$?
17 answers
Please delete this question please. It is a duplicate. Thank you!!!!!! I cannot delete the question.
Thanks!
algebra-precalculus
asked Jan 25 at 0:27
M. C.M. C.
449
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marked as duplicate by Chris Culter, T. Bongers, Trevor Gunn, David K, Shailesh Jan 25 at 2:12
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
How to prove Euler's formula: $e^{ivarphi}=cos(varphi) +isin(varphi)$?
17 answers
Please delete this question please. It is a duplicate. Thank you!!!!!! I cannot delete the question.
Thanks!
algebra-precalculus
asked Jan 25 at 0:27
M. C.M. C.
449
$endgroup$
marked as duplicate by Chris Culter, T. Bongers, Trevor Gunn, David K, Shailesh Jan 25 at 2:12
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
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Are you asking for a proof of Euler's formula?
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– clathratus
Jan 25 at 0:31
1
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Have you done any research at all on how this is shown? Wikipedia, multiple questions here, any complex analysis book, most calculus books..... Please clarify what it is that you aren't sure about, rather than just asking for a rehashing of something that's quite available.
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– T. Bongers
Jan 25 at 0:44
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If anyone can delete this question, that would be great! It is a duplicate.
$endgroup$
– M. C.
Jan 28 at 21:36
add a comment |
$begingroup$
This question already has an answer here:
How to prove Euler's formula: $e^{ivarphi}=cos(varphi) +isin(varphi)$?
17 answers
Please delete this question please. It is a duplicate. Thank you!!!!!! I cannot delete the question.
Thanks!
algebra-precalculus
asked Jan 25 at 0:27
M. C.M. C.
449
$endgroup$
This question already has an answer here:
How to prove Euler's formula: $e^{ivarphi}=cos(varphi) +isin(varphi)$?
17 answers
Please delete this question please. It is a duplicate. Thank you!!!!!! I cannot delete the question.
Thanks!
This question already has an answer here:
How to prove Euler's formula: $e^{ivarphi}=cos(varphi) +isin(varphi)$?
17 answers
algebra-precalculus
algebra-precalculus
asked Jan 25 at 0:27
M. C.M. C.
449
asked Jan 25 at 0:27
M. C.M. C.
449
edited Jan 28 at 21:36
M. C.
asked Jan 25 at 0:27
M. C.M. C.
449
asked Jan 25 at 0:27
M. C.M. C.
449
asked Jan 25 at 0:27
M. C.M. C.
449
449
marked as duplicate by Chris Culter, T. Bongers, Trevor Gunn, David K, Shailesh Jan 25 at 2:12
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Chris Culter, T. Bongers, Trevor Gunn, David K, Shailesh Jan 25 at 2:12
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
$begingroup$
Are you asking for a proof of Euler's formula?
$endgroup$
– clathratus
Jan 25 at 0:31
1
$begingroup$
Have you done any research at all on how this is shown? Wikipedia, multiple questions here, any complex analysis book, most calculus books..... Please clarify what it is that you aren't sure about, rather than just asking for a rehashing of something that's quite available.
$endgroup$
– T. Bongers
Jan 25 at 0:44
$begingroup$
If anyone can delete this question, that would be great! It is a duplicate.
$endgroup$
– M. C.
Jan 28 at 21:36
add a comment |
1
$begingroup$
Are you asking for a proof of Euler's formula?
$endgroup$
– clathratus
Jan 25 at 0:31
1
$begingroup$
Have you done any research at all on how this is shown? Wikipedia, multiple questions here, any complex analysis book, most calculus books..... Please clarify what it is that you aren't sure about, rather than just asking for a rehashing of something that's quite available.
$endgroup$
– T. Bongers
Jan 25 at 0:44
$begingroup$
If anyone can delete this question, that would be great! It is a duplicate.
$endgroup$
– M. C.
Jan 28 at 21:36
1
1
$begingroup$
Are you asking for a proof of Euler's formula?
$endgroup$
– clathratus
Jan 25 at 0:31
$begingroup$
Are you asking for a proof of Euler's formula?
$endgroup$
– clathratus
Jan 25 at 0:31
1
1
$begingroup$
Have you done any research at all on how this is shown? Wikipedia, multiple questions here, any complex analysis book, most calculus books..... Please clarify what it is that you aren't sure about, rather than just asking for a rehashing of something that's quite available.
$endgroup$
– T. Bongers
Jan 25 at 0:44
$begingroup$
Have you done any research at all on how this is shown? Wikipedia, multiple questions here, any complex analysis book, most calculus books..... Please clarify what it is that you aren't sure about, rather than just asking for a rehashing of something that's quite available.
$endgroup$
– T. Bongers
Jan 25 at 0:44
$begingroup$
If anyone can delete this question, that would be great! It is a duplicate.
$endgroup$
– M. C.
Jan 28 at 21:36
$begingroup$
If anyone can delete this question, that would be great! It is a duplicate.
$endgroup$
– M. C.
Jan 28 at 21:36
add a comment |
1 Answer
1
active
oldest
votes
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If you remember series, notice that
$$ e^{i x } = sum_{n geq 0} frac{ i^n x^n }{n!} $$
Now, notice that $i^2 = -1 $ but $i^{3} = -i$, and $i^4 = 1 $ and $i^5 = i$ so on, and since
$$ sin x = sum_{n geq 0} frac{ (-1)^n x^{2n+1 } }{(2n+1)!} ; ; text{and} ; ;cos x = sum_{n geq 0} frac{ (-1)^n x^{2n } }{(2n)!} $$
after breaking the $n$ in the first summation for even cases and odd cases and seeing in the third line how the $i's$ alternate, one obtains the result
answered Jan 25 at 0:32
Jimmy SabaterJimmy Sabater
3,023325
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If you remember series, notice that
$$ e^{i x } = sum_{n geq 0} frac{ i^n x^n }{n!} $$
Now, notice that $i^2 = -1 $ but $i^{3} = -i$, and $i^4 = 1 $ and $i^5 = i$ so on, and since
$$ sin x = sum_{n geq 0} frac{ (-1)^n x^{2n+1 } }{(2n+1)!} ; ; text{and} ; ;cos x = sum_{n geq 0} frac{ (-1)^n x^{2n } }{(2n)!} $$
after breaking the $n$ in the first summation for even cases and odd cases and seeing in the third line how the $i's$ alternate, one obtains the result
answered Jan 25 at 0:32
Jimmy SabaterJimmy Sabater
3,023325
$endgroup$
add a comment |
$begingroup$
If you remember series, notice that
$$ e^{i x } = sum_{n geq 0} frac{ i^n x^n }{n!} $$
Now, notice that $i^2 = -1 $ but $i^{3} = -i$, and $i^4 = 1 $ and $i^5 = i$ so on, and since
$$ sin x = sum_{n geq 0} frac{ (-1)^n x^{2n+1 } }{(2n+1)!} ; ; text{and} ; ;cos x = sum_{n geq 0} frac{ (-1)^n x^{2n } }{(2n)!} $$
after breaking the $n$ in the first summation for even cases and odd cases and seeing in the third line how the $i's$ alternate, one obtains the result
answered Jan 25 at 0:32
Jimmy SabaterJimmy Sabater
3,023325
$endgroup$
add a comment |
$begingroup$
If you remember series, notice that
$$ e^{i x } = sum_{n geq 0} frac{ i^n x^n }{n!} $$
Now, notice that $i^2 = -1 $ but $i^{3} = -i$, and $i^4 = 1 $ and $i^5 = i$ so on, and since
$$ sin x = sum_{n geq 0} frac{ (-1)^n x^{2n+1 } }{(2n+1)!} ; ; text{and} ; ;cos x = sum_{n geq 0} frac{ (-1)^n x^{2n } }{(2n)!} $$
after breaking the $n$ in the first summation for even cases and odd cases and seeing in the third line how the $i's$ alternate, one obtains the result
answered Jan 25 at 0:32
Jimmy SabaterJimmy Sabater
3,023325
$endgroup$
If you remember series, notice that
$$ e^{i x } = sum_{n geq 0} frac{ i^n x^n }{n!} $$
Now, notice that $i^2 = -1 $ but $i^{3} = -i$, and $i^4 = 1 $ and $i^5 = i$ so on, and since
$$ sin x = sum_{n geq 0} frac{ (-1)^n x^{2n+1 } }{(2n+1)!} ; ; text{and} ; ;cos x = sum_{n geq 0} frac{ (-1)^n x^{2n } }{(2n)!} $$
after breaking the $n$ in the first summation for even cases and odd cases and seeing in the third line how the $i's$ alternate, one obtains the result
answered Jan 25 at 0:32
Jimmy SabaterJimmy Sabater
3,023325
answered Jan 25 at 0:32
Jimmy SabaterJimmy Sabater
3,023325
answered Jan 25 at 0:32
Jimmy SabaterJimmy Sabater
3,023325
answered Jan 25 at 0:32
Jimmy SabaterJimmy Sabater
3,023325
3,023325
add a comment |
add a comment |
1
$begingroup$
Are you asking for a proof of Euler's formula?
$endgroup$
– clathratus
Jan 25 at 0:31
1
$begingroup$
Have you done any research at all on how this is shown? Wikipedia, multiple questions here, any complex analysis book, most calculus books..... Please clarify what it is that you aren't sure about, rather than just asking for a rehashing of something that's quite available.
$endgroup$
– T. Bongers
Jan 25 at 0:44
$begingroup$
If anyone can delete this question, that would be great! It is a duplicate.
$endgroup$
– M. C.
Jan 28 at 21:36