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Mesaj |
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Fie
\[
\varepsilon_{k}=\cos\frac{2k\pi}{n}+i\sin\frac{2k\pi}{n},~k=0,1,\ldots,n-1.
\]
S\u{a} se arate c\u{a}
\[
\overline{\varepsilon_{k}}=\varepsilon_{n-k}.
\]
[/equation]
Va roog
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[Citat]
Fie
\[
\varepsilon_{k}=\cos\frac{2k\pi}{n}+i\sin\frac{2k\pi}{n},~k=0,1,\ldots,n-1.
\]
S\u{a} se arate c\u{a}
\[
\overline{\varepsilon_{k}}=\varepsilon_{n-k}.
\]
[/equation]
|
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|
|
\[
\varepsilon_{k}=\cos\frac{2k\pi}{n}+i\sin\frac{2k\pi}{n},~k=0,1,\ldots,n-1.
\]
S\u{a} se arate c\u{a}
\[
\overline{\varepsilon_{k}}=\varepsilon_{n-k}.
\]
[/equation]
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---
C.Telteu
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Iar
Calculam acum raportul:
---
C.Telteu
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Access general
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