Lösung 3.1:1f
Aus Online Mathematik Brückenkurs 2
(Unterschied zwischen Versionen)
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Let's begin by calculating some powers of ''i'', | Let's begin by calculating some powers of ''i'', | ||
- | {{ | + | {{Abgesetzte Formel||<math>\begin{align} |
i^2 &= i\cdot i = -1\,,\\[5pt] | i^2 &= i\cdot i = -1\,,\\[5pt] | ||
i^3 &= i^2\cdot i = (-1)\cdot i = -i\,,\\[5pt] | i^3 &= i^2\cdot i = (-1)\cdot i = -i\,,\\[5pt] | ||
Zeile 9: | Zeile 9: | ||
Now, we observe that because <math>i^4=1</math>, we can try to factorize <math>i^{11}</math> and <math>i^{20}</math> in terms of <math>i^4</math>, | Now, we observe that because <math>i^4=1</math>, we can try to factorize <math>i^{11}</math> and <math>i^{20}</math> in terms of <math>i^4</math>, | ||
- | {{ | + | {{Abgesetzte Formel||<math>\begin{align} |
i^{11} &= i^{4+4+3} = i^4\cdot i^4\cdot i^3 = 1\cdot 1 \cdot (-i) = -i\,,\\[5pt] | i^{11} &= i^{4+4+3} = i^4\cdot i^4\cdot i^3 = 1\cdot 1 \cdot (-i) = -i\,,\\[5pt] | ||
i^{20} &= i^{4+4+4+4+4} = i^4\cdot i^4\cdot i^4\cdot i^4\cdot i^4 = 1\cdot 1 \cdot 1\cdot 1 \cdot 1 = 1\,\textrm{.} | i^{20} &= i^{4+4+4+4+4} = i^4\cdot i^4\cdot i^4\cdot i^4\cdot i^4 = 1\cdot 1 \cdot 1\cdot 1 \cdot 1 = 1\,\textrm{.} | ||
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The answer becomes | The answer becomes | ||
- | {{ | + | {{Abgesetzte Formel||<math>i^{20}+i^{11}=1-i\,\textrm{.}</math>}} |
Version vom 13:05, 10. Mär. 2009
Let's begin by calculating some powers of i,
\displaystyle \begin{align}
i^2 &= i\cdot i = -1\,,\\[5pt] i^3 &= i^2\cdot i = (-1)\cdot i = -i\,,\\[5pt] i^4 &= i^2\cdot i^2 = (-1)\cdot (-1) = 1\,\textrm{.} \end{align} |
Now, we observe that because \displaystyle i^4=1, we can try to factorize \displaystyle i^{11} and \displaystyle i^{20} in terms of \displaystyle i^4,
\displaystyle \begin{align}
i^{11} &= i^{4+4+3} = i^4\cdot i^4\cdot i^3 = 1\cdot 1 \cdot (-i) = -i\,,\\[5pt] i^{20} &= i^{4+4+4+4+4} = i^4\cdot i^4\cdot i^4\cdot i^4\cdot i^4 = 1\cdot 1 \cdot 1\cdot 1 \cdot 1 = 1\,\textrm{.} \end{align} |
The answer becomes
\displaystyle i^{20}+i^{11}=1-i\,\textrm{.} |