Lösung 3.1:1f

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<center> [[Image:3_1_1f.gif]] </center>
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Let's begin by calculating some powers of i:
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<math>\begin{align}i^2&=i\cdot i=-1,\\
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i^3&=i^2\cdot i = (-1)\cdot i = -i,\\
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i^4&=i^2\cdot i^2 = (-1)\cdot (-1) = 1.\end{align}</math>
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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>,
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<math>\begin{align}i^2&=i\cdot i=-1,\\
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i^{11}&=i^{4+4+3} = i^4\cdot i^4\cdot i^3 = 1\cdot 1 \cdot (-i)=-i\\
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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\end{align}</math>
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The answer becomes
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<math>i^{20}+i^{11}=1-i</math>
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Version vom 12:05, 18. Sep. 2008

Let's begin by calculating some powers of i:

\displaystyle \begin{align}i^2&=i\cdot i=-1,\\ i^3&=i^2\cdot i = (-1)\cdot i = -i,\\ i^4&=i^2\cdot i^2 = (-1)\cdot (-1) = 1.\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^2&=i\cdot i=-1,\\ i^{11}&=i^{4+4+3} = i^4\cdot i^4\cdot i^3 = 1\cdot 1 \cdot (-i)=-i\\ 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\end{align}

The answer becomes

\displaystyle i^{20}+i^{11}=1-i

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_3.1:1f“