Lösung 3.1:1f

Aus Online Mathematik Brückenkurs 2

(Unterschied zwischen Versionen)
Wechseln zu: Navigation, Suche
K
Zeile 1: Zeile 1:
-
{{NAVCONTENT_START}}
+
Let's begin by calculating some powers of ''i'',
-
Let's begin by calculating some powers of i:
+
-
<math>\begin{align}i^2&=i\cdot i=-1,\\
+
{{Displayed math||<math>\begin{align}
-
i^3&=i^2\cdot i = (-1)\cdot i = -i,\\
+
i^2 &= i\cdot i = -1\,,\\[5pt]
-
i^4&=i^2\cdot i^2 = (-1)\cdot (-1) = 1.\end{align}</math>
+
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}</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>,
+
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>,
-
<math>\begin{align}i^2&=i\cdot i=-1,\\
+
{{Displayed math||<math>\begin{align}
-
i^{11}&=i^{4+4+3} = i^4\cdot i^4\cdot i^3 = 1\cdot 1 \cdot (-i)=-i\\
+
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\end{align}</math>
+
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}</math>}}
The answer becomes
The answer becomes
-
<math>i^{20}+i^{11}=1-i</math>
+
{{Displayed math||<math>i^{20}+i^{11}=1-i\,\textrm{.}</math>}}
-
 
+
-
{{NAVCONTENT_STOP}}
+

Version vom 15:07, 29. Okt. 2008

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