Lösung 3.1:1d

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K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
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We expand the expression by multiplying each term in the first bracket with every term in the second bracket,
We expand the expression by multiplying each term in the first bracket with every term in the second bracket,
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{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
(3-2i)(7+5i)
(3-2i)(7+5i)
&= 3\cdot 7 + 3 \cdot 5i + \cdots\\[5pt]
&= 3\cdot 7 + 3 \cdot 5i + \cdots\\[5pt]
Zeile 9: Zeile 9:
Then, we use that <math>i^2=-1</math> and write the real and imaginary parts together,
Then, we use that <math>i^2=-1</math> and write the real and imaginary parts together,
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{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
(3-2i)(7+5i)
(3-2i)(7+5i)
&=21+15i-14i-10i^2\\[5pt]
&=21+15i-14i-10i^2\\[5pt]

Version vom 13:05, 10. Mär. 2009

We expand the expression by multiplying each term in the first bracket with every term in the second bracket,

\displaystyle \begin{align}

(3-2i)(7+5i) &= 3\cdot 7 + 3 \cdot 5i + \cdots\\[5pt] &= 3\cdot 7 + 3 \cdot 5i - 2i\cdot 7 - 2i \cdot 5i\,\textrm{.} \end{align}

Then, we use that \displaystyle i^2=-1 and write the real and imaginary parts together,

\displaystyle \begin{align}

(3-2i)(7+5i) &=21+15i-14i-10i^2\\[5pt] &=21+15i-14i-10\cdot(-1)\\[5pt] &=(21+10)+(15i-14i)\\[5pt] &=31+(15-14)i\\[5pt] &=31+i\,\textrm{.} \end{align}

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