Lösung 3.1:1e
Aus Online Mathematik Brückenkurs 2
(Unterschied zwischen Versionen)
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| Zeile 1: | Zeile 1: | ||
A suitable first step can be to work out the square term, <math>(2-i)^2</math>, by expanding it, | A suitable first step can be to work out the square term, <math>(2-i)^2</math>, by expanding it, | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\begin{align} |
(2-i)^2 &= 2^2 - 2\cdot 2i + i^2\\[5pt] | (2-i)^2 &= 2^2 - 2\cdot 2i + i^2\\[5pt] | ||
&= 4-4i+i^2\\[5pt] | &= 4-4i+i^2\\[5pt] | ||
| Zeile 10: | Zeile 10: | ||
After that, we calculate the remaining product, | After that, we calculate the remaining product, | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\begin{align} |
(1+i)(3-4i) &= 1\cdot3 - 1\cdot 4i + i\cdot 3 - i\cdot 4i\\[5pt] | (1+i)(3-4i) &= 1\cdot3 - 1\cdot 4i + i\cdot 3 - i\cdot 4i\\[5pt] | ||
&= 3-4i+3i-4i^2\\[5pt] | &= 3-4i+3i-4i^2\\[5pt] | ||
Version vom 13:05, 10. Mär. 2009
A suitable first step can be to work out the square term, \displaystyle (2-i)^2, by expanding it,
| \displaystyle \begin{align}
(2-i)^2 &= 2^2 - 2\cdot 2i + i^2\\[5pt] &= 4-4i+i^2\\[5pt] &= 4-4i-1\\[5pt] &= 3-4i\,\textrm{.} \end{align} |
After that, we calculate the remaining product,
| \displaystyle \begin{align}
(1+i)(3-4i) &= 1\cdot3 - 1\cdot 4i + i\cdot 3 - i\cdot 4i\\[5pt] &= 3-4i+3i-4i^2\\[5pt] &= 3+(-4+3)i-4\cdot (-1)\\[5pt] &= 3-i+4\\[5pt] &= 7-i\,\textrm{.} \end{align} |
