Processing Math: Done
Lösung 3.1:2d
Aus Online Mathematik Brückenkurs 2
(Unterschied zwischen Versionen)
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Because the expression is rather large, we work step by step. We start by making the numerator and denominator each have the same denominator, | Because the expression is rather large, we work step by step. We start by making the numerator and denominator each have the same denominator, | ||
- | {{ | + | {{Abgesetzte Formel||<math>\begin{align} |
5-\frac{1}{1+i} | 5-\frac{1}{1+i} | ||
&=\frac{5\cdot (1+i)}{1+i}-\frac{1}{1+i} | &=\frac{5\cdot (1+i)}{1+i}-\frac{1}{1+i} | ||
Zeile 14: | Zeile 14: | ||
Hence, | Hence, | ||
- | {{ | + | {{Abgesetzte Formel||<math>\frac{5-\dfrac{1}{1+i}}{3i+\dfrac{i}{2-3i}} =\ \frac{\dfrac{4+5i}{1+i}}{\dfrac{9+7i}{2-3i}} = \frac{(4+5i)(2-3i)}{(9+7i)(1+i)}\,\textrm{.}</math>}} |
We multiply out the numerator and denominator | We multiply out the numerator and denominator | ||
- | {{ | + | {{Abgesetzte Formel||<math>\begin{align} |
\frac{(4+5i)(2-3i)}{(9+7i)(1+i)} | \frac{(4+5i)(2-3i)}{(9+7i)(1+i)} | ||
&= \frac{4\cdot 2 -4 \cdot 3i +5i\cdot 2 - 5i\cdot 3i}{9\cdot1+9\cdot i +7i\cdot 1 +7i \cdot i}\\[5pt] | &= \frac{4\cdot 2 -4 \cdot 3i +5i\cdot 2 - 5i\cdot 3i}{9\cdot1+9\cdot i +7i\cdot 1 +7i \cdot i}\\[5pt] | ||
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This is an ordinary quotient of two complex numbers which we compute by multiplying top and bottom by the complex conjugate of the numerator, | This is an ordinary quotient of two complex numbers which we compute by multiplying top and bottom by the complex conjugate of the numerator, | ||
- | {{ | + | {{Abgesetzte Formel||<math>\begin{align} |
\frac{23-2i}{2+16i} | \frac{23-2i}{2+16i} | ||
&= \frac{(23-2i)(2-16i)}{(2+16i)(2-16i)}\\[5pt] | &= \frac{(23-2i)(2-16i)}{(2+16i)(2-16i)}\\[5pt] | ||
Zeile 37: | Zeile 37: | ||
If we divide up the numbers into factors, | If we divide up the numbers into factors, | ||
- | {{ | + | {{Abgesetzte Formel||<math>\begin{align} |
14 &= 2\cdot 7\,,\\[5pt] | 14 &= 2\cdot 7\,,\\[5pt] | ||
372 &= 2\cdot 186=2\cdot 2\cdot 93=2\cdot 2\cdot 3\cdot 31\,,\\[5pt] | 372 &= 2\cdot 186=2\cdot 2\cdot 93=2\cdot 2\cdot 3\cdot 31\,,\\[5pt] | ||
Zeile 45: | Zeile 45: | ||
we can simplify the answers | we can simplify the answers | ||
- | {{ | + | {{Abgesetzte Formel||<math>\begin{align} |
\frac{14}{260}-\frac{372}{260}\,i | \frac{14}{260}-\frac{372}{260}\,i | ||
&= \frac{2\cdot 7}{2\cdot 2\cdot 5\cdot 13}-\frac{2\cdot 2\cdot 3\cdot 31}{2\cdot 2\cdot 5\cdot 13}\,i\\[5pt] | &= \frac{2\cdot 7}{2\cdot 2\cdot 5\cdot 13}-\frac{2\cdot 2\cdot 3\cdot 31}{2\cdot 2\cdot 5\cdot 13}\,i\\[5pt] |
Version vom 13:06, 10. Mär. 2009
Because the expression is rather large, we work step by step. We start by making the numerator and denominator each have the same denominator,
![]() ![]() |
Hence,
We multiply out the numerator and denominator
![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() |
This is an ordinary quotient of two complex numbers which we compute by multiplying top and bottom by the complex conjugate of the numerator,
![]() ![]() ![]() ![]() |
If we divide up the numbers into factors,
![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() |
we can simplify the answers
![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() |