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Lösung 3.1:2d

Aus Online Mathematik Brückenkurs 2

(Unterschied zwischen Versionen)
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K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
Zeile 1: Zeile 1:
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,
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{{Displayed math||<math>\begin{align}
+
{{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,
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{{Displayed math||<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>}}
+
{{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
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{{Displayed math||<math>\begin{align}
+
{{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]
Zeile 27: Zeile 27:
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,
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{{Displayed math||<math>\begin{align}
+
{{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,
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{{Displayed math||<math>\begin{align}
+
{{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
-
{{Displayed math||<math>\begin{align}
+
{{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,

5−11+i3i+i2−3i=1+i5(1+i)−11+i=1+i5+5i−1=1+i4+5i=2−3i3i(2−3i)+i2−3i=2−3i6i−9i2+i=2−3i9+7i.

Hence,

5−11+i3i+i2−3i= 2−3i9+7i1+i4+5i=(9+7i)(1+i)(4+5i)(2−3i).

We multiply out the numerator and denominator

(9+7i)(1+i)(4+5i)(2−3i)=91+9i+7i1+7ii42−43i+5i2−5i3i=9+9i+7i−78−12i+10i+15=2+16i23−2i.

This is an ordinary quotient of two complex numbers which we compute by multiplying top and bottom by the complex conjugate of the numerator,

2+16i23−2i=(2+16i)(2−16i)(23−2i)(2−16i)=22−(16i)2232−2316i−2i2+2i16i=4+25646−368i−4i−32=26014−372i.

If we divide up the numbers into factors,

14372260=27=2186=2293=22331=1026=25132

we can simplify the answers

14260−260372i=2722513−2251322331i=72513−513331i=7130−6593i.
Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_3.1:2d“