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Lösung 3.1:4b

Aus Online Mathematik Brückenkurs 2

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K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
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If we divide both sides by <math>2-i</math>, we obtain <math>z</math> by itself on the left-hand side,
If we divide both sides by <math>2-i</math>, we obtain <math>z</math> by itself on the left-hand side,
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{{Displayed math||<math>z=\frac{3+2i}{2-i}\,\textrm{.}</math>}}
+
{{Abgesetzte Formel||<math>z=\frac{3+2i}{2-i}\,\textrm{.}</math>}}
It remains to calculate the quotient on the right-hand side. We multiply top and bottom by the complex conjugate of the denominator,
It remains to calculate the quotient on the right-hand side. We multiply top and bottom by the complex conjugate of the denominator,
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{{Displayed math||<math>\begin{align}
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{{Abgesetzte Formel||<math>\begin{align}
z &= \frac{(3+2i)(2+i)}{(2-i)(2+i)}\\[5pt]
z &= \frac{(3+2i)(2+i)}{(2-i)(2+i)}\\[5pt]
&= \frac{3\cdot 2+3\cdot i +2i\cdot 2+2i\cdot i}{2^2-i^2}\\[5pt]
&= \frac{3\cdot 2+3\cdot i +2i\cdot 2+2i\cdot i}{2^2-i^2}\\[5pt]
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Also, we substitute <math>z=\tfrac{4}{5}+\tfrac{7}{5}i</math> into the original equation to assure ourselves that we have calculated correctly,
Also, we substitute <math>z=\tfrac{4}{5}+\tfrac{7}{5}i</math> into the original equation to assure ourselves that we have calculated correctly,
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{{Displayed math||<math>\begin{align}
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{{Abgesetzte Formel||<math>\begin{align}
\text{LHS}
\text{LHS}
&= (2-i)z\\[5pt]
&= (2-i)z\\[5pt]

Version vom 13:06, 10. Mär. 2009

If we divide both sides by 2i, we obtain z by itself on the left-hand side,

z=2i3+2i.

It remains to calculate the quotient on the right-hand side. We multiply top and bottom by the complex conjugate of the denominator,

z=(2i)(2+i)(3+2i)(2+i)=22i232+3i+2i2+2ii=4+16+3i+4i2=54+7i=54+57i.

Also, we substitute z=54+57i into the original equation to assure ourselves that we have calculated correctly,

LHS=(2i)z=(2i)54+57i=254+257ii54i57i=58+514i54i+57=58+7+5144i=515+510i=3+2i=RHS.