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Lösung 3.1:4b

Aus Online Mathematik Brückenkurs 2

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Also, we substitute into the original equation to assure ourselves that we have calculated correctly:
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Also, we substitute <math>z=\frac{4}{5}+\frac{7}{5}i</math> into the original equation to assure ourselves that we have calculated correctly:

Version vom 10:53, 23. Sep. 2008

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


z=2−i3+2i


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


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


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


LHS=(2−i)z=(2−i)(54+57i)=254−i54+257i−i57i=58−54i+514i+57=58+7+514−4i=515+510i=3+2i=RHS

Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_3.1:4b“