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Lösung 3.3:4c

Aus Online Mathematik Brückenkurs 2

(Unterschied zwischen Versionen)
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K (Lösning 3.3:4c moved to Solution 3.3:4c: Robot: moved page)
Zeile 1: Zeile 1:
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{{NAVCONTENT_START}}
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We complete the square on the left-hand side:
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<center> [[Image:3_3_4c.gif]] </center>
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{{NAVCONTENT_STOP}}
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<math>\begin{align}
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& \left( z+1 \right)^{\text{2}}-1^{2}+3=0 \\
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& \left( z+1 \right)^{\text{2}}+2=0 \\
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\end{align}</math>
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Taking the root now gives
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<math>z+1=\pm i\sqrt{2}</math>
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i.e.
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<math>z=-1+i\sqrt{2}</math>
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and
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<math>z=-1-i\sqrt{2}</math>.
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We test the solutions in the equation to ascertain that we have calculated correctly.
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<math>\begin{align}
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& z=-1+i\sqrt{2}:\quad z^{2}+2z+3=\left( -1+i\sqrt{2} \right)^{2}+2\left( -1+i\sqrt{2} \right)+3 \\
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& =\left( -1 \right)^{2}-2\centerdot i\sqrt{2}+i^{2}\left( \sqrt{2} \right)^{2}-2+2i\sqrt{2}+3 \\
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& =1-2\centerdot i\sqrt{2}-2-2+2i\sqrt{2}+3=0, \\
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\end{align}</math>
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<math>\begin{align}
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& z=-1-i\sqrt{2}:\quad z^{2}+2z+3=\left( -1-i\sqrt{2} \right)^{2}+2\left( -1-i\sqrt{2} \right)+3 \\
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& =\left( -1 \right)^{2}+2\centerdot i\sqrt{2}+i^{2}\left( \sqrt{2} \right)^{2}-2-2i\sqrt{2}+3 \\
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& =1+2\centerdot i\sqrt{2}-2-2-2\sqrt{2}i+3=0, \\
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\end{align}</math>

Version vom 10:32, 25. Okt. 2008

We complete the square on the left-hand side:


z+12−12+3=0z+12+2=0


Taking the root now gives z+1=i2  i.e. z=−1+i2  and z=−1−i2 .

We test the solutions in the equation to ascertain that we have calculated correctly.


z=−1+i2:z2+2z+3=−1+i22+2−1+i2+3=−12−2i2+i222−2+2i2+3=1−2i2−2−2+2i2+3=0


z=−1−i2:z2+2z+3=−1−i22+2−1−i2+3=−12+2i2+i222−2−2i2+3=1+2i2−2−2−22i+3=0

Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_3.3:4c“