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Lösung 3.3:3d

Aus Online Mathematik Brückenkurs 2

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K (Lösning 3.3:3d moved to Solution 3.3:3d: Robot: moved page)
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{{NAVCONTENT_START}}
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Before we can complete the square of the expression, we need to take out the factor
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<center> [[Image:3_3_3d.gif]] </center>
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<math>i</math>
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{{NAVCONTENT_STOP}}
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in front of
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<math>z^{2}</math>
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<math>i\left( z^{2}+\frac{2+3i}{i}z-\frac{1}{i} \right).</math>
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Then, simplify the complex fractions by multiplying top and bottom by
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<math>-i</math>
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(the denominator's complex conjugate):
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<math>\begin{align}
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& i\left( z^{2}+\frac{\left( 2+3i \right)\centerdot \left( -i \right)}{i\centerdot \left( -i \right)}z-\frac{1\centerdot \left( -i \right)}{i\centerdot \left( -i \right)} \right) \\
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& =i\left( z^{2}+\frac{-2i+3}{1}z-\frac{-i}{1} \right) \\
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& =i\left( z^{2}+\left( 3-2i \right)z+i \right). \\
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\end{align}</math>
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Now we are ready to complete the square of the second-degree expression inside the bracket:
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<math>\begin{align}
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& i\left( z^{2}+\left( 3-2i \right)z+i \right)=i\left( \left( z+\frac{3-2i}{2} \right)^{2}-\left( \frac{3-2i}{2} \right)^{2}+i \right) \\
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& =i\left( \left( z+\frac{3}{2}-i \right)^{2}-\left( \frac{3}{2}-i \right)^{2}+i \right) \\
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& =i\left( \left( z+\frac{3}{2}-i \right)^{2}-\frac{9}{4}+3i+1+i \right) \\
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& =i\left( \left( z+\frac{3}{2}-i \right)^{2}-\frac{5}{4}+4i \right) \\
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& =\left( z+\frac{3}{2}-i \right)^{2}-\frac{5}{4}i+4i^{2} \\
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& =\left( z+\frac{3}{2}-i \right)^{2}-4-\frac{5}{4}i. \\
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\end{align}</math>

Version vom 09:05, 25. Okt. 2008

Before we can complete the square of the expression, we need to take out the factor i in front of z2


iz2+i2+3izi1 


Then, simplify the complex fractions by multiplying top and bottom by i (the denominator's complex conjugate):


iz2+ii2+3iizii1i=iz2+12i+3z1i=iz2+32iz+i


Now we are ready to complete the square of the second-degree expression inside the bracket:


iz2+32iz+i=iz+232i2232i2+i=iz+23i223i2+i=iz+23i249+3i+1+i=iz+23i245+4i=z+23i245i+4i2=z+23i2445i