Lösung 3.3:3d

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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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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 \displaystyle i in front of \displaystyle z^{2}


\displaystyle i\left( z^{2}+\frac{2+3i}{i}z-\frac{1}{i} \right).


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


\displaystyle \begin{align} & 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) \\ & =i\left( z^{2}+\frac{-2i+3}{1}z-\frac{-i}{1} \right) \\ & =i\left( z^{2}+\left( 3-2i \right)z+i \right). \\ \end{align}


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


\displaystyle \begin{align} & 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) \\ & =i\left( \left( z+\frac{3}{2}-i \right)^{2}-\left( \frac{3}{2}-i \right)^{2}+i \right) \\ & =i\left( \left( z+\frac{3}{2}-i \right)^{2}-\frac{9}{4}+3i+1+i \right) \\ & =i\left( \left( z+\frac{3}{2}-i \right)^{2}-\frac{5}{4}+4i \right) \\ & =\left( z+\frac{3}{2}-i \right)^{2}-\frac{5}{4}i+4i^{2} \\ & =\left( z+\frac{3}{2}-i \right)^{2}-4-\frac{5}{4}i. \\ \end{align}

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