Lösung 3.1:4d

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In the equation, z occurs only as and, to begin with, we can therefore treatas unknown.
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In the equation, <math>z</math> occurs only as <math>\bar{z}</math> and, to begin with, we can therefore treat <math>\bar{z}</math> as unknown.
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Divide both sides by 2+i,
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Divide both sides by <math>2+i</math>,
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<math>\bar{z}=\frac{1+i}{2+i}</math>
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EQ1
 
and calculate the quotient on the right-hand side by multiplying top and bottom by the complex conjugate of the numerator:
and calculate the quotient on the right-hand side by multiplying top and bottom by the complex conjugate of the numerator:
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EQ2
 
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This means that .
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<math>\begin{align}\bar{z}&=\frac{(1+i)(2-i)}{(2+i)(2-i)}=\frac{1\cdot 2-1\cdot i +i \cdot 2 - i\cdot i}{2^2-i^2}\\
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&=\frac{2-i+2i+1}{4+1}=\frac{3+i}{5}=\frac{3}{5}+\frac{1}{5}i.\end{align}</math>
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This means that <math>z=\frac{3}{5}-\frac{1}{5}i.</math>
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We check that satisfies the original equation:
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We check that <math>z=\frac{3}{5}-\frac{1}{5}i</math> satisfies the original equation:
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EQ3
 
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<math>\begin{align}LHS &= (2+i)\bar{z} = (2+i)\overline{(\frac{3}{5}-\frac{1}{5}i)}=(2+i)(\frac{3}{5}+\frac{1}{5}i)\\
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&=2\cdot\frac{3}{5}+2\cdot\frac{1}{5}i+i\cdot\frac{3}{5}+i\cdot\frac{1}{5}i=\frac{6}{5}+\frac{2}{5}i+\frac{3}{5}i-\frac{1}{5}\\
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&=\frac{6-1}{5}+\frac{2+3}{5}i=1+i= RHS\end{align}</math>
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Version vom 11:03, 23. Sep. 2008

In the equation, \displaystyle z occurs only as \displaystyle \bar{z} and, to begin with, we can therefore treat \displaystyle \bar{z} as unknown.

Divide both sides by \displaystyle 2+i,


\displaystyle \bar{z}=\frac{1+i}{2+i}


and calculate the quotient on the right-hand side by multiplying top and bottom by the complex conjugate of the numerator:


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


This means that \displaystyle z=\frac{3}{5}-\frac{1}{5}i.

We check that \displaystyle z=\frac{3}{5}-\frac{1}{5}i satisfies the original equation:


\displaystyle \begin{align}LHS &= (2+i)\bar{z} = (2+i)\overline{(\frac{3}{5}-\frac{1}{5}i)}=(2+i)(\frac{3}{5}+\frac{1}{5}i)\\ &=2\cdot\frac{3}{5}+2\cdot\frac{1}{5}i+i\cdot\frac{3}{5}+i\cdot\frac{1}{5}i=\frac{6}{5}+\frac{2}{5}i+\frac{3}{5}i-\frac{1}{5}\\ &=\frac{6-1}{5}+\frac{2+3}{5}i=1+i= RHS\end{align}

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