Lösung 3.1:4d
Aus Online Mathematik Brückenkurs 2
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Version vom 10:33, 11. Mär. 2009
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 denominator,
| \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}\\[5pt] &= \frac{2-i+2i+1}{4+1} = \frac{3+i}{5} = \frac{3}{5}+\frac{1}{5}\,i\,\textrm{.} \end{align} |
This means that \displaystyle z=\tfrac{3}{5}-\tfrac{1}{5}i\,.
We check that \displaystyle z=\tfrac{3}{5}-\tfrac{1}{5}i satisfies the original equation,
| \displaystyle \begin{align}
\text{LHS} &= (2+i)\bar{z} = (2+i)\overline{\Bigl(\frac{3}{5}-\frac{1}{5}\,i\Bigr)} = (2+i)\Bigl(\frac{3}{5}+\frac{1}{5}\,i\Bigr)\\[5pt] &= 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}\\[5pt] &=\frac{6-1}{5}+\frac{2+3}{5}\,i = 1+i = \text{RHS}\,\textrm{.} \end{align} |
