Processing Math: Done
Lösung 3.1:4d
Aus Online Mathematik Brückenkurs 2
(Unterschied zwischen Versionen)
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- | In the equation, z occurs only as and, to begin with, we can therefore | + | 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. |
- | Divide both sides by 2+i, | + | Divide both sides by <math>2+i</math>, |
+ | |||
+ | |||
+ | <math>\bar{z}=\frac{1+i}{2+i}</math> | ||
- | 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: | ||
- | EQ2 | ||
- | This means that | + | <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}\\ |
+ | &=\frac{2-i+2i+1}{4+1}=\frac{3+i}{5}=\frac{3}{5}+\frac{1}{5}i.\end{align}</math> | ||
+ | |||
+ | |||
+ | This means that <math>z=\frac{3}{5}-\frac{1}{5}i.</math> | ||
- | We check that satisfies the original equation: | + | We check that <math>z=\frac{3}{5}-\frac{1}{5}i</math> satisfies the original equation: |
- | EQ3 | ||
+ | <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)\\ | ||
+ | &=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}</math> | ||
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Version vom 11:03, 23. Sep. 2008
In the equation,
Divide both sides by
=2+i1+i
and calculate the quotient on the right-hand side by multiplying top and bottom by the complex conjugate of the numerator:
=(2+i)(2−i)(1+i)(2−i)=22−i21
2−1
i+i
2−i
i=4+12−i+2i+1=53+i=53+51i
This means that
We check that
=(2+i)(53−51i)=(2+i)(53+51i)=2
53+2
51i+i
53+i
51i=56+52i+53i−51=56−1+52+3i=1+i=RHS