Lösung 3.2:1d

Aus Online Mathematik Brückenkurs 2

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If we calculate the expression, we get the answer at once,
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If we calculate the expression, we get the answer at once:
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<math>\begin{align}z-\bar{w}+u &= (2+i)-(2-3i)+(-1-2i)\\
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{{Displayed math||<math>\begin{align}
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&= 2-2-1+(1+3-2)i=-1+2i.\end{align}</math>
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z-\bar{w}+u
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&= (2+i)-(2-3i)+(-1-2i)\\[5pt]
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&= 2-2-1+(1+3-2)i\\[5pt]
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&= -1+2i\,\textrm{.}
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\end{align}</math>}}
If, on the other hand, we interpret the expression in terms of vectors, we must first understand the vector <math>\bar{w}</math> geometrically. When we take the complex conjugate of <math>w</math>, we change the sign of the imaginary part, which is the same as reflecting <math>w</math> in the real axis.
If, on the other hand, we interpret the expression in terms of vectors, we must first understand the vector <math>\bar{w}</math> geometrically. When we take the complex conjugate of <math>w</math>, we change the sign of the imaginary part, which is the same as reflecting <math>w</math> in the real axis.
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[[Image:3_2_1d-2(2).gif|center]]
[[Image:3_2_1d-2(2).gif|center]]
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Version vom 09:31, 29. Okt. 2008

If we calculate the expression, we get the answer at once,

\displaystyle \begin{align}

z-\bar{w}+u &= (2+i)-(2-3i)+(-1-2i)\\[5pt] &= 2-2-1+(1+3-2)i\\[5pt] &= -1+2i\,\textrm{.} \end{align}

If, on the other hand, we interpret the expression in terms of vectors, we must first understand the vector \displaystyle \bar{w} geometrically. When we take the complex conjugate of \displaystyle w, we change the sign of the imaginary part, which is the same as reflecting \displaystyle w in the real axis.

We can then construct the expression \displaystyle z-\bar{w}+u one term at a time.