Lösung 3.2:2e

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Because the expression contains both <math>z</math> and <math>\bar{z}</math>, we write out <math>z=x+iy</math>, where <math>x</math> is the real part of <math>z</math> and <math>y</math> is the imaginary part. Thus, we have
Because the expression contains both <math>z</math> and <math>\bar{z}</math>, we write out <math>z=x+iy</math>, where <math>x</math> is the real part of <math>z</math> and <math>y</math> is the imaginary part. Thus, we have
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<math>\mathrm{Re}z=x</math>
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:*<math>\mathop{\rm Re}z = x</math>
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:*<math>i+\bar{z} = i+(x-iy) = x+(1-y)i</math>
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<math>i+\bar{z}=i+(x-iy)=x+(1-y)i</math>
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and the condition becomes
and the condition becomes
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<math>x=x+(1-y)i \iff 0=(1-y)i</math>
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{{Displayed math||<math>x=x+(1-y)i\quad\Leftrightarrow\quad 0=(1-y)i</math>}}
which means that <math>y=1</math>.
which means that <math>y=1</math>.
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The set therefore consists of all complex numbers with imaginary part <math>1</math>.
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The set therefore consists of all complex numbers with imaginary part 1.
[[Image:3_2_2_e.gif|center]]
[[Image:3_2_2_e.gif|center]]
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Version vom 09:44, 29. Okt. 2008

Because the expression contains both \displaystyle z and \displaystyle \bar{z}, we write out \displaystyle z=x+iy, where \displaystyle x is the real part of \displaystyle z and \displaystyle y is the imaginary part. Thus, we have

  • \displaystyle \mathop{\rm Re}z = x
  • \displaystyle i+\bar{z} = i+(x-iy) = x+(1-y)i

and the condition becomes

\displaystyle x=x+(1-y)i\quad\Leftrightarrow\quad 0=(1-y)i

which means that \displaystyle y=1.

The set therefore consists of all complex numbers with imaginary part 1.

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_3.2:2e“