Lösung 3.1:4f

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{{NAVCONTENT_START}}
 
Because the equation contains both <math>z</math> and <math>\bar{z}</math>, we cannot use <math>z</math> (or <math>\bar{z}</math>) alone as an unknown, so we are forced to set
Because the equation contains both <math>z</math> and <math>\bar{z}</math>, we cannot use <math>z</math> (or <math>\bar{z}</math>) alone as an unknown, so we are forced to set
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{{Displayed math||<math>z=x+iy</math>}}
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<math>z=x+iy</math>
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and use the real part <math>x</math> and the imaginary part <math>y</math> as unknowns.
and use the real part <math>x</math> and the imaginary part <math>y</math> as unknowns.
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With this approach, the left-hand side of the equation becomes
With this approach, the left-hand side of the equation becomes
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{{Displayed math||<math>\begin{align}
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<math>(1+i)(x-iy)+i(x+iy)</math>
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(1+i)(x-iy)+i(x+iy)
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&= 1\cdot x -1\cdot iy +i\cdot x -i^2y + i\cdot x + i^2y\\[5pt]
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<math>
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&= x-iy+ix+y+ix-y\\[5pt]
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\begin{align} &=1\cdot x -1\cdot iy +i\cdot x -i^2y + i\cdot x + i^2y\\
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&= x+(2x-y)i
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&=x-iy+ix+y+ix-y\\
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\end{align}</math>}}
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&=x+(2x-y)i\end{align}</math>
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and the whole equation becomes
and the whole equation becomes
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{{Displayed math||<math>x+(2x-y)i=3+5i\,\textrm{.}</math>}}
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<math>x+(2x-y)i=3+5i</math>.
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The two complex numbers on the left- and right-hand sides are equal when both real and imaginary parts are equal, i.e.
The two complex numbers on the left- and right-hand sides are equal when both real and imaginary parts are equal, i.e.
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{{Displayed math||<math>\left\{\begin{align}x\phantom{{}-y}{}&=3\,,\\[5pt] 2x-y&=5\,\textrm{.}\end{align}\right.</math>}}
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<math>\begin{cases}x=3\\2x-y=5.\end{cases}</math>
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This gives <math>x=3</math> and <math>y=2x-5=2\cdot 3-5=1</math>. Thus, the equation has the solution <math>z=3+i</math>.
This gives <math>x=3</math> and <math>y=2x-5=2\cdot 3-5=1</math>. Thus, the equation has the solution <math>z=3+i</math>.
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A quick check shows that <math>z=3+i</math> satisfies the equation in the exercise:
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A quick check shows that <math>z=3+i</math> satisfies the equation in the exercise,
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<math>\begin{align}LHS &= (1+i)\bar{z}+iz=(1+i)(3-i)+i(3+i)\\
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&= 3-i+3i+1+3i-1 = 3+5i = RHS\end{align}</math>
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{{NAVCONTENT_STOP}}
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{{Displayed math||<math>\begin{align}
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\text{LHS}
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&= (1+i)\bar{z}+iz\\[5pt]
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&= (1+i)(3-i)+i(3+i)\\[5pt]
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&= 3-i+3i+1+3i-1\\[5pt]
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&= 3+5i\\[5pt]
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&= \text{RHS}\,\textrm{.}
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\end{align}</math>}}

Version vom 08:30, 30. Okt. 2008

Because the equation contains both \displaystyle z and \displaystyle \bar{z}, we cannot use \displaystyle z (or \displaystyle \bar{z}) alone as an unknown, so we are forced to set

\displaystyle z=x+iy

and use the real part \displaystyle x and the imaginary part \displaystyle y as unknowns.

With this approach, the left-hand side of the equation becomes

\displaystyle \begin{align}

(1+i)(x-iy)+i(x+iy) &= 1\cdot x -1\cdot iy +i\cdot x -i^2y + i\cdot x + i^2y\\[5pt] &= x-iy+ix+y+ix-y\\[5pt] &= x+(2x-y)i \end{align}

and the whole equation becomes

\displaystyle x+(2x-y)i=3+5i\,\textrm{.}

The two complex numbers on the left- and right-hand sides are equal when both real and imaginary parts are equal, i.e.

\displaystyle \left\{\begin{align}x\phantom{{}-y}{}&=3\,,\\[5pt] 2x-y&=5\,\textrm{.}\end{align}\right.

This gives \displaystyle x=3 and \displaystyle y=2x-5=2\cdot 3-5=1. Thus, the equation has the solution \displaystyle z=3+i.

A quick check shows that \displaystyle z=3+i satisfies the equation in the exercise,

\displaystyle \begin{align}

\text{LHS} &= (1+i)\bar{z}+iz\\[5pt] &= (1+i)(3-i)+i(3+i)\\[5pt] &= 3-i+3i+1+3i-1\\[5pt] &= 3+5i\\[5pt] &= \text{RHS}\,\textrm{.} \end{align}

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_3.1:4f“