Lösung 3.1:4f

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<center> [[Image:3_1_4f-1(2).gif]] </center>
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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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<center> [[Image:3_1_4f-2(2).gif]] </center>
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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.
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With this approach, the left-hand side of the equation becomes
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<math>(1+i)(x-iy)+i(x+iy)
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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-iy+ix+y+ix-y\\
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&=x+(2x-y)i\end{align}</math>
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and the whole equation becomes
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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.
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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>.
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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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Version vom 11:45, 23. Sep. 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 (1+i)(x-iy)+i(x+iy) \begin{align} &=1\cdot x -1\cdot iy +i\cdot x -i^2y + i\cdot x + i^2y\\ &=x-iy+ix+y+ix-y\\ &=x+(2x-y)i\end{align}


and the whole equation becomes


\displaystyle x+(2x-y)i=3+5i.


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


\displaystyle \begin{cases}x=3\\2x-y=5\end{cases}.


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}LHS &= (1+i)\bar{z}+iz=(1+i)(3-i)+i(3+i)\\ &= 3-i+3i+1+3i-1 = 3+5i = RHS\end{align}

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