Lösung 3.1:4c

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<center> [[Image:3_1_4c.gif]] </center>
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If we subtract <math>2z</math> from both sides,
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<math>iz+2-2z=-3</math>
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and then subtract <math>2</math> from both sides, we have <math>z</math> terms left only on the left-hand side,
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<math>iz-2z=-3-2</math>
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After taking out a factor <math>z</math> from the left-hand side,
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<math>(i-2)z=-5</math>
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we obtain, after dividing by <math>-2+i</math>,
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<math>\begin{align}z&=\frac{-5}{-2+i}=\frac{-5(-2-i)}{(-2+i)(-2-i)}=\frac{(-5)\cdot(-2)-5\cdot(-i)}{(-2)^2-i^2}\\
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&=\frac{10+5i}{4+1}=\frac{10+5i}{5}=2+i.\end{align}</math>
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A quick check shows that <math>z=2+i</math> satisfies the original equation
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<math>\begin{align}LHS &= iz+2=i(2+i)+2=2i-1+2=1+2i\\
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RHS &= 2z-3 = 2(2+i)-3=4+2i-3=1+2i.\end{align}</math>
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Version vom 10:51, 23. Sep. 2008

If we subtract \displaystyle 2z from both sides,


\displaystyle iz+2-2z=-3


and then subtract \displaystyle 2 from both sides, we have \displaystyle z terms left only on the left-hand side,


\displaystyle iz-2z=-3-2


After taking out a factor \displaystyle z from the left-hand side,


\displaystyle (i-2)z=-5


we obtain, after dividing by \displaystyle -2+i,


\displaystyle \begin{align}z&=\frac{-5}{-2+i}=\frac{-5(-2-i)}{(-2+i)(-2-i)}=\frac{(-5)\cdot(-2)-5\cdot(-i)}{(-2)^2-i^2}\\ &=\frac{10+5i}{4+1}=\frac{10+5i}{5}=2+i.\end{align}


A quick check shows that \displaystyle z=2+i satisfies the original equation


\displaystyle \begin{align}LHS &= iz+2=i(2+i)+2=2i-1+2=1+2i\\ RHS &= 2z-3 = 2(2+i)-3=4+2i-3=1+2i.\end{align}

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