Lösung 3.1:4c

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If we subtract <math>2z</math> from both sides,
If we subtract <math>2z</math> from both sides,
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{{Displayed math||<math>iz+2-2z=-3</math>}}
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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,
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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{{Displayed math||<math>iz-2z=-3-2\,\textrm{.}</math>}}
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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,
After taking out a factor <math>z</math> from the left-hand side,
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{{Displayed math||<math>(i-2)z=-5\,,</math>}}
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<math>(i-2)z=-5</math>
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we obtain, after dividing by <math>-2+i</math>,
we obtain, after dividing by <math>-2+i</math>,
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{{Displayed math||<math>\begin{align}
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z &= \frac{-5}{-2+i}
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= \frac{-5(-2-i)}{(-2+i)(-2-i)}
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= \frac{(-5)\cdot(-2)-5\cdot(-i)}{(-2)^2-i^2}\\[5pt]
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&= \frac{10+5i}{4+1}
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= \frac{10+5i}{5}
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= 2+i\,\textrm{.}\end{align}</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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A quick check shows that <math>z=2+i</math> satisfies the original equation,
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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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{{Displayed math||<math>\begin{align}
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\text{LHS} &= iz+2 = i(2+i)+2 = 2i-1+2 = 1+2i\,,\\[5pt]
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\text{RHS} &= 2z-3 = 2(2+i)-3 = 4+2i-3 = 1+2i\,\textrm{.}
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\end{align}</math>}}

Version vom 07:55, 30. Okt. 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\,\textrm{.}

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}\\[5pt] &= \frac{10+5i}{4+1} = \frac{10+5i}{5} = 2+i\,\textrm{.}\end{align}

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

\displaystyle \begin{align}

\text{LHS} &= iz+2 = i(2+i)+2 = 2i-1+2 = 1+2i\,,\\[5pt] \text{RHS} &= 2z-3 = 2(2+i)-3 = 4+2i-3 = 1+2i\,\textrm{.} \end{align}

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