Lösung 3.1:4e

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If we multiply both sides by <math>z+i</math>, we avoid having <math>z</math> in the denominator,
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If we multiply both sides by z+i, we avoid having z in the denominator:
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{{Displayed math||<math>iz+1=(3+i)(z+i)\,\textrm{.}</math>}}
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At the same time, this means that we are now working with a new equation which is not necessarily entirely equivalent with the original equation. Should the new equation show itself to have <math>z=-i</math> as a solution, then we must ignore that solution, because our initial equation cannot possibly have <math>z=-i</math> as a solution (the denominator of the left-hand side becomes zero).
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<math>iz+1=(3+i)(z+i).</math>
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At the same time, this means that we are now working with a new equation which is not necessarily entirely equivalent with the original equation. Should the new equation show itself to have z=-i as a solution, then we must ignore that solution, because our initial equation cannot possibly have <math>z=-i</math> as a solution (the denominator of the left-hand side becomes zero).
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We expand the right-hand side in the new equation,
We expand the right-hand side in the new equation,
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{{Displayed math||<math>iz+1 = 3z+3i+iz-1\,,</math>}}
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<math>iz+1=3z+3i+iz-1</math>
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and move all the terms in <math>z</math> over to the left-hand side and the constants to the right-hand side,
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and move all the terms in z over to the left-hand side and the constants to the right-hand side:
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<math>\begin{align}iz-3z-iz &= 3i-1-1,\\
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-3z &= -2+3i.\end{align}</math>
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{{Displayed math||<math>\begin{align}
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iz-3z-iz &= 3i-1-1\,,\\[5pt]
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-3z &= -2+3i\,\textrm{.}
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\end{align}</math>}}
Then, we obtain
Then, we obtain
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{{Displayed math||<math>z = \frac{-2+3i}{-3} = \frac{2}{3}-i\,\textrm{.}</math>}}
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<math>z=\frac{-2+3i}{-3}=\frac{2}{3}-i.</math>
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It is a little troublesome to divide two complex numbers, so we will therefore not check whether <math>z=\tfrac{2}{3}-i</math> is a solution to the original equation, but satisfy ourselves with substituting into the equation <math>iz+1=(3+i)(z+i)</math>,
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It is a little troublesome to divide two complex numbers, so we will therefore not check whether <math>z=\frac{2}{3}-i</math> is a solution to the original equation, but satisfy ourselves with substituting into the equation <math>iz+1=(3+i)(z+i)</math>:
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<math>\begin{align}LHS &=iz+1=i(\frac{2}{3}-i)+1=\frac{2}{3}\cdot i+1+1=2+\frac{2}{3}i,\\
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RHS &= (3+i)(z+i)=(3+i)(\frac{2}{3}-i+i)=(3+i)\frac{2}{3}=2+\frac{2}{3}i.\end{align}</math>
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{{Displayed math||<math>\begin{align}
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\text{LHS} &= iz+1 = i(\tfrac{2}{3}-i)+1 = \tfrac{2}{3}\cdot i+1+1 = 2+\tfrac{2}{3}i,\\[5pt]
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\text{RHS} &= (3+i)(z+i) = (3+i)(\tfrac{2}{3}-i+i) = (3+i)\tfrac{2}{3} = 2+\tfrac{2}{3}i\,\textrm{.}
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\end{align}</math>}}

Version vom 08:20, 30. Okt. 2008

If we multiply both sides by \displaystyle z+i, we avoid having \displaystyle z in the denominator,

\displaystyle iz+1=(3+i)(z+i)\,\textrm{.}

At the same time, this means that we are now working with a new equation which is not necessarily entirely equivalent with the original equation. Should the new equation show itself to have \displaystyle z=-i as a solution, then we must ignore that solution, because our initial equation cannot possibly have \displaystyle z=-i as a solution (the denominator of the left-hand side becomes zero).

We expand the right-hand side in the new equation,

\displaystyle iz+1 = 3z+3i+iz-1\,,

and move all the terms in \displaystyle z over to the left-hand side and the constants to the right-hand side,

\displaystyle \begin{align}

iz-3z-iz &= 3i-1-1\,,\\[5pt] -3z &= -2+3i\,\textrm{.} \end{align}

Then, we obtain

\displaystyle z = \frac{-2+3i}{-3} = \frac{2}{3}-i\,\textrm{.}

It is a little troublesome to divide two complex numbers, so we will therefore not check whether \displaystyle z=\tfrac{2}{3}-i is a solution to the original equation, but satisfy ourselves with substituting into the equation \displaystyle iz+1=(3+i)(z+i),

\displaystyle \begin{align}

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

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