Lösung 3.1:4e

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<center> [[Image:3_1_4e-1(2).gif]] </center>
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If we multiply both sides by z+i, we avoid having z in the denominator:
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<center> [[Image:3_1_4e-2(2).gif]] </center>
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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,
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<math>iz+1=3z+3i+iz-1</math>
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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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Then, we obtain
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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=\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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Version vom 11:35, 23. Sep. 2008

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


\displaystyle iz+1=(3+i)(z+i).


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 \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 z over to the left-hand side and the constants to the right-hand side:


\displaystyle \begin{align}iz-3z-iz &= 3i-1-1,\\ -3z &= -2+3i.\end{align}


Then, we obtain


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

It is a little troublesome to divide two complex numbers, so we will therefore not check whether \displaystyle z=\frac{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}LHS &=iz+1=i(\frac{2}{3}-i)+1=\frac{2}{3}\cdot i+1+1=2+\frac{2}{3}i,\\ RHS &= (3+i)(z+i)=(3+i)(\frac{2}{3}-i+i)=(3+i)\frac{2}{3}=2+\frac{2}{3}i.\end{align}

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