Lösung 3.3:5d

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Zeile 1: Zeile 1:
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Let us first divide both sides by
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Let us first divide both sides by <math>4+i</math>, so that the coefficient in front of <math>z^2</math> becomes <math>1</math>,
-
<math>4+i</math>, so that the coefficient in front of
+
-
<math>z^{2}</math>
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-
becomes
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-
<math>\text{1}</math>,
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-
 
+
-
 
+
-
<math>z^{2}+\frac{1-21i}{4+i}z=\frac{17}{4+i}</math>
+
 +
{{Displayed math||<math>z^2 + \frac{1-21i}{4+i}z = \frac{17}{4+i}\,\textrm{.}</math>}}
The two complex quotients become
The two complex quotients become
-
 
+
{{Displayed math||<math>\begin{align}
-
<math>\begin{align}
+
\frac{1-21i}{4+i}
-
& \frac{\left( 1-21i \right)\left( 4-i \right)}{\left( 4+i \right)\left( 4-i \right)}=\frac{4-i-84i+21i^{2}}{4^{2}-i^{2}} \\
+
&= \frac{(1-21i)(4-i)}{(4+i)(4-i)}
-
& =\frac{-17-85i}{16+1}=\frac{-17-85i}{17}=-1-5i \\
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= \frac{4-i-84i+21i^2}{4^2-i^2}\\[5pt]
-
\end{align}</math>
+
&= \frac{-17-85i}{16+1}
-
 
+
= \frac{-17-85i}{17}
-
 
+
= -1-5i\,,\\[10pt]
-
 
+
\frac{17}{4+i}
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<math>\begin{align}
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&= \frac{17(4-i)}{(4+i)(4-i)}
-
& \frac{17}{4+i}=\frac{17\left( 4-i \right)}{\left( 4+i \right)\left( 4-i \right)}=\frac{17\left( 4-i \right)}{4^{2}-i^{2}} \\
+
= \frac{17(4-i)}{4^2-i^2}\\[5pt]
-
& =\frac{17\left( 4-i \right)}{17}=4-i \\
+
&= \frac{17(4-i)}{17}
-
\end{align}</math>
+
= 4-i\,\textrm{.}
-
 
+
\end{align}</math>}}
Thus, the equation becomes
Thus, the equation becomes
 +
{{Displayed math||<math>z^2 - (1+5i)z = 4-i\,\textrm{.}</math>}}
-
<math>z^{2}-\left( 1+5i \right)z=4-i</math>
+
Now, we complete the square of the left-hand side,
 +
{{Displayed math||<math>\begin{align}
 +
\Bigl(z-\frac{1+5i}{2}\Bigr)^2 - \Bigl(\frac{1+5i}{2}\Bigr)^2 &= 4-i\,,\\[5pt]
 +
\Bigl(z-\frac{1+5i}{2}\Bigr)^2 - \Bigl(\frac{1}{4}+\frac{5}{2}\,i+\frac{25}{4}i^2 \Bigr) &= 4-i\,,\\[5pt]
 +
\Bigl(z-\frac{1+5i}{2}\Bigr)^2 - \frac{1}{4} - \frac{5}{2}i + \frac{25}{4} &= 4-i\,, \\[5pt]
 +
\Bigl(z-\frac{1+5i}{2}\Bigr)^2 &= -2+\frac{3}{2}\,i\,\textrm{.}
 +
\end{align}</math>}}
-
Now, we complete the square of the left-hand side:
+
If we set <math>w=z-\frac{1+5i}{2}</math>, we have a binomial equation in <math>w</math>,
 +
{{Displayed math||<math>w^2 = -2+\frac{3}{2}\,i</math>}}
-
<math>\begin{align}
+
which we solve by putting <math>w=x+iy</math>,
-
& \left( z-\frac{1+5i}{2} \right)^{2}-\left( \frac{1+5i}{2} \right)^{2}=4-i \\
+
-
& \left( z-\frac{1+5i}{2} \right)^{2}-\left( \frac{1}{4}+\frac{5}{2}i+\frac{25}{4}i^{2} \right)=4-i \\
+
-
& \left( z-\frac{1+5i}{2} \right)^{2}-\frac{1}{4}-\frac{5}{2}i+\frac{25}{4}=4-i \\
+
-
& \left( z-\frac{1+5i}{2} \right)^{2}=-2+\frac{3}{2}i \\
+
-
\end{align}</math>
+
-
 
+
-
 
+
-
If we set
+
-
<math>w=z-\frac{1+5i}{2}</math>, we have a binomial equation in
+
-
<math>w</math>,
+
-
 
+
-
 
+
-
<math>w^{2}=-2+\frac{3}{2}i</math>
+
-
 
+
-
+
-
which we solve by putting
+
-
<math>w=x+iy\text{ }</math>,
+
-
 
+
-
 
+
-
<math>\left( x+iy \right)^{2}=-2+\frac{3}{2}i</math>
+
 +
{{Displayed math||<math>(x+iy)^2 = -2+\frac{3}{2}\,i</math>}}
or, if the left-hand side is expanded,
or, if the left-hand side is expanded,
-
 
+
{{Displayed math||<math>x^2-y^2+2xyi = -2+\frac{3}{2}\,i\,\textrm{.}</math>}}
-
<math>x^{2}-y^{2}+2xyi=-2+\frac{3}{2}i</math>
+
-
 
+
If we identify the real and imaginary parts on both sides, we get
If we identify the real and imaginary parts on both sides, we get
-
 
+
{{Displayed math||<math>\begin{align}
-
<math>\begin{align}
+
x^2-y^2 &= -2\,,\\[5pt]
-
& x^{2}-y^{2}=-2 \\
+
2xy &= \frac{3}{2}\,,
-
& 2xy=\frac{3}{2} \\
+
\end{align}</math>}}
-
\end{align}</math>
+
-
 
+
and if we take the magnitude of both sides and put them equal to each other, we obtain a third relation:
and if we take the magnitude of both sides and put them equal to each other, we obtain a third relation:
-
 
+
{{Displayed math||<math>x^2 + y^2 = \sqrt{(-2)^2+\bigl(\tfrac{3}{2}\bigr)^2} = \tfrac{5}{2}\,\textrm{.}</math>}}
-
<math>x^{2}+y^{2}=\sqrt{\left( -2 \right)^{2}+\left( \frac{3}{2} \right)^{2}}=\frac{5}{2}</math>
+
-
 
+
Strictly speaking, this last relation is contained within the first two and we include it because makes the calculations easier.
Strictly speaking, this last relation is contained within the first two and we include it because makes the calculations easier.
Zeile 82: Zeile 59:
Together, the three relations constitute the following system of equations:
Together, the three relations constitute the following system of equations:
 +
{{Displayed math||<math>\left\{\begin{align}
 +
x^2-y^2 &= -2\,,\\[5pt]
 +
2xy &= \frac{3}{2}\,,\\[5pt]
 +
x^2 + y^2 &= \frac{5}{2}\,\textrm{.}
 +
\end{align} \right.</math>}}
-
 
+
From the first and the third equations, we can relatively easily obtain the values that <math>x</math> and <math>y</math> can take.
-
<math>\left\{ \begin{array}{*{35}l}
+
-
\ x^{2}-y^{2}=2 \\
+
-
\ 2xy=-\frac{3}{2} \\
+
-
\ x^{2}+y^{2}=\frac{5}{2} \\
+
-
\end{array} \right.</math>
+
-
 
+
-
 
+
-
 
+
-
From the first and the third equations, we can relatively easily obtain the values that
+
-
<math>x</math>
+
-
and
+
-
<math>y</math>
+
-
can take.
+
Add the first and third equations,
Add the first and third equations,
 +
{| align="center" style="padding:10px 0px 10px 0px;"
 +
||
 +
|align="right"|<math>x^2</math>
 +
||<math>{}-{}</math>
 +
|align="right"|<math>y^2</math>
 +
||<math>{}={}</math>
 +
|align="right"|<math>-2</math>
 +
|-
 +
|align="left"|<math>+\ \ </math>
 +
|align="right"|<math>x^2</math>
 +
||<math>{}+{}</math>
 +
||<math>y^2</math>
 +
||<math>{}={}</math>
 +
|align="right"|<math>\tfrac{5}{2}</math>
 +
|-
 +
|colspan="6"|<hr>
 +
|-
 +
||
 +
|align="right"|<math>2x^2</math>
 +
||
 +
||
 +
||<math>{}={}</math>
 +
|align="right"|<math>\tfrac{1}{2}</math>
 +
|}
-
EQ1
+
which gives that <math>x=\pm \tfrac{1}{2}</math>.
-
 
+
-
which gives that
+
-
<math>x=\pm \frac{1}{2}</math>.
+
Then, subtract the first equation from the third equation,
Then, subtract the first equation from the third equation,
 +
{| align="center" style="padding:10px 0px 10px 0px;"
 +
||
 +
|align="right"|<math>x^2</math>
 +
||<math>{}+{}</math>
 +
|align="right"|<math>y^2</math>
 +
||<math>{}={}</math>
 +
|align="right"|<math>\tfrac{5}{2}</math>
 +
|-
 +
|align="left"|<math>-\ \ </math>
 +
|align="right"|<math>\bigl(x^2</math>
 +
||<math>{}-{}</math>
 +
|align="right"|<math>y^2</math>
 +
||<math>{}={}</math>
 +
|align="right"|<math>-2\rlap{\bigr)}</math>
 +
|-
 +
|colspan="6"|<hr>
 +
|-
 +
||
 +
||
 +
||
 +
|align="right"|<math>2y^2</math>
 +
||<math>{}={}</math>
 +
|align="right"|<math>\tfrac{9}{2}</math>
 +
|}
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EQ13
+
i.e. <math>y=\pm\tfrac{3}{2}</math>.
-
 
+
-
i.e.
+
-
<math>y=\pm \frac{3}{2}</math>.
+
This gives four possible combinations,
This gives four possible combinations,
-
 
+
{{Displayed math||<math>\left\{\begin{align}
-
<math>\left\{ \begin{array}{*{35}l}
+
x &= \tfrac{1}{2}\\[5pt]
-
x=\frac{1}{2} \\
+
y &= \tfrac{3}{2}
-
y=\frac{3}{2} \\
+
\end{align}\right.
-
\end{array} \right.\quad \left\{ \begin{array}{*{35}l}
+
\qquad
-
x=\frac{1}{2} \\
+
\left\{\begin{align}
-
y=-\frac{3}{2} \\
+
x &= \tfrac{1}{2}\\[5pt]
-
\end{array} \right.\quad \left\{ \begin{array}{*{35}l}
+
y &= -\tfrac{3}{2}
-
x=-\frac{1}{2} \\
+
\end{align}\right.
-
y=\frac{3}{2} \\
+
\qquad
-
\end{array} \right.\quad \left\{ \begin{array}{*{35}l}
+
\left\{\begin{align}
-
x=-\frac{1}{2} \\
+
x &= -\tfrac{1}{2}\\[5pt]
-
y=-\frac{3}{2} \\
+
y &= \tfrac{3}{2}
-
\end{array} \right.</math>
+
\end{align}\right.
-
 
+
\qquad
 +
\left\{\begin{align}
 +
x &= -\tfrac{1}{2}\\[5pt]
 +
y &= -\tfrac{3}{2}
 +
\end{align} \right.</math>}}
of which only two also satisfy the second equation.
of which only two also satisfy the second equation.
-
 
+
{{Displayed math||<math>\left\{\begin{align}
-
<math>\left\{ \begin{array}{*{35}l}
+
x &= \tfrac{1}{2}\\[5pt]
-
x=\frac{1}{2} \\
+
y &= \tfrac{3}{2}
-
y=\frac{3}{2} \\
+
\end{align}\right.
-
\end{array} \right.\quad \quad \text{and}\quad \quad \left\{ \begin{array}{*{35}l}
+
\qquad\text{and}\qquad
-
x=-\frac{1}{2} \\
+
\left\{\begin{align}
-
y=-\frac{3}{2} \\
+
x &= -\tfrac{1}{2}\\[5pt]
-
\end{array} \right.</math>
+
y &= -\tfrac{3}{2}
-
 
+
\end{align}\right.</math>}}
This means that the binomial equation has the two solutions,
This means that the binomial equation has the two solutions,
-
 
+
{{Displayed math||<math>w=\frac{1}{2}+\frac{3}{2}\,i\qquad</math> and <math>\qquad w=-\frac{1}{2}-\frac{3}{2}\,i\,,</math>}}
-
 
+
-
<math>w=\frac{1}{2}+\frac{3}{2}i</math>
+
-
and
+
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<math>w=\frac{1}{-2}-\frac{3}{2}i</math>
+
-
 
+
and that the original equation has the solutions
and that the original equation has the solutions
 +
{{Displayed math||<math>z=1+4i\qquad</math> and <math>\qquad z=i</math>}}
-
<math>z=1+4i</math>
+
according to the relation <math>w=z-\frac{1+5i}{2}</math>.
-
and
+
-
<math>z=i</math>
+
-
 
+
-
according to the relation
+
-
<math>w=z-\frac{1+5i}{2}</math>.
+
Finally, we check that the solutions really do satisfy the equation.
Finally, we check that the solutions really do satisfy the equation.
- 
- 
- 
-
<math>\begin{align}
 
-
& 1+4i:\quad \left( 4+i \right)z^{2}+\left( 1-21i \right)z \\
 
-
& =\left( 4+i \right)\left( 1+4i \right)^{2}+\left( 1-21i \right)\left( 1+4i \right) \\
 
-
& =\left( 4+i \right)\left( 1+8i+16i^{2} \right)+\left( 1+4i-21i-84i^{2} \right) \\
 
-
& =\left( 4+i \right)\left( -15+8i \right)+1-17i+84 \\
 
-
& =-60+32i-15i+8i^{2}+1-17i+84 \\
 
-
& =-60+32i-15i-8+1-17i+84=17 \\
 
-
\end{align}</math>
 
- 
- 
<math>\begin{align}
<math>\begin{align}
-
& z=i:\quad \left( 4+i \right)z^{2}+\left( 1-21i \right)z \\
+
z=1+4i:\quad (4+i)z^2+(1-21i)z
-
& =\left( 4+i \right)i^{2}+\left( 1-21i \right)i \\
+
&= (4+i)(1+4i)^2+(1-21i)(1+4i)\\[5pt]
-
& =\left( 4+i \right)\left( -1 \right)+i-21i^{2} \\
+
&= (4+i)(1+8i+16i^2)+(1+4i-21i-84i^2)\\[5pt]
-
& =-4-i+i+21=17 \\
+
&= (4+i)(-15+8i)+1-17i+84\\[5pt]
 +
&= -60+32i-15i+8i^2+1-17i+84\\[5pt]
 +
&= -60+32i-15i-8+1-17i+84\\[5pt]
 +
&= 17\,,\\[10pt]
 +
z={}\rlap{i:}\phantom{1+4i:}{}\quad (4+i)z^2+(1-21i)z
 +
&= (4+i)i^2 + (1-21i)i\\[5pt]
 +
&= (4+i)(-1)+i-21i^2\\[5pt]
 +
&= -4-i+i+21\\[5pt]
 +
&= 17\,\textrm{.}
\end{align}</math>
\end{align}</math>

Version vom 11:38, 31. Okt. 2008

Let us first divide both sides by \displaystyle 4+i, so that the coefficient in front of \displaystyle z^2 becomes \displaystyle 1,

\displaystyle z^2 + \frac{1-21i}{4+i}z = \frac{17}{4+i}\,\textrm{.}

The two complex quotients become

\displaystyle \begin{align}

\frac{1-21i}{4+i} &= \frac{(1-21i)(4-i)}{(4+i)(4-i)} = \frac{4-i-84i+21i^2}{4^2-i^2}\\[5pt] &= \frac{-17-85i}{16+1} = \frac{-17-85i}{17} = -1-5i\,,\\[10pt] \frac{17}{4+i} &= \frac{17(4-i)}{(4+i)(4-i)} = \frac{17(4-i)}{4^2-i^2}\\[5pt] &= \frac{17(4-i)}{17} = 4-i\,\textrm{.} \end{align}

Thus, the equation becomes

\displaystyle z^2 - (1+5i)z = 4-i\,\textrm{.}

Now, we complete the square of the left-hand side,

\displaystyle \begin{align}

\Bigl(z-\frac{1+5i}{2}\Bigr)^2 - \Bigl(\frac{1+5i}{2}\Bigr)^2 &= 4-i\,,\\[5pt] \Bigl(z-\frac{1+5i}{2}\Bigr)^2 - \Bigl(\frac{1}{4}+\frac{5}{2}\,i+\frac{25}{4}i^2 \Bigr) &= 4-i\,,\\[5pt] \Bigl(z-\frac{1+5i}{2}\Bigr)^2 - \frac{1}{4} - \frac{5}{2}i + \frac{25}{4} &= 4-i\,, \\[5pt] \Bigl(z-\frac{1+5i}{2}\Bigr)^2 &= -2+\frac{3}{2}\,i\,\textrm{.} \end{align}

If we set \displaystyle w=z-\frac{1+5i}{2}, we have a binomial equation in \displaystyle w,

\displaystyle w^2 = -2+\frac{3}{2}\,i

which we solve by putting \displaystyle w=x+iy,

\displaystyle (x+iy)^2 = -2+\frac{3}{2}\,i

or, if the left-hand side is expanded,

\displaystyle x^2-y^2+2xyi = -2+\frac{3}{2}\,i\,\textrm{.}

If we identify the real and imaginary parts on both sides, we get

\displaystyle \begin{align}

x^2-y^2 &= -2\,,\\[5pt] 2xy &= \frac{3}{2}\,, \end{align}

and if we take the magnitude of both sides and put them equal to each other, we obtain a third relation:

\displaystyle x^2 + y^2 = \sqrt{(-2)^2+\bigl(\tfrac{3}{2}\bigr)^2} = \tfrac{5}{2}\,\textrm{.}

Strictly speaking, this last relation is contained within the first two and we include it because makes the calculations easier.

Together, the three relations constitute the following system of equations:

\displaystyle \left\{\begin{align}

x^2-y^2 &= -2\,,\\[5pt] 2xy &= \frac{3}{2}\,,\\[5pt] x^2 + y^2 &= \frac{5}{2}\,\textrm{.} \end{align} \right.

From the first and the third equations, we can relatively easily obtain the values that \displaystyle x and \displaystyle y can take.

Add the first and third equations,

\displaystyle x^2 \displaystyle {}-{} \displaystyle y^2 \displaystyle {}={} \displaystyle -2
\displaystyle +\ \ \displaystyle x^2 \displaystyle {}+{} \displaystyle y^2 \displaystyle {}={} \displaystyle \tfrac{5}{2}
\displaystyle 2x^2 \displaystyle {}={} \displaystyle \tfrac{1}{2}

which gives that \displaystyle x=\pm \tfrac{1}{2}.

Then, subtract the first equation from the third equation,

\displaystyle x^2 \displaystyle {}+{} \displaystyle y^2 \displaystyle {}={} \displaystyle \tfrac{5}{2}
\displaystyle -\ \ \displaystyle \bigl(x^2 \displaystyle {}-{} \displaystyle y^2 \displaystyle {}={} \displaystyle -2\rlap{\bigr)}
\displaystyle 2y^2 \displaystyle {}={} \displaystyle \tfrac{9}{2}

i.e. \displaystyle y=\pm\tfrac{3}{2}.

This gives four possible combinations,

\displaystyle \left\{\begin{align}

x &= \tfrac{1}{2}\\[5pt] y &= \tfrac{3}{2} \end{align}\right. \qquad \left\{\begin{align} x &= \tfrac{1}{2}\\[5pt] y &= -\tfrac{3}{2} \end{align}\right. \qquad \left\{\begin{align} x &= -\tfrac{1}{2}\\[5pt] y &= \tfrac{3}{2} \end{align}\right. \qquad \left\{\begin{align} x &= -\tfrac{1}{2}\\[5pt] y &= -\tfrac{3}{2} \end{align} \right.

of which only two also satisfy the second equation.

\displaystyle \left\{\begin{align}

x &= \tfrac{1}{2}\\[5pt] y &= \tfrac{3}{2} \end{align}\right. \qquad\text{and}\qquad \left\{\begin{align} x &= -\tfrac{1}{2}\\[5pt] y &= -\tfrac{3}{2} \end{align}\right.

This means that the binomial equation has the two solutions,

\displaystyle w=\frac{1}{2}+\frac{3}{2}\,i\qquad and \displaystyle \qquad w=-\frac{1}{2}-\frac{3}{2}\,i\,,

and that the original equation has the solutions

\displaystyle z=1+4i\qquad and \displaystyle \qquad z=i

according to the relation \displaystyle w=z-\frac{1+5i}{2}.

Finally, we check that the solutions really do satisfy the equation.

\displaystyle \begin{align} z=1+4i:\quad (4+i)z^2+(1-21i)z &= (4+i)(1+4i)^2+(1-21i)(1+4i)\\[5pt] &= (4+i)(1+8i+16i^2)+(1+4i-21i-84i^2)\\[5pt] &= (4+i)(-15+8i)+1-17i+84\\[5pt] &= -60+32i-15i+8i^2+1-17i+84\\[5pt] &= -60+32i-15i-8+1-17i+84\\[5pt] &= 17\,,\\[10pt] z={}\rlap{i:}\phantom{1+4i:}{}\quad (4+i)z^2+(1-21i)z &= (4+i)i^2 + (1-21i)i\\[5pt] &= (4+i)(-1)+i-21i^2\\[5pt] &= -4-i+i+21\\[5pt] &= 17\,\textrm{.} \end{align}

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_3.3:5d“