Lösung 3.3:5c

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As usual we begin by completing the square,
As usual we begin by completing the square,
 +
{{Displayed math||<math>\begin{align}
 +
\Bigl(z-\frac{1+3i}{2}\Bigr)^2 - \Bigl(\frac{1+3i}{2}\Bigr)^2-4+3i &= 0\,,\\[5pt]
 +
\Bigl(z-\frac{1+3i}{2}\Bigr)^2 - \Bigl(\frac{1}{4}+\frac{3}{2}\,i+\frac{9}{4}\,i^2 \Bigr) - 4 + 3i &= 0\,,\\[5pt]
 +
\Bigl(z-\frac{1+3i}{2}\Bigr)^2 - \frac{1}{4} - \frac{3}{2}\,i + \frac{9}{4} - 4 + 3i &= 0\,,\\[5pt]
 +
\Bigl(z-\frac{1+3i}{2}\Bigr)^2 - 2 + \frac{3}{2}\,i &= 0\,,
 +
\end{align}</math>}}
-
<math>\begin{align}
+
and if we treat <math>w=z-\frac{1+3i}{2}</math> as unknown, we have the equation
-
& \left( z-\frac{1+3i}{2} \right)^{2}-\left( \frac{1+3i}{2} \right)^{2}-4+3i=0 \\
+
-
& \left( z-\frac{1+3i}{2} \right)^{2}-\left( \frac{1}{4}+\frac{3}{2}i+\frac{9}{4}i^{2} \right)-4+3i=0 \\
+
-
& \left( z-\frac{1+3i}{2} \right)^{2}-\frac{1}{4}-\frac{3}{2}i+\frac{9}{4}-4+3i=0 \\
+
-
& \left( z-\frac{1+3i}{2} \right)^{2}-2+\frac{3}{2}i=0 \\
+
-
\end{align}</math>
+
 +
{{Displayed math||<math>w=2-\frac{3}{2}\,i\,\textrm{.}</math>}}
-
and if we treat as unknown, we have the equation's
+
Up until now, we have solved binomial equations of this type by going over to polar form, but if we were to do that in this case, we would have a problem with determining the exact value of the argument of the right-hand side. Instead, we put <math>w=x+iy</math> and try to obtain <math>x</math> and <math>y</math> from the equation.
 +
With <math>w=x+iy</math>, the equation becomes
-
<math>w=z-\frac{1+3i}{2}</math>
+
{{Displayed math||<math>(x+iy)^2 = 2-\frac{3}{2}\,i</math>}}
-
 
+
-
 
+
-
Up until now, we have solved binomial equations of this type by going over to polar form, but if we were to do that in this case, we would have a problem with determining the exact value of the argument of the right-hand side. Instead, we put
+
-
<math>w=x+iy</math>
+
-
and try to obtain
+
-
<math>x</math>
+
-
and
+
-
<math>y</math>
+
-
from the equation.
+
-
 
+
-
With
+
-
<math>w=x+iy</math>, the equation becomes
+
-
 
+
-
 
+
-
<math>\left( x+iy \right)^{2}=2-\frac{3}{2}i</math>
+
-
 
+
and, with the left-hand side expanded,
and, with the left-hand side 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 set the real and imaginary part of both sides equal, we obtain the equation system
If we set the real and imaginary part of both sides equal, we obtain the equation system
-
 
+
{{Displayed math||<math>\left\{\begin{align}
-
<math>\left\{ \begin{array}{*{35}l}
+
x^2-y^2 &= 2\\[5pt]
-
x^{2}-y^{2}=2 \\
+
2xy &= -\frac{3}{2}\,\textrm{.}
-
2xy=-\frac{3}{2} \\
+
\end{align}\right.</math>}}
-
\end{array} \right.</math>
+
-
 
+
We would very well be able to solve this system, but there is a further relation that we can obtain which will simplify the calculations. If we go back to the relation
We would very well be able to solve this system, but there is a further relation that we can obtain which will simplify the calculations. If we go back to the relation
-
 
+
{{Displayed math||<math>(x+iy)^2 = 2-\frac{3}{2}\,i</math>}}
-
<math>\left( x+iy \right)^{2}=2-\frac{3}{2}i</math>
+
-
 
+
and take the modulus of both sides, we obtain
and take the modulus of both sides, we obtain
-
 
+
{{Displayed math||<math>x^2 + y^2 = \sqrt{2^2+\bigl(-\tfrac{3}{2}\bigr)^2}</math>}}
-
<math>x^{2}+y^{2}=\sqrt{2^{2}+\left( -\frac{3}{2} \right)^{2}}</math>
+
-
 
+
-
 
+
i.e.
i.e.
 +
{{Displayed math||<math>x^2 + y^2 = \frac{5}{2}\,\textrm{.}</math>}}
-
<math>x^{2}+y^{2}=\frac{5}{2}</math>
+
We add this relation to our two other equations,
-
 
+
-
 
+
-
We add this relation to our two other equations:
+
-
 
+
-
 
+
-
<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>
+
 +
{{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>}}
Now, add the first and third equations
Now, 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{9}{2}</math>
 +
|}
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EQU1
+
which gives that <math>x=\pm\tfrac{3}{2}</math>. Then, subtracting the first equation from the third,
 +
{| 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{1}{2}</math>
 +
|}
-
which gives that
+
we get <math>y=\pm\tfrac{1}{2}</math>. This gives potentially four solutions to the equation system,
-
<math>y=\pm \frac{3}{2}</math>. Then, subtracting the first equation from the third,
+
-
EQU2
+
{{Displayed math||<math>\left\{\begin{align}
 +
x &= \tfrac{3}{2}\\[5pt]
 +
y &= \tfrac{1}{2}
 +
\end{align}\right.
 +
\qquad
 +
\left\{\begin{align}
 +
x &= \tfrac{3}{2}\\[5pt]
 +
y &= -\tfrac{1}{2}
 +
\end{align}\right.
 +
\qquad
 +
\left\{\begin{align}
 +
x &= -\tfrac{3}{2}\\[5pt]
 +
y &= \tfrac{1}{2}
 +
\end{align}\right.
 +
\qquad
 +
\left\{\begin{align}
 +
x &= -\tfrac{3}{2}\\[5pt]
 +
y &= -\tfrac{1}{2}
 +
\end{align}\right.</math>}}
 +
but only two of these satisfy the second equation <math>2xy=-\tfrac{3}{2}</math>, namely
-
we get
+
{{Displayed math||<math>\left\{\begin{align}
-
<math>y=\pm \frac{1}{2}</math>. This gives potentially four solutions to the equation system,
+
x &= \tfrac{3}{2}\\[5pt]
-
 
+
y &= -\tfrac{1}{2}
-
 
+
\end{align}\right.
-
<math>\left\{ \begin{array}{*{35}l}
+
\qquad\text{and}\qquad
-
x=\frac{3}{2} \\
+
\left\{\begin{align}
-
y=\frac{1}{2} \\
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x &= -\tfrac{3}{2}\\[5pt]
-
\end{array} \right.\quad \left\{ \begin{array}{*{35}l}
+
y &= \tfrac{1}{2}
-
x=\frac{3}{2} \\
+
\end{align}\right.</math>}}
-
y=-\frac{1}{2} \\
+
-
\end{array} \right.\quad \left\{ \begin{array}{*{35}l}
+
-
x=-\frac{3}{2} \\
+
-
y=\frac{1}{2} \\
+
-
\end{array} \right.\quad \left\{ \begin{array}{*{35}l}
+
-
x=-\frac{3}{2} \\
+
-
y=-\frac{1}{2} \\
+
-
\end{array} \right.</math>
+
-
 
+
-
 
+
-
but only two of these satisfy the second equation
+
-
<math>2xy=-\frac{3}{2}</math>, namely
+
-
 
+
-
 
+
-
<math>\left\{ \begin{array}{*{35}l}
+
-
x=\frac{3}{2} \\
+
-
y=-\frac{1}{2} \\
+
-
\end{array} \right.\quad \quad \text{and}\quad \left\{ \begin{array}{*{35}l}
+
-
x=-\frac{3}{2} \\
+
-
y=\frac{1}{2} \\
+
-
\end{array} \right.</math>
+
-
 
+
Hence, we have that the solutions are
Hence, we have that the solutions are
 +
{{Displayed math||<math>w=\frac{3-i}{2}\quad</math> and <math>\quad w=\frac{-3+i}{2}\,,</math>}}
-
<math>w=\frac{3-i}{2}</math>
+
or, expressed in <math>z</math>,
-
and
+
-
<math>w=\frac{-3+i}{2}</math>
+
-
 
+
-
 
+
-
or, expressed in
+
-
<math>z</math>,
+
-
 
+
-
 
+
-
<math>z=2+i</math>
+
-
and
+
-
<math>z=-1+2i</math>
+
-
 
+
-
 
+
-
Because the calculation has been rather long, there is the risk that we have calculated incorrectly somewhere and we therefore check that the solutions satisfy the equation in the exercise:
+
-
 
+
-
 
+
-
<math>\begin{align}
+
-
& z=2+i:\quad z^{2}-\left( 1+3i \right)z-4+3i=\left( 2+i \right)^{2}-\left( 1+3i \right)\left( 2+i \right)-4+3i \\
+
-
& =4+4i+i^{2}-\left( 2+i+6i+3i^{2} \right)-4+3i \\
+
-
& =4+4i-1-2-7i+3-4+3i=0 \\
+
-
\end{align}</math>
+
 +
{{Displayed math||<math>z=2+i\quad</math> and <math>\quad z=-1+2i\,\textrm{.}</math>}}
 +
Because the calculation has been rather long, there is the risk that we have calculated incorrectly somewhere and we therefore check that the solutions satisfy the equation in the exercise,
<math>\begin{align}
<math>\begin{align}
-
& z=-1+2i:\quad z^{2}-\left( 1+3i \right)z-4+3i=\left( -1+2i \right)^{2}-\left( 1+3i \right)\left( -1+2i \right)-4+3i \\
+
z={}\rlap{2+i:}\phantom{-1+2i:}{}\quad z^2-(1+3i)z-4+3i
-
& =\left( -1 \right)^{2}-4i+4i^{2}-\left( -1+2i-3i+6i^{2} \right)-4+3i \\
+
&= (2+i)^2 - (1+3i)(2+i) - 4 + 3i\\[5pt]
-
& =1-4i-4+1+i+6-4+3i=0 \\
+
&= 4+4i+i^2-(2+i+6i+3i^2)-4+3i\\[5pt]
 +
&= 4+4i-1-2-7i+3-4+3i\\[5pt]
 +
&= 0\,,\\[10pt]
 +
z=-1+2i:\quad z^2-(1+3i)z-4+3i
 +
&= (-1+2i)^2-(1+3i)(-1+2i)-4+3i\\[5pt]
 +
&= (-1)^2-4i+4i^2-(-1+2i-3i+6i^2)-4+3i\\[5pt]
 +
&= 1-4i-4+1+i+6-4+3i\\[5pt]
 +
&= 0\,\textrm{.}
\end{align}</math>
\end{align}</math>

Version vom 07:56, 31. Okt. 2008

As usual we begin by completing the square,

\displaystyle \begin{align}

\Bigl(z-\frac{1+3i}{2}\Bigr)^2 - \Bigl(\frac{1+3i}{2}\Bigr)^2-4+3i &= 0\,,\\[5pt] \Bigl(z-\frac{1+3i}{2}\Bigr)^2 - \Bigl(\frac{1}{4}+\frac{3}{2}\,i+\frac{9}{4}\,i^2 \Bigr) - 4 + 3i &= 0\,,\\[5pt] \Bigl(z-\frac{1+3i}{2}\Bigr)^2 - \frac{1}{4} - \frac{3}{2}\,i + \frac{9}{4} - 4 + 3i &= 0\,,\\[5pt] \Bigl(z-\frac{1+3i}{2}\Bigr)^2 - 2 + \frac{3}{2}\,i &= 0\,, \end{align}

and if we treat \displaystyle w=z-\frac{1+3i}{2} as unknown, we have the equation

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

Up until now, we have solved binomial equations of this type by going over to polar form, but if we were to do that in this case, we would have a problem with determining the exact value of the argument of the right-hand side. Instead, we put \displaystyle w=x+iy and try to obtain \displaystyle x and \displaystyle y from the equation.

With \displaystyle w=x+iy, the equation becomes

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

and, with the left-hand side expanded,

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

If we set the real and imaginary part of both sides equal, we obtain the equation system

\displaystyle \left\{\begin{align}

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

We would very well be able to solve this system, but there is a further relation that we can obtain which will simplify the calculations. If we go back to the relation

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

and take the modulus of both sides, we obtain

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

i.e.

\displaystyle x^2 + y^2 = \frac{5}{2}\,\textrm{.}

We add this relation to our two other 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.

Now, 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{9}{2}

which gives that \displaystyle x=\pm\tfrac{3}{2}. Then, subtracting the first equation from the third,

\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{1}{2}

we get \displaystyle y=\pm\tfrac{1}{2}. This gives potentially four solutions to the equation system,

\displaystyle \left\{\begin{align}

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

but only two of these satisfy the second equation \displaystyle 2xy=-\tfrac{3}{2}, namely

\displaystyle \left\{\begin{align}

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

Hence, we have that the solutions are

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

or, expressed in \displaystyle z,

\displaystyle z=2+i\quad and \displaystyle \quad z=-1+2i\,\textrm{.}

Because the calculation has been rather long, there is the risk that we have calculated incorrectly somewhere and we therefore check that the solutions satisfy the equation in the exercise,

\displaystyle \begin{align} z={}\rlap{2+i:}\phantom{-1+2i:}{}\quad z^2-(1+3i)z-4+3i &= (2+i)^2 - (1+3i)(2+i) - 4 + 3i\\[5pt] &= 4+4i+i^2-(2+i+6i+3i^2)-4+3i\\[5pt] &= 4+4i-1-2-7i+3-4+3i\\[5pt] &= 0\,,\\[10pt] z=-1+2i:\quad z^2-(1+3i)z-4+3i &= (-1+2i)^2-(1+3i)(-1+2i)-4+3i\\[5pt] &= (-1)^2-4i+4i^2-(-1+2i-3i+6i^2)-4+3i\\[5pt] &= 1-4i-4+1+i+6-4+3i\\[5pt] &= 0\,\textrm{.} \end{align}

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