Lösung 3.4:4
Aus Online Mathematik Brückenkurs 2
Because \displaystyle z=\text{1}-\text{2}i should be a root of the equation, we can substitute \displaystyle z=\text{1}-\text{2}i in and the equation should be satisfied:
\displaystyle \left( \text{1}-\text{2}i \right)^{3}+a\left( \text{1}-\text{2}i \right)+b=0
We will therefore adjust the constants
\displaystyle a
and
\displaystyle b
so that the relation above holds. We simplify the left-hand side,
\displaystyle -11+2i+a\left( \text{1}-\text{2}i \right)+b=0
and collect together the real and imaginary parts:
\displaystyle \left( -11+a+b \right)+\left( 2-2a \right)i=0
If the left-hand side is to equal the right-hand side, the left-hand side's real and imaginary parts must be equal to zero, i.e.
\displaystyle \left\{ \begin{array}{*{35}l} -11+a+b=0 \\ 2-2a=0 \\ \end{array} \right.
This gives
\displaystyle a=\text{1}
and
\displaystyle b=\text{1}0.
The equation is thus
\displaystyle z^{2}+z+10=0
and has the prescribed root
\displaystyle z=\text{1}-\text{2}i.
What we have is a polynomial with real coefficients and we therefore know that the equation has, in addition, the complex conjugate root \displaystyle z=\text{1}+\text{2}i.
Hence, we know two of the equation's three roots and we can obtain the third root with help of the factor theorem. According to the factor theorem, the equation's left-hand side contains the factor
\displaystyle \left( z-\left( 1-2i \right) \right)\left( z-\left( 1+2i \right) \right)=z^{2}-2z+5
and this means that we can write
\displaystyle z^{3}+z+10=\left( z-A \right)\left( z^{2}-2z+5 \right)
where
\displaystyle z-A
is the factor which corresponds to the third root
\displaystyle z=A. Using polynomial division, we obtain the factor
\displaystyle \begin{align}
& z-A=\frac{z^{3}+z+10}{z^{2}-2z+5} \\
& =\frac{z^{3}-2z^{2}+5z+2z^{2}-5z+z+10}{z^{2}-2z+5} \\
& =\frac{z\left( z^{2}-2z+5 \right)+2z^{2}-4z+10}{z^{2}-2z+5} \\
& =z+\frac{2z^{2}-4z+10}{z^{2}-2z+5} \\
& =z+\frac{2\left( z^{2}-2z+5 \right)}{z^{2}-2z+5} \\
& =z+2 \\
\end{align}
Thus, the remaining root is
\displaystyle z=-\text{2}.