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Aus Online Mathematik Brückenkurs 2

If the equation has the root z=1, this means, according to the factor rule, that the equation must contain the factor z−1, i.e. the polynomial on the left-hand side can be written as

z3−3z2+4z−2=(z2+Az+B)(z−1)

for some constants A and B. We can determine the second unknown factor using polynomial division,

z2+Az+B=z−1z3−3z2+4z−2=z−1z3−z2+z2−3z2+4z−2=z−1z2(z−1)−2z2+4z−2=z2+z−1−2z2+4z−2=z2+z−1−2z2+2z−2z+4z−2=z2+z−1−2z(z−1)+2z−2=z2−2z+z−12z−2=z2−2z+z−12(z−1)=z2−2z+2.

Thus, the equation can be written as

(z−1)(z2−2z+2)=0.

The advantage of writing the equation in this factorized form is that we can now conclude that the equation's two other roots must be zeros of the factor z2−2z+2. This is because the left-hand side is zero only when at least one of the factors z−1 or z2−2z+2 is zero, and we see directly that z−1 is zero only when z=1.

Hence, we determine the roots by solving the equation

z2−2z+2=0.

Completing the square gives

(z−1)2−12+2(z−1)2=0=−1

and taking the root gives that z−1=i, i.e. z=1−i and \displaystyle z=1+i\,.

The equation's other roots are \displaystyle z=1-i and \displaystyle z=1+i\,.

As an extra check, we investigate whether \displaystyle z = 1 \pm i really are roots of the equation.

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

Note: Writing

\displaystyle z^3-3z^2+4z-2 = \bigl((z-3)z+4\bigr)z-2

is known as the Horner scheme and is used to reduce the amount of the arithmetical work.