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Lösung 3.4:2

Aus Online Mathematik Brückenkurs 2

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Version vom 10:51, 11. Mär. 2009

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 z=1+i.

The equation's other roots are z=1−i and z=1+i.


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

z=1+i:z3−3z2+4z−2z=1−i:z3−3z2+4z−2=(z−3)z+4z−2=(1+i−3)(1+i)+4(1+i)−2=(−2+i)(1+i)+4(1+i)−2=(−2+i−2i−1+4)(1+i)−2=(1−i)(1+i)−2=12−i2−2=1+1−2=0=(z−3)z+4z−2=(1−i−3)(1−i)+4(1−i)−2=(−2−i)(1−i)+4(1−i)−2=(−2−i+2i−1+4)(1−i)−2=(1+i)(1−i)−2=12−i2−2=1+1−2=0.


Note: Writing

z3−3z2+4z−2=(z−3)z+4z−2 

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

Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_3.4:2“