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Aus Online Mathematik Brückenkurs 2

Because z=1−2i should be a root of the equation, we can substitute z=1−2i in and the equation should be satisfied:

1−2i3+a1−2i+b=0 

We will therefore adjust the constants a and b so that the relation above holds. We simplify the left-hand side,

−11+2i+a1−2i+b=0 

and collect together the real and imaginary parts:

−11+a+b+2−2ai=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.

−11+a+b=02−2a=0 

This gives a=1 and b=10.

The equation is thus

z2+z+10=0

and has the prescribed root z=1−2i.

What we have is a polynomial with real coefficients and we therefore know that the equation has, in addition, the complex conjugate root z=1+2i.

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

z−1−2iz−1+2i=z2−2z+5 

and this means that we can write

z3+z+10=z−Az2−2z+5 

where z−A is the factor which corresponds to the third root z=A. Using polynomial division, we obtain the factor

z−A=z2−2z+5z3+z+10=z2−2z+5z3−2z2+5z+2z2−5z+z+10=z2−2z+5zz2−2z+5+2z2−4z+10=z+z2−2z+52z2−4z+10=z+z2−2z+52z2−2z+5=z+2

Thus, the remaining root is z=−2.