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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,

(1−2i)3+a(1−2i)+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+a(1−2i)+b=0

and collect together the real and imaginary parts,

(−11+a+b)+(2−2a)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.

−11+a+b2−2a=0=0.

This gives a=1 and b=10.

The equation is thus

z3+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−(1−2i)z−(1+2i)=z2−2z+5

and this means that we can write

z3+z+10=(z−A)(z2−2z+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+5z(z2−2z+5)+2z2−4z+10=z+z2−2z+52z2−4z+10=z+z2−2z+52(z2−2z+5)=z+2.

Thus, the remaining root is z=−2.