1. Identify the second complex root using the conjugate roots theorem.
Since the coefficients A, B, and C of the polynomial f(z) are real numbers, the complex roots must occur in conjugate pairs.
Therefore, 2−i is also a root of f(z)=0.
2. Form the quadratic factor corresponding to the complex conjugate roots.
Sum of complex roots =(2+i)+(2−i)=4
Product of complex roots =(2+i)(2−i)=22−i2=4−(−1)=5
The corresponding quadratic factor is given by z2−(sum of roots)z+(product of roots), which is:
z2−4z+5
3. Express the polynomial as a product of its quadratic factors and compare constant terms to find $k$.
Given that f(z) is a polynomial of degree 4 with a leading coefficient of 1, and (z−k)2 is a factor, we can express f(z) as:
f(z)=(z2−4z+5)(z−k)2
f(z)=(z2−4z+5)(z2−2kz+k2)
Comparing the constant terms of both sides of the identity:
5k2=45
k2=9
Since k is given as a positive real number, k=3.
4. Expand the factorised form to find the values of $A$, $B$, and $C$.
Substitute k=3 back into the expression:
f(z)=(z2−4z+5)(z2−6z+9)
Expanding the expression:
f(z)=z2(z2−6z+9)−4z(z2−6z+9)+5(z2−6z+9)
f(z)=z4−6z3+9z2−4z3+24z2−36z+5z2−30z+45
f(z)=z4−10z3+38z2−66z+45
Comparing the coefficients with the original equation f(z)=z4+Az3+Bz2+Cz+45:
A=−10
B=38
C=−66
答案A=−10, B=38, C=−66, k=3