Answer to Question #116819 in Analytic Geometry for desmond

Question

(a) Find the real root of the equation z3 + z + 10 = 0 given that one root is 1 − 2i.
(b) Given that 3 + i is a root of the equation z3 − 3z2 − 8z + 30 = 0, find the
remaining roots.
(c) Given that 1 + i is a root of the equation z3 − 2z + k = 0, find the other two
roots and the value of the real constant k.
(d) Given that 2 − 3i is a root of the equation z3 + pz2 + qz + 13 = 0, find the other
two roots and the values of the real constants p and q.
(e) Show that z = i is a root of the equation z4 + z3 + z − 1 = 0. Find the three
other roots.

Answer 1

a) let f(z) = z³+z+10=0

If z=1-2i is a root of f , then, z=1+2i is also a root of f as complex roots occur in conjugate pairs.

thus, z = 1-2i and z= 1+2i

z-1 = $\displaystyle \mp$ i

(z-1)²+4 divides f(z)

f(z) = [(z-1)²+4] [ z+2]

the real root of f(z) is z= -2

verification: f(-2) = 0

b) let f(z) = z³ − 3z² − 8z + 30 = 0

If z=3+i is a root of f , then, z=3-i is also a root of f as complex roots occur in conjugate pairs.

thus, z = 3+i and z= 3-i

(z-3)² + 1 divides f(z)

f(z) = [(z-3)²+1] [ z+3]

the real root of f(z) is z= -3

verification: f(-3) = 0

c) let f(z) =  z³ − 2z + k = 0

If z=1+i is a root of f , then, z=1-i is also a root of f as complex roots occur in conjugate pairs.

thus, z = 1+i and z= 1-i

(z-1)² + 1 divides f(z), so remainder will be equal to zero

f(z) = [(z-1)²+1] [ z+2] + (k-4)

here, the remainder is (k-4) = 0

thus, k=4

f(z) = z³ − 2z + 4

the real root of f(z) is z= -2

verification: f(-2) = 0

d) let f(z) =  z³ + pz² + qz + 13 = 0

If z=2-3i is a root of f , then, z=2+3i is also a root of f as complex roots occur in conjugate pairs.

thus, z = 2-3i and z= 2+3i

(z-2)² + 9 divides f(z), so remainder will be equal to zero

f(z) = [(z-2)²+9] [ z+(4+p)] + [(q-13)z-4(4+p)z+13-13(4+p)]

here, the remainder is [(q-13)z-4(4+p)z+13-13(4+p)]= 0

thus, p=-3 , q=9

f(z) = z³-3z²+9z+13

the real root of f(z) is z= -1

verification: f(-1) = 0

e) let f(z) = z⁴+z³+z-1=0

If z=i is a root of f , then, z=-i is also a root of f as complex roots occur in conjugate pairs.

thus, z = i and z= -i

(z)²+1 divides f(z)

f(z) = [(z)²+1] [ z²+z-1]

the real roots of f(z) is z= $\frac{-1\displaystyle \mp \sqrt{5}}{2}$

verification: f($\frac{-1+ \sqrt{5}}{2}$ ) = 0 and f($\frac{-1- \sqrt{5}}{2}$ ) =0