Answer to Question #117071 in Complex Analysis for Amoah Henry

Question

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.
(f) Show that z = −1+i is a root of the equation z4 −2z3 −z2 +2z +10 = 0. Find the remaining roots.
(g) Solve the equation 6z4 − 47z3 + 148z2 − 167z + 52 = 0

Answer 1

a)

$z_1=1-2i, z_2=1+2i$

$z^3+z+10=(z-(1-2i))(z-(1+2i))(Az+B)$

$z^3+z+10=((z-1)^2+4)(Az+B)$

$z^3+z+10=(z^2-2z+5)(Az+B)$

$z^3+z+10=Az^3+Bz^2-2Az^2-2Bz+5Az+5B$

$z^3+z+10=Az^3+(B-2A)z^2+(5A-2B)z+5B$

$A=1, B=2$

$Az+B=0$

$z+2=0$

Answer:

$z_3=-2$

b)

$z_1=3+i, z_2=3-i$

$z^3 − 3z^2 − 8z + 30=(z-(3+i))(z-(3-i))(Az+B)$

$z^3 − 3z^2 − 8z + 30=((z-3)^2+1)(Az+B)$

$z^3 − 3z^2 − 8z + 30=(z^2-6z+10)(Az+B)$

$z^3 − 3z^2 − 8z + 30=Az^3+Bz^2-6Az^2-6Bz+10Az+10B$

$z^3 − 3z^2 − 8z + 30=Az^3+(B-6A)z^2+(10A-6B)z+10B$

$A=1, B=3$

$Az+B=0$

$z+3=0$

Answer:

$z_3=-3$

c)

$z_1=1+i, z_2=1-i$

$z^3 − 2z + k=(z-(1+i))(z-(1-i))(Az+B)$

$z^3 − 2z + k=((z-1)^2+1)(Az+B)$

$z^3 − 2z + k=(z^2-2z+2)(Az+B)$

$z^3 − 2z + k=Az^3+Bz^2-2Az^2-2Bz+2Az+2B$

$z^3 − 2z + k=Az^3+(B-2A)z^2+(2A-2B)z+2B$

$A=1, B=2$

$2B=k$

Answer:

$z_3=-2, k=4$

d)

$z_1=2-3i. z_2=2+3i$

$z^3 +pz^2 +qz +13=(z-(2-3i))(z-(2+3i))(Az+B)$

$z^3 +pz^2 +qz +13=((z-2)^2+9)(Az+B)$

$z^3 +pz^2 +qz +13=(z^2-4z+13)(Az+B)$

$z^3 +pz^2 +qz +13=Az^3+Bz^2-4Az^2-4Bz+13Az+13B$

$A=1, B=1$

Answer:

$z_3=-1$

$p=B-4A=1-4=-3$

$q=13A-4B=13-4=9$

e)

$i^4 + i^3 + i −1=1-i+i-1=0$

$z_1=i, z_2=-i$

$z^4 + z^3 + z − 1=(z-i)(z+i)(Az^2+Bz+C)$

$z^4 + z^3 + z − 1=(z^2+1)(Az^2+Bz+C)$

$z^4 + z^3 + z − 1=Az^4+Az^2+Bz^3+Bz+Cz^2+C$

$A=1, C=-1, B=1$

$z^2+z-1=0$

Answer:

$z_3=\frac {-1-\sqrt{1+4}}{2}=\frac {-1-\sqrt{5}}{2}$

$z_4=\frac {-1+\sqrt{5}}{2}$

f)

$(-1+i)^4 −2(-1+i)^3 −(-1+i)^2 +2(-1+i) +10=$

$=(1-2i-1)^2-2(1-2i-1)(-1+i)-(1-2i-1)-2+2i+10=$

$=-4+4i(-1+i)+4i+8=4-4i-4+4i=0$

$z_1=-1+i, z_2=-1-i$

$z^4 −2z^3 −z^2 +2z +10=(z-(-1+i))(z-(-1-i))(Az^2+Bz+C)$

$z^4 −2z^3 −z^2 +2z +10=((z+1)^2+1)(Az^2+Bz+C)$

$z^4 −2z^3 −z^2 +2z +10=(z^2+2z+2)(Az^2+Bz+C)$

$z^4 −2z^3 −z^2 +2z +10=Az^4+Bz^3+Cz^2+2Az^3+2Bz^2+2Cz+2Az^2+2Bz+2C$

$A=1, C=5$

$2C+2B=2$

$C+B=1$

$B=-C=-5$

$z^2-5z+5=0$

Answer:

$z_3=\frac {5-\sqrt{25-20}}{2}=\frac {5-\sqrt{5}}{2}$

$z_4=\frac {5+\sqrt{5}}{2}$

g)

The Rational Root Theorem states that if a polynomial zeroes for a rational number  P/Q   then  P  is a factor of the Trailing Constant and  Q  is a factor of the Leading Coefficient.

In this case, the Leading Coefficient is  6  and the Trailing Constant is  52.

The factor(s) are:

of the Leading Coefficient :  1,2 ,3 ,6

of the Trailing Constant :  1 ,2 ,4 ,13 ,26 ,52

After the test we have:

$z_1=\frac {1}{2}, z_2=\frac {4}{3}$

Then:

$\frac {6z^4 − 47z^3 + 148z^2 − 167z + 52}{(2z-1)(3z-4)}=z^2-6z+13$

$z^2-6z+13=0$

$z_3=\frac {6-\sqrt{36-52}}{2}=3-2i$

$z_4=3+2i$