Now substitute these into the equation.
P3 11
7 24i a( 2 11i) b(3 4i) 10( 2 i) 25 0 ( 7 2a 3b 20 25) ( 24 11a 4b 10)i 0
Complex numbers
Equating real and imaginary parts gives 2a 3b 2 0 11a 4b 14 0 Solving these equations simultaneously gives a The equation is z 4
2z 3
2z 2
Since 2
i is one root, 2
So (z
i)(z
2
2
10z
25
2, b
2.
0.
i is another root.
(z 2)2 1 z 2 4z 5
i)
is a factor. Using polynomial division or by inspection z4
2z 3
2z 2
10z
(z 2
25
4z
5)(z 2
2z
5).
The other two roots are the solutions of the quadratic equation z 2
2z
5
0.
Using the quadratic formula z
2±
4−4×5 2
2±
− 16
2 2 ± 4i 2 1 ± 2i
The roots of the equation are 2 ± i and 1 ± 2i. EXERCISE 11H
304
i is a root of z 3
z2
1
Check that 2
2
One root of z 3 15z 2 Solve the equation.
3
Given that 1 i is a root of z 3 p and q, and the other roots.
4
One root of z 4 10z 3 Solve the equation.
42z 2
82z
65
0 is 3
5
The equation z 4 8z 3 Solve the equation.
20z 2
72z
99
0 has a pure imaginary root.
76z
140 pz 2
7z
15
0, and find the other roots.
0 is an integer. qz
12
0, find the real numbers 2i.