Use conjugacy only with real coefficients

A graphic calculator is not to be used in answering this question.

(a) w3+pw2+qw+30=0w^3+pw^2+qw+30=0, with p,qp,q real constants, has root w=2−iw=2-i. Find p,qp,q, showing working. marks

(b) z2+(−5+2i)z+(21−i)=0z^2+(-5+2i)z+(21-i)=0 has root z=3+uiz=3+ui, where uu is real. Find uu and hence the second root in Cartesian form a+bia+bi, showing working. marks

[2017/IJC/Prelim/I/4]

Total marks: 8

Solution

(a)

Find p and q from a conjugate pair

w3+pw2+qw+30=0w^3+pw^2+qw+30=0, with p,qp,q real, has root 2−i2-i. Find p,qp,q.

Explore this idea

Hint

  1. Use the conjugate root to form a real quadratic factor.

  2. Find the third root from the constant term.

Answer · (a)

p=2,q=−19p=2,q=-19.

Explore the idea

Slide the third root

The pair 2±i2\pm i is fixed. Move the real root rr and read the coefficients. Only r=−6r=-6 gives constant term 3030.

Real root r-6
Try an example
p = −(4 + r)
2
q = 5 + 4r
-19
constant term −5r
30
  • Moves as the slider changes; a faint path shows where it goes
Key idea

A real cubic is pinned down by its conjugate pair and one real root. The constant term then fixes the real root.

Ideas to try
Slide until the constant term is 3030. Then read pp and qq.
(b)

A complex quadratic with a root of the form 3 + ui

z2+(−5+2i)z+(21−i)=0z^2+(-5+2i)z+(21-i)=0 has root z=3+uiz=3+ui, where uu is real. Find uu and the second root.

Explore this idea

Hint

  1. Substitute z=3+uiz=3+ui and equate real and imaginary parts.

  2. Use the sum of roots for the second root.

Answer · (b)

u=−5u=-5; second root 2+3i2+3i.

Explore the idea

Slide along the line Re z = 3

The point 3+ui3+ui moves vertically. The two readouts are the real and imaginary parts of the quadratic. Both are zero only at one height.

Value of u-5
Try an example
real part of the quadratic
0
imaginary part
0
z
3 − 5i
  • Moves as the slider changes; a faint path shows where it goes
Key idea

With a complex coefficient, the conjugate of a root need not be a root. The second root here is 2+3i2+3i, not 3+5i3+5i.

Ideas to try
  1. At u=3u=3 the real part is 00 but the imaginary part is 88: not a root.
  2. At u=−5u=-5 both parts vanish. The conjugate 3+5i3+5i (red) is not a root.
Takeaway

Check the coefficients first

Only polynomials with all-real coefficients have conjugate pairs. Otherwise rely on sums and products of roots.