A purely imaginary root unlocks a cubic

(i) Solve z2−(1+2i)z+1+7i=0z^2-(1+2i)z+1+7i=0. marks

(ii) One root of z3−(1+4i)z2−(3−9i)z+(14−2i)=0z^3-(1+4i)z^2-(3-9i)z+(14-2i)=0 is purely imaginary. Find all roots in Cartesian form a+iba+ib. marks

[2018/RI/Promo/11a]

Total marks: 8

Solution

(i)

Solve a complex quadratic

Solve z2−(1+2i)z+1+7i=0z^2-(1+2i)z+1+7i=0.

Explore this idea

Hint

  1. Find two numbers with sum 1+2i1+2i and product 1+7i1+7i.

Answer · (i)

2−i,−1+3i2-i,-1+3i.

Explore the idea

The roots and their sum

Drag to look around the Argand plane. Real parts run along the horizontal axis and imaginary parts along the vertical axis.

Reset the controls
Key idea

The sum of the roots is the vector sum, the diagonal of the parallelogram they span.

Ideas to try
The two dashed edges complete a parallelogram. Its far corner is the sum of the roots, the coefficient 1+2i1+2i.
(ii)

Purely imaginary root of a cubic

One root of z3−(1+4i)z2−(3−9i)z+(14−2i)=0z^3-(1+4i)z^2-(3-9i)z+(14-2i)=0 is purely imaginary. Find all roots in the form a+iba+ib.

Explore this idea

Hint

  1. Set z=itz=it with tt real and separate parts.

  2. Use both equations to find tt.

  3. Divide out the factor and solve the quadratic.

Answer · (ii)

2i,2−i,−1+3i2i,2-i,-1+3i.

Explore the idea

Slide up the imaginary axis

The point z=itz=it moves along the imaginary axis. Both readouts must be zero for it to be a root.

Value of t in z = it2
Try an example
real part t² − 9t + 14
0
imaginary part −t³ + 4t² − 3t − 2
0
  • Moves as the slider changes; a faint path shows where it goes
Key idea

Setting z=itz=it turns one complex equation into two real ones. A genuine root must satisfy both at the same tt.

Ideas to try
  1. At t=7t=7 the real part is 00 but the imaginary part is −170-170: not a root.
  2. At t=2t=2 both parts vanish: z=2iz=2i is a root.
Takeaway

Two real equations, one unknown

When one root has a known form, such as purely imaginary, split into real and imaginary parts. Divide out the factor to find the remaining roots.