Reduce a complex system to a quadratic

The complex numbers zz and ww satisfy iw2+2wz=2iiw^2+2wz=2i and z+iz=2+iwz+iz=2+iw. Find zz and ww, giving your answers in the form a+iba+ib, where aa and bb are real numbers. marks

[2024/ACJC/Prelim/I/1]

Solution

Find z and w

Substitute, then solve a complex quadratic

iw2+2wz=2iiw^2+2wz=2i and z+iz=2+iwz+iz=2+iw. Find zz and ww in the form a+iba+ib.

Explore this idea

Hint

  1. Solve the second equation for zz.

  2. Substitute into the first to get a quadratic in ww.

  3. Solve it and find each matching zz.

Answer · Find z and w

(z,w)=(1,1+i)(z,w)=(1,1+i) or (3/5−6i/5,−3/5+i/5)(3/5-6i/5,-3/5+i/5).

Explore the idea

Two solution pairs

Choose a solution and read how well each equation is satisfied. Both residuals are zero.

Solution number1
Try an example
z
1
w
1 + i
residual of eq. 1
0
residual of eq. 2
0
  • Moves as the slider changes; a faint path shows where it goes
Key idea

A quadratic in ww means two answers, and each gives its own zz. Always return both pairs.

Ideas to try
Switch between the two solutions. Both give a residual of 00 in each equation.
Takeaway

Reduce to something familiar

Eliminate one unknown until a quadratic remains, then use the quadratic formula. Complex coefficients only change the arithmetic, not the method.