Eliminate with complex coefficients

Do not use a calculator in answering this question. Showing your working, find the complex numbers zz and ww which satisfy the simultaneous equations

4iz−3w=1+5i,2z+(1+i)w=2+6i.4iz-3w=1+5i,\qquad2z+(1+i)w=2+6i.
marks

[2017/MJC/Prelim/I/3]

Solution

Solve for z and w

Eliminate, then divide by a complex number

Without a calculator, find zz and ww given 4iz−3w=1+5i4iz-3w=1+5i and 2z+(1+i)w=2+6i2z+(1+i)w=2+6i.

Explore this idea

Hint

  1. Multiply the second equation by 2i2i so that the zz terms match.

  2. Subtract to eliminate zz and solve for ww by multiplying by the conjugate.

  3. Substitute back to find zz.

Answer · Solve for z and w

z=5+2i,w=−3+5iz=5+2i,w=-3+5i.

Explore the idea

Multiplying by i is a quarter-turn

4iz4iz is 4z4z turned through 90∘90^\circ anticlockwise. Slide to apply the turn, and read how far 4iz4iz lands from 3w3w.

Turn applied to 4z0°
Try an example
turned 4z
20 + 8i
turned 4z − 3w
29 − 7i
  • Moves as the slider changes; a faint path shows where it goes
Key idea

Equation (1) says 4iz4iz and 3w3w differ by exactly 1+5i1+5i. The quarter-turn turns 4z4z into 4iz4iz and closes that gap.

Ideas to try
  1. Slide to 90∘90^\circ. The amber point is 4iz=−8+20i4iz=-8+20i, and the red dashed segment from 3w3w is the vector 1+5i1+5i.
  2. Slide back to 0∘0^\circ: the same segment no longer equals 1+5i1+5i, which is why the rotation matters.
Takeaway

Real methods, complex numbers

Elimination works as usual. Division is the one new step: multiply by the conjugate so that the denominator becomes real.