Conjugation and component equations

(a) Given arg⁡(a+ib)=θ\arg(a+ib)=\theta, where a,ba,b are real and positive, find in terms of θ\theta and/or π\pi:

(i) arg⁡(a−ib)\arg(a-ib);

(ii) arg⁡(b+ia)\arg(b+ia). marks

(b) Solve iz+2w=0iz+2w=0 and z−w∗=3z-w^*=3, giving z,wz,w in the form a+iba+ib, where a,ba,b are real. marks

[2015/JJC/Prelim/II/1]

Total marks: 8

Solution

(a)(i)

Argument of a conjugate

arg⁡(a+ib)=θ\arg(a+ib)=\theta where a,b>0a,b>0. Find arg⁡(a−ib)\arg(a-ib).

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Hint

  1. Conjugation reflects in the real axis.

Answer · (a)(i)

−θ-\theta.

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Reflect in the real axis

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

Angle θ of z35°
Try an example
arg(a + ib)
35°
arg(a − ib)
-35°
  • Moves as the slider changes; a faint path shows where it goes
Key idea

A conjugate has the same modulus and the opposite argument.

Ideas to try
Change θ\theta. The two points stay mirror images, and their arguments are always opposite.
(a)(ii)

Argument after swapping parts

Find arg⁡(b+ia)\arg(b+ia).

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Hint

  1. Relate b+iab+ia to the conjugate by a quarter-turn.

Answer · (a)(ii)

π/2−θ\pi/2-\theta.

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Swapping a and b reflects in y = x

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

Angle θ of a + ib30°
Try an example
arg(a + ib)
30°
arg(b + ia)
60° = 90° − θ
  • Moves as the slider changes; a faint path shows where it goes
Key idea

The two arguments always add to π2\frac\pi2, since the points are mirror images in the line y=xy=x.

Ideas to try
Slide to 45∘45^\circ: the two points coincide on the dashed line y=xy=x, because a=ba=b there.
(b)

Components when a conjugate appears

Solve iz+2w=0iz+2w=0 and z−w∗=3z-w^*=3, giving z,wz,w in the form a+iba+ib.

Explore this idea

Hint

  1. Write w=x+iyw=x+iy, so w∗=x−iyw^*=x-iy.

  2. Eliminate zz and equate real and imaginary parts.

Answer · (b)

w=1−2i,z=4+2iw=1-2i,z=4+2i.

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The solution on the Argand diagram

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

Conjugation mirrors ww in the real axis. Then zz is w∗w^* shifted 33 units to the right.

Ideas to try
  1. The red segment from w∗w^* to zz is horizontal with length 33: that is z−w∗=3z-w^*=3.
  2. iziz is zz turned a quarter-turn anticlockwise, and it equals −2w-2w.
Takeaway

When a conjugate appears, split into parts

Equations containing both ww and w∗w^* are not linear in ww alone. Writing w=x+iyw=x+iy turns them into two real equations.