(© Tony Kelly & Kieran Mills)
COMPLEX NUMBERS & MATRICES (Q 3, PAPER 1)
2002
3 (a) Express −1 + 3i in the form r (cos θ + i sin θ ), where i 2 = −1.
3 (b) (i) Given that z = 2 − i 3, find the real number t such that z 2 + tz is real.
(ii) w is a complex number such that ww − 2iw = 7 − 4i, where w is the complex
conjugate of w.
Find two possible values of w. Express each in the form p + qi, where p, q ∈ R.
3 (c) The following three statements are true whenever x and y are real numbers:
• x+y=y+x
• xy = yx
• If xy = 0 then either x = 0 or y = 0.
⎛ 3 −1⎞
Investigate whether the statements are also true when x is the matrix ⎜
⎟ and
⎝ −6 2 ⎠
⎛ 2 3⎞
y is the matrix ⎜
⎟.
⎝ 6 9⎠
SOLUTION
3 (a)
STEPS
1. Find r = z = Re 2 + Im 2 first.
2. Draw a free-hand picture to see what quadrant θ is in.
3. Find θ from tan θ =
Im
Re
and by looking at the picture.
4. Write z = r (cos θ + i sin θ ).
2
2
1. z = −1 + 3i ⇒ r = z = (−1) + ( 3 ) = 2
2. Draw a picture.
3. tan θ =
z
3
= 3 ⇒ θ = 60o (First quadrant)
−1
∴θ = 120o =
2π
(Second quadrant)
3
4. ∴ z = 2(cos 23π + i sin 23π )
Im
3
1
θ
-1
Re
(© Tony Kelly & Kieran Mills)
3 (b) (i)
z = 2 − i 3 ⇒ z 2 + tz = (2 − i 3 ) 2 + t (2 − i 3 )
= 4 − 4 3i + 3i 2 + 2t − it 3 = (1 + 2t ) − (4 3 + t 3 )i
As this number is real ⇒ 4 3 + t 3 = 0 ⇒ t = −4
3 (b) (ii)
Let w = p + qi and w = p − qi.
ww − 2iw = 7 − 4i ⇒ ( p + qi )( p − qi ) − 2i ( p + qi ) = 7 − 4i
⇒ p 2 + q 2 − 2ip − 2qi 2 = 7 − 4i ⇒ ( p 2 + q 2 + 2q) − 2ip = 7 − 4i
Equate the real and imaginary parts ⇒ ( p 2 + q 2 + 2q ) = 7 and −2 p = −4 ⇒ p = 2
⇒ (2) 2 + q 2 + 2q = 7 ⇒ q 2 + 2q − 3 = 0 ⇒ (q + 3)(q − 1) = 0
⇒ q = −3, 1
Answers: 2 − 3i, 2 + i
3 (c)
•
•
⎛ 3 −1⎞ ⎛ 2 3 ⎞ ⎛ 5 2 ⎞
⎛ 2 3 ⎞ ⎛ 3 −1⎞ ⎛ 5 2 ⎞
x+ y =⎜
⎟ and y + x = ⎜
⎟=⎜
⎟+⎜
⎟
⎟=⎜
⎟+⎜
⎝ −6 2 ⎠ ⎝ 6 9 ⎠ ⎝ 0 11⎠
⎝ 6 9 ⎠ ⎝ −6 2 ⎠ ⎝ 0 11⎠
∴ x + y = y + x (True)
⎛ 3 −1⎞ ⎛ 2 3 ⎞ ⎛ 0 0 ⎞
⎛ 2 3 ⎞ ⎛ 3 −1⎞ ⎛ −12 4 ⎞
xy = ⎜
⎟ and yx = ⎜
⎟=⎜
⎟⎜
⎟
⎟=⎜
⎟⎜
⎝ −6 2 ⎠ ⎝ 6 9 ⎠ ⎝ 0 0 ⎠
⎝ 6 9 ⎠ ⎝ −6 2 ⎠ ⎝ −36 12 ⎠
∴ xy ≠ yx (False)
• If xy = 0 then either x = 0 or y = 0. This is false as it was already shown in the previous
part xy = 0 even though x ≠ 0 and y ≠ 0.