(© Tony Kelly & Kieran Mills)
COMPLEX NUMBERS & MATRICES (Q 3, PAPER 1)
2004
3 (a) Find the real numbers p and q such that 2( p + iq ) + i ( p − iq ) = 5 + i, where i 2 = −1.
π
π
4π
4π
and z2 = cos + i sin . Evaluate z1 z2 , giving your
+ i sin
3
3
3
3
answer in the form x + iy.
3 (b) (i) z1 = cos
(ii) w1 = a + ib and w2 = c + id . Prove that ( w1w2 ) = ( w1 )( w2 ), where w is the
complex conjugate w.
⎛ 1 −3 ⎞
⎛ 4 3⎞
3 (c) Let A = ⎜
⎟ and P = ⎜
⎟.
⎝ −1 2 ⎠
⎝ −2 −1⎠
Evaluate A−1 PA and hence ( A−1 PA)10.
(i)
−1
10
−1 10
(ii) Use the fact that ( A PA) = A P A to evaluate P10.
SOLUTION
3 (a)
2( p + iq ) + i ( p − iq ) = 5 + i ⇒ (2 p − i 2 q ) + ( p + 2q ) = 5 + i
⇒ (2 p + q ) + ( p + 2q ) = 5 + i [Equate the real parts and the imaginary parts.]
⇒ 2 p + q = 5 and p + 2q = 1
Solving simultaneously: p = 3, q = −1
3 (b) (i)
RULES FOR COMBINING OBJECTS IN POLAR FORM
1. (cos A ⊕ i sin A)(cos B ⊕ i sin B) = cos( A + B) + i sin( A + B)
z1 = cos
π
π
4π
4π
and z2 = cos + i sin .
+ i sin
3
3
3
3
⎛ 4π π ⎞
⎛ 4π π ⎞
⎛ 5π
⇒ z1 z2 = cos ⎜
+ ⎟ + i sin ⎜
+ ⎟ = cos ⎜
⎝ 3 3⎠
⎝ 3 3⎠
⎝ 3
⎞
⎛ 5π ⎞
⎟ + i sin ⎜
⎟
⎠
⎝ 3 ⎠
= cos 300o + i sin 300o = cos 60o − i sin 60o = 12 −
i
3
2
3 (b) (ii)
LHS
RHS
( w1w2 ) = (a + ib)(c + id )
( w1 )( w2 ) = (a + ib)(c + id )
= (ac − bd ) + (ad + bc)i
= (a − ib)(c − id )
= (ac − bd ) − (ad + bc)i
= (ac − bd ) − (ad + bc)i
(© Tony Kelly & Kieran Mills)
3 (c) (i)
⎛ 1 −3 ⎞
⎛ 4 3⎞
A=⎜
⎟ and P = ⎜
⎟.
⎝ −1 2 ⎠
⎝ −2 −1⎠
n
⎛ a 0⎞
⎛ a 0 ⎞ ⎛ an
n
D=⎜
⇒
D
=
⎟
⎜
⎟ =⎜
⎝0 b⎠
⎝0 b⎠ ⎝ 0
0⎞
⎟
bn ⎠
....... 7
Evaluate A−1 PA and hence ( A−1 PA)10.
⎛ d −b ⎞
⎛a b⎞
1
−1
A=⎜
⎜
⎟
⎟⇒ A =
c
d
−
ad
bc
(
)
⎝ −c a ⎠
⎝
⎠
⎛ 1 −3 ⎞
1 ⎛ 2 3 ⎞ 1 ⎛ 2 3 ⎞ ⎛ −2 −3 ⎞
−1
A=⎜
⎟=⎜
⎟= ⎜
⎟⇒ A =
⎜
⎟
2 − 3 ⎝ 1 1 ⎠ −1 ⎝ 1 1 ⎠ ⎝ −1 −1 ⎠
⎝ −1 2 ⎠
⎛ −2 −3 ⎞ ⎛ 4 3 ⎞ ⎛ 1 −3 ⎞ ⎛ −2 −3 ⎞ ⎛ 1 −3 ⎞ ⎛ 1 0 ⎞
A−1 PA = ⎜
⎟=⎜
⎟⎜
⎟
⎟=⎜
⎟⎜
⎟⎜
⎝ −1 −1 ⎠ ⎝ −2 −1⎠ ⎝ −1 2 ⎠ ⎝ −2 −2 ⎠ ⎝ −1 2 ⎠ ⎝ 0 2 ⎠
The result is a diagonal matrix which means you can use formula 7.
10
⎛110
⎛1 0⎞
=
( A PA) = ⎜
⎜
⎟
⎝0 2⎠
⎝0
−1
10
0 ⎞
0 ⎞ ⎛1
=⎜
⎟
10 ⎟
2 ⎠ ⎝ 0 1024 ⎠
3 (c) (ii)
You are told that ( A−1 PA)10 = A−1 P10 A and are asked to find P10.
0 ⎞
0 ⎞ ⎛ −2 −3 ⎞ 10 ⎛ 1 −3 ⎞
⎛1
⎛1
−1 10
⇒⎜
⎟P ⎜
⎟=⎜
⎟ = A P A⇒⎜
⎟
⎝ 0 1024 ⎠ ⎝ −1 −1 ⎠
⎝ 0 1024 ⎠
⎝ −1 2 ⎠
To isolate P10, multiply each side by matrix A at the front and A−1 at the end.
Therefore, ( A−1 PA)10 = A−1 P10 A ⇒ A( A−1 PA)10 A−1 = AA−1 P10 AA−1
⇒ A( A−1 PA)10 A−1 = P10
0 ⎞ ⎛ −2 − 3 ⎞
⎛ 1 −3 ⎞ ⎛ 1
⇒ P10 = ⎜
⎟
⎟⎜
⎟⎜
⎝ −1 2 ⎠ ⎝ 0 1024 ⎠ ⎝ −1 −1 ⎠
⎛ 1 −3072 ⎞ ⎛ −2 −3 ⎞ ⎛ 3070 3069 ⎞
=⎜
⎟
⎟=⎜
⎟⎜
⎝ −1 2048 ⎠ ⎝ −1 −1 ⎠ ⎝ −2046 −2045 ⎠
....... 8