complex numbers xii – standard mathematics. if n is a positive integer, prove that
TRANSCRIPT
Complex Numbers
XII – STANDARD
MATHEMATICS
PREPARED BY:R.RAJENDRAN. M.A., M. Sc., M. Ed.,K.C.SANKARALINGA NADAR HR. SEC. SCHOOL,CHENNAI-21
If n is a positive integer, prove that
θ2
πnisin θ
2
πn cos
θ icosθsin 1
θ icosθsin 1
n
θ2
πisin θ
2
π cos θ icosθsin zLet
θ2π
isin θ2π
cos
1
z
1Then
θ2
πisin θ
2
π cos
θ icosθsin
n
n
z
11
z1
θ icosθsin 1
θ icosθsin 1
n
z
1zz1
nz
θ2
πisin θ
2
π cos
n
θ2
πnisin θ
2
πn cos
If a = cos 2 + isin 2, b = cos2 + isin 2, c = cos 2 + isin 2. Prove that (i) γβα2cos
abc
1abc
a = cos 2 + isin 2
b = cos 2 + isin 2
c = cos 2 + isin 2
abc = (cos 2 + isin 2 )(cos 2 + isin 2) (cos 2 + isin 2)
= cos (2 + 2 + 2) + isin (2 + 2 + 2)
2
1
γ2β22αsinγ2β22αcosabc i
= cos ( + + ) + isin ( + + ) ……..(1)
2
γ2β22αsini
2
γ2β22αcos
11γβαsiniγβαcosabc
abc
1
= cos ( + + ) – i sin ( + + ) …….(2)
adding (1) and (2) we get
γβαcos2abc
1abc
= (cos 2 + isin 2 )(cos 2 + isin 2)(cos 2 + isin 2)– 1
= (cos 2 + isin 2 )(cos 2 + isin 2)(cos 2 – isin 2)= (cos 2 + isin 2 )(cos 2 + isin 2)(cos (–2) + isin (–2))= cos(2 + 2 – 2) + isin(2 + 2 – 2)= cos 2( + – ) + isin 2( + – )………….(1)
If a = cos 2 + isin 2, b = cos2 + isin 2, c = cos 2 + isin 2. Prove that (ii)
γβα2cos2abc
cba 222
abc
cbaLHS
222
abc
c
abc
ba 222
ab
c
c
ab
1abcc
ab
-1
c
ab
ab
c
= [cos 2( + – ) + i sin 2( + – )]-1
= cos 2( + – ) – i sin 2( + – )………(2)
Adding (1) and (2), we get,
)γβα(2cos2abc
cba 222
γβα2cos2ab
c
c
ab
Simplify: 32
43
)4θisin θ4 (cos3θisin θ3cos
3θisin θ3cosθ2siniθ2cos
32
43
)4θisin θ4 (cos3θisin θ3cos
3θisin θ3cosθ2siniθ2cos
3423
432
)θisin θ (cosθisin θcos
3θ -isinθ)3cos(θsiniθcos
4323
4332
)θisin θ (cosθisin θcos
θisin θcosθsiniθcos
126
126
θisin θ cos θisin θcos
θisin θcos θisin θcos
(cos + isin )0
= 1
Simplify:6
6
πcos i
6
πsin
6
6
πcos i
6
πsin
6
6
π
2
πsin i
6
π
2
πcos
6
6
2πsin i
6
2πcos
6
2π6sin i
6
2π6 cos
2πsin i2π cos
101 i
Prove that: 1
8
πcos i
8
πsin1
8
πcos i
8
πsin1
8
8
πcos i
8
πsinzLet
8
π
2
πsin i
8
π
2
πcos
8
π4πsin i
8
π4πcos
8
3πsin i
8
π3 cosz
1
8
3πsin i
8
π3cos
z
1
8
3πsin i
8
π3cos
z
1
88
z
11
z1
8
πcos i
8
πsin1
8
πcos i
8
πsin1
8
8
z
z
z1z1
8
8
3πsin i
8
π3cos
3πsin i3π cos
101 i
If cos + cos + cos = 0 = sin + sin + sin : prove that(i) cos 3 + cos 3 + cos 3 = 3cos( + + )(ii) sin 3 + sin 3 + sin 3 = 3sin( + + )
Given: cos + cos + cos = 0 and
sin + sin + sin = 0
(cos + cos + cos ) + i(sin + sin + sin ) = 0
(cos + isin )(cos + isin )(cos + isin ) = 0
Let a = cos + isin
b = cos + isin
c = cos + isin
If a + b + c = 0, then a3 + b3 + c3 = 3abc
(cos + isin )3 + (cos + isin )3 + (cos + isin )3
= 3(cos + isin )(cos + isin )(cos + isin )
(cos 3 + isin 3) + (cos 3 + isin 3) + (cos 3 + isin 3)
= 3[cos ( + + ) + isin ( + + )]
(cos 3 + cos 3 + cos 3) + i(sin 3 + sin 3 + sin 3)
= 3cos ( + + ) + 3isin ( + + )
Equating real and imaginary parts
cos 3 + cos 3 + cos 3 = 3cos ( + + )
sin 3 + sin 3 + sin 3 = 3sin ( + + )
If n is a positive integer, prove that
(i) (1 + i)n + (1 – i)n =
Let 1 + i = r(cos + isin )
4
nπcos2 2
2n
22 11 r 211
θisin θ cos2i1
2
1θ cos
2
1θsin
4
πθ
4
πisin
4
π cos2i1
n
nn
4
πisin
4
π cos2i)(1
)1(..........4
nπisin
4
nπ cos)2(i)1( 2
nn
Substituting i = - i
n
nn
4
πisin
4
π cos2i)(1
)2.(..........4
nπisin
4
nπ cos)2(i)(1 2
nn
Adding (1) and (2), we get
4
nπisin
4
nπ cos
4
nπisin
4
nπ cos)2(
i)(1i)(1
2
n
nn
4
nπ 2cos)2( 2
n
4
nπ cos)2(
12
n
4
nπ cos)2(i)(1i)(1 2
2nnn
If n is a positive integer, prove that
(ii) (3 + i)n + (3 – i)n = 6
nπcos2 1n
Let 3 + i = r(cos + isin )
22 1)3( r
2413
θisin θ cos2i3
2
3θ cos
2
1θsin
6
πθ
6
πisin
6
π cos2i3
Substituting i = –i
n
nn
6
πisin
6
π cos2i)3(
)2.(..........6
nπisin
6
nπ cos)2(i)3( nn
Adding (1) and (2), we get
)1......(6
nπisin
6
nπ cos2i)3( nn
n
nn
6
πisin
6
π cos2i)3(
6
nπ 2cos)2( n
4
nπ cos)2( 1n
6
nπ cos)2(i)3(i)3( 1nnn
6
nπisin
6
nπ cos
6
nπisin
6
nπ cos)2(
i)3(i)3(
n
nn
If n is a positive integer, prove that(1 + cos + isin )n + (1 + cos - isin )n
(1 + cos + isin )n + (1 + cos - isin )n2
nθcos
2
θcos2 n1n
n2
n2
2
θcos
2
θi2sin
2
θcos2
2
θcos
2
θi2sin
2
θcos2
nnn
nnn
2
θisin
2
θcos
2
θcos2
2
θisin
2
θcos
2
θcos2
2
nθisin
2
nθcos
2
θcos2
2
nθisin
2
nθcos
2
θcos2 nnnn
2
nθisin
2
nθcos
2
nθisin
2
nθcos
2
θcos2 nn
2
nθcos2
2
θcos2 nn
2
nθcos
2
θcos2 n1n
2
nθcos
2
θcos2 n1n
If and are the roots of x2 – 2x + 4 = 0 prove that n – n = i2n+1 sin (n/3) and deduce 9 – 9
x2 – 2x + 4 = 0 a = 1, b = –2, c = 4
a2
ac4bb-x
2
12
414(-2)(-2)- 2
2
1642
2
122
2
i 342 2
2
3i22
3 i1
The two roots are
3 i1α 3 i1β
Let 1 + i 3 = r(cos + isin )
22 )3(1 r 2431
θisin θ cos23i1
2
1θ cos
2
3θsin
3
πθ
3
πisin
3
π cos23i1
n
nnn
3
πisin
3
π cos23i1 α
)1....(..........3
nπisin
3
nπ cos2 α nn
Substituting i = –i we get
)2....(..........3
nπisin
3
nπ cos2 β nn
Subtracting (2) from (1) we get
nnnn 3i13i1 βα
3
nπisin
3
nπ cos2
3
nπisin
3
nπ cos2 nn
3
nπisin
3
nπ cos
3
nπisin
3
nπ cos2n
3
nπsin 2i2n
3
nπsin 2 1ni
If n = 9
9 – 9
3
9πsin 2 19i
3sin 210i
sin 210i
0)0(210 i
If x + 1/x = 2cos prove that (i)xn + 1/xn = 2cos n (ii) xn – 1/xn = 2isin n
θ 2cosx
1x
θ 2cosx
1x2
θ cos2x 1x2
0 1θ cos2x x2
a2
ac4bb-x
2
12
1142cosθ-2cosθ--x
2
2
4θ4cos2cosθ 2
2
θcos-14-θ 2cos 2
2
θsin4-θ 2cos 2
2
θsin4-θ 2cos 2
2
θsini2θ 2cos θsiniθ cos
θsiniθ cosx θsiniθ cosx
1
nn
nn θsiniθ cosθsiniθ cos
x
1x
nθsininθ cosnθsininθ cos
nθ cos2
nn
nn θsiniθ cosθsiniθ cos
x
1x
nθsininθ cosnθsininθ cos
nθsin i2
If x + 1/x = 2cos, y + 1/y = 2cos, prove that one of the values of
nm2cos x
y
y
x (i)
m
n
n
m
2cosx
1x
2cosx
1x2
cos2x 1x2
0 1 cos2x x2
a2
ac4bb-x
2
12
1142cos-2cos--x
2
2
44cos 2cos 2
nm2isin x
y
y
x (i)
m
n
n
m
2
sin i2 2cos
2
sin4- cos2 2
2
cos142cos 2
sin i cos
sin i cosx
sin i cosx
1
mm sin i cosx
msin im cosxm
similarly
sin i cosy
sin i cosy
1
nn sin i cosy
nsin in cosyn
Adding (1) and (2)
nm2cos x
y
y
x (2)(1)
m
n
n
m
1nmn
m
nsin in cosmsin im cosyxy
x
nsin in cosmsin im cos
(1).......nmsin inm cos
(2).....nmsin inm cos
yx
1
x
y
n
mm
n
nm2isin x
y
y
x(2)(1)
m
n
n
m
Solve: x8 + x5 – x3 – 1 = 0
x8 + x5 – x3 – 1 = 0
x5(x3 + 1) – (x3 + 1) = 0
(x3 + 1)(x5 – 1) = 0
x3 + 1 = 0, x5 – 1 = 0
Case (1) x3 + 1 = 0
x3 = – 1
= cos + isin
= cos(2k + ) + isin(2k + )
x = {cos(2k + ) + isin(2k + )}1/3
0,1,2k,3
π2kπsini
3
π2kπcos
The three values are
,3
πsini
3
πcos , πsiniπcos
3
5πsini
3
5πcos
Case (2) x5 – 1 = 0
x5 = 1
= cos 0 + isin 0
= cos(2k) + isin(2k)
x = {cos(2k) + isin(2k)}1/5
0,1,2,3,4k,5
2kπsini
5
2kπcos
The five values are
1 0sini0cos 0,kfor
,5
2πsini
5
2πcos 1, k for
,5
4πsini
5
4πcos 2, k for
,5
6πsini
5
6πcos 3, k for
5
8πsini
5
8πcos 4, k for
Solve: x7 + x4 + x3 + 1 = 0
x7 + x4 + x3 + 1 = 0
x4(x3 + 1) + (x3 + 1) = 0
(x3 + 1)(x4 + 1) = 0
x3 + 1 = 0, x4 + 1 = 0
Case (1) x3 + 1 = 0
x3 = – 1
= cos + isin
= cos(2k + ) + isin(2k + )
x = {cos(2k + ) + isin(2k + )}1/3
0,1,2k,3
π2kπsini
3
π2kπcos
The three values are
,3
πsini
3
πcos , πsiniπcos
3
5πsini
3
5πcos
Case (2) x4 + 1 = 0
x4 = – 1
= cos + isin
= cos(2k + ) + isin(2k + )
x = {cos(2k + ) + isin(2k + )}1/4
0,1,2,3k,4
1)π (2k sini
4
1)π (2k cos
The four values are
2
1i
2
1
4sini
4cos 0,kfor
2
1
2
1
4
3πsini
4
3πcos 1, k for i
2
1
2
1
4
5πsini
4
5πcos 2, k for i
2
1
2
1
4
7πsini
4
7πcos 3, k for i
Find all the values of (1 + i)1/4
22 11 r 211
θisin θ cos2i1
2
1θ cos
2
1θsin
4
πθ
4
πisin
4
π cos2i1
Let 1 + i = r(cos + isin )
4
π2kπisin
4
π2kπ cos2i1
4
1
4
1
4
1
4
π2kπisin
4
π2kπ cos2i1
4
π2kπ
4
1isin
4
π2kπ
4
1cos2 4
1
For k = 0,1,2,3,4 we get all the values
Find all the values of
22
2
3
2
1 r
1
4
4
2
1θ cos
2
3θsin
3
πθ
And hence prove that the product of the values is 1
4
3
2
3
2
1
i
)sin(cos2
3
2
1Let iri
)sin(cos2
3
2
1 ii
Since sin is – ve and cos is +ve
lies in the fourth quadrant
3sin
3cos
2
3
2
1
ii
4
34
3
3
πisin
3
π cos
2
3
2
1
i
For k = 0,1,2,3, we get all the values
4
1
πisinπ cos
4
1
π2kπisinπ2kπ cos
π2kπ
4
1isinπ2kπ
4
1cos
4
sini4
cos 0,kfor
4
πsini
4
πcos 1, k for
4
3πsini
4
3πcos 2, k for
4
5πsini
4
5πcos 3, k for
Their product =
4
5sini
4
5cos
4
3sini
4
3cos
4sini
4cos
4sini
4cos
4
5
4
3
44sini
4
5
4
3
44cos
4
8sini
4
8cos
2sini2cos
= 1
x2 – 2px + (p2 + q2) = 0 a = 1, b = –2p, c = p2 + q2
a2
ac4bb-x
2
12
qp14(-2p)(-2p)- 222
2
)qp(44p2p 222
2
q42p 2
2
i q 42p 22
2
iq22p q ip
The two roots are
q ipα q ipβ
If and are the roots of x2 – 2px + (p2 + q2) = 0 and tan = q/ y+p prove that
n1
nn
sin
nsin )(y)(y
nq
py
qtan
py
q
cos
sin
qsinθ
cosθpy
pqsinθ
cosθy
iqppqsinθ
cosθy
iqqsinθ
cosθy
sinθ
)sin iq(cosθy
θsin
)sin i(cosθq)(y
n
nn n
iqppqsinθ
cosθy
iqqsinθ
cosθy
sinθ
)sin iq(cosθy
θsin
)sin i(cosθq)(y
n
nn n
)1....(θsin
)sin(cosnq)(y
n
n nin
)2.....(θsin
)sinn i(cosnθq)(y
n
n n
nn )(y)(y
)()(sin
)sinn i(cosnθq)sinn i(cosnθqn
nn
iqpiqp
2iq θsin
nsin i2qn
n
θsin
nsin qn
1-n
n1
nn
sin
nsin )(y)(y
nq
P represents the variable complex number z, find the locus of P if
Let z = x + iy be the variable complex number
1 iz
1zRe
iiyx
1iyx
iz
1z
)1i(yx
iy)1(x
)1i(yx
)1i(yx
)1i(yx
iy)1(x
)1(yix
1)y(yi1)1)(yi(xixy)1x(x222
2
)1(yx
1)1)(y(xxyi1)y(y)1x(x22
1 iz
1zRe
1 )1(yx
1)y(y)1x(x22
x(x + 1) + y(y + 1) = x2 + (y + 1)2
x2 + x + y2 + y = x2 + y2 + 2y + 1
x + y = 2y + 1
x – 2y + y – 1 = 0
x – y – 1 = 0
The locus of P is
x – y – 1 = 0
Find the square root of – 8 – 6i.Let square root of – 8 – 6i be x + iy
– 8 – 6i = (x + iy)2
= x2 + 2ixy + i2y2
= x2 + 2ixy – y2
– 8 – 6i = x2 – y2 + 2ixy
Equating the real and imaginary parts
x2 – y2 = – 8 ……..(1)
2xy = –6 ……..(2)
xy = –3
y = –3/x
Sub y = –3/x in (1)
83
22
x
x
89
22
x
x
89
2
4
x
x
x4 – 9 = – 8x
x4 + 8x2 – 9 = 0
x4 + 9x2 – x2 – 9 = 0
x2(x2 + 9) – 1(x2 + 9) = 0
(x2 + 9)(x2 – 1) = 0
x2 = –9, 1
x = ±3i, ±1
Since is real x = ±1
If x = 1, then y = –3
If x = –1, then y = 3
The square root = 1 – 3i and –1 + 3i