1.Algebra Booster

Complex Numbers 4.85
X¢
O
Y
C
P
(3, 4)
Y¢
47. Given, z p + q – z p – z q + 1 = 0
fi z p (z q – 1) – 1(z q – 1) = 0
fi (z q – 1)(z p – 1) = 0
either(z p – 1) = 0 or (z q – 1) = 0
either(a p – 1) = 0 or (a q – 1) = 0
p
q
( a -1) ( a -1)
either = 0or = 0
( a -1) ( a -1)
either1 + a + a 2 + … + a p – 1 = 0
or 1 + a + a 2 + … + a q – 1 = 0
z 1
48. Given |z| = 1 and w = -
z + 1
fi z – 1 = wz + w
fi (1 – w)z = 1 + w
1 + w
fi z =
1 - w
fi
1 + w |1 + w|
| z|
= =
1 -w
|1 -w|
fi |1 + w | = |1 – w |
fi |1 + w | 2 = |1 – w | 2
fi (1 + w)(1 + w) = (1 -w)(1 -w)
fi 1 + w + w + | w| = 1 -w - w + | w|
fi w + w = -w -w
fi 2( w + w) = 0
fi ( w + w) = 0
fi 2Re(w) = 0
fi Re(w) = 0
49. Given, |z 1
| < 1, |z 2
| > 1
2
1– zz 1 2
We have, 1–
z - z
1 2
Q
X
2 2
2 2
1- 2 - 1 2
2
| z1-
z2|
| z z | |1– z z |
=
( z - z )( z - z )-(1– z z )(1– z z )
=
1 2 1 2 1 2 1 2
2
| z1-
z2|
1
= [(| | + | | - - )
| z |
- + ◊ - -
(| | | | 1 | | | | )
=
2 2
z1 z2 zz
2
1 2 zz 1 2
1-
z2
2 2
(1 | z1| | z2| zz 1 2 zz 1 2)]
2 2 2 2
z1 + z2 - - z1 ◊ z2
2
| z1-
z2|
2 2
z1 - - 2
0
2
| z1-
z2|
(| | 1)(1 |z | )
= >
Therefore,
1
z
1 - zz
1 2
fi
z1-
z2
50. We have
fi
fi
n
Â
r = 1
n
r = 1
2
- zz 1 2
1-
z2
< 1
r
( az) = 1
r
1 = Â( az r )
r
< 1
n
n
r
 r Â
r= 1 r=
1
1 = ( az) £ | a|| z|
n
n
Ê1ˆ
Ê1ˆ
fi 1< Â2Á < 2
Ë
˜
r= 1
3¯
 Á
Ë
˜
r=
1
3¯
2
3
fi 1< = 1
1
1 -
3
It is a contradiction.
Thus, there is no complex number z such that
n
1
r
|| z < and Â( az r ) = 1
3
r = 1
51. Given
(1 + w 2 ) n = (1 + w 4 ) n
2
n
Ê1
+ w ˆ
fi Á 1
4 ˜ =
Ë1
+ w ¯
n
Ê – w ˆ
fi Á 1
2 =
Ë–
w
˜
¯
n
Ê 1 ˆ
fi Á = 1
Ë
˜
w ¯
fi (w 2 ) n = 1
Thus, the least value of n is 3.
52. Given z - a
= k
z - b
2
| z - a|
2
fi = k
2
| z - b|
fi
( z -a)( z -a)
= k
( z - b)( z - b)
fi
2
fi
( z -a)( z - a) = k ( z - b)( z - b)
2
r
n
2 2 2 2 2
|| z -az - az + | a| = k (|| z -bz - bz
+ | b|)
fi
2 2 2 2
(1 -k ) | z| -( a -k b) z -( a -bk ) z
2 2
+ (| a| - k | b| ) = 0
n