1.Algebra Booster

Complex Numbers 4.57
24. Given curves are |z – 3| = 2 and |z| = 2.
X¢
O
Y
P
Q
Y¢
Let z = x + iy.
Given |z| = 2
fi
2 2
x + y = 2
fi x 2 + y 2 = 4 …(i)
Also, |z – 3| = 2
fi |x + iy – 3| = 2
fi |(x – 3) + iy| = 2
fi
2 2
( x- 3) + y = 2
fi (x – 3) 2 + y 2 = 4
fi x 2 – 6x + 9 + y 2 = 4
fi x 2 + y 2 – 6x + 9 = 4 …(ii)
From Eqs (i) and (ii), we get
4 – 6x + 9 = 4
fi –6x = –9 fi x = 3/2
O¢
7
When x = 3/2, y =±
2
Hence, the points of intersection are
Ê 3 7 3 7
Á + i
ˆ ,
Ê -i
ˆ
Ë
˜
2 2 ¯
Á
Ë
˜
2 2 ¯
Thus, the length of the common chord,
=PQ
Ê 7 ˆ
= 2 ◊Á
Ë
˜
2 ¯
= 7
25. We have,
t 2 + t + 1 = 0
fi t = w, w 2
When t = w,
2
Ê 1ˆ
2 2
Át
+ = ( w + w ) = 1
Ë
˜
t ¯
2
Ê 2 1 ˆ
2 2
Át
+ = ( w + w ) = 1
Ë 2 ˜
t ¯
2
Ê 3 1 ˆ
2
Át
+ = (1 + 1) = 4
Ë 3 ˜
t ¯
2
Ê 4 1 ˆ
2 2
Át
+ = ( w + w ) = 1
Ë 4 ˜
t ¯
2
Ê 5 1 ˆ 2 2
Át
+ = ( w + w) = 1
Ë 5 ˜
t ¯
2
Ê 6 1 ˆ
2
Át
+ = (1 + 1) = 4
Ë 6 ˜
t ¯
X
Ê
Át
Ë
2
2014 1 ˆ
2 2
w w
2014 ˜
+ = ( + ) = 1
t ¯
2 2 2
1 2 1 2014 1
Now, Á Ê t + ˆ + Ê t + ˆ + … + Ê t +
ˆ
Ë
˜ 2 Á 2014 ˜
t ¯
Á
Ë
˜
t ¯ Ë t ¯
2 2 2
3 1 6 1 9 1
3 Á 6˜
9
Ê ˆ Ê ˆ Ê ˆ
= Át + + t + + t + +º
Ë
˜
t ¯ Ë t ¯
Á
Ë
˜
t ¯
Ê 2013 1
+ Át
+
Ë t
2 2 2
2 4
2 4
2013
Ê 1ˆ Ê 1ˆ Ê 1ˆ
+ Át+ + t + + t + +º
Ë
˜
t ¯
Á
Ë
˜
t ¯
Á
Ë
˜
t ¯
Ê 2014 1
+ Át
+
Ë t
2014
= (4 + 4 + 4 + … + 4) (671 times)
+ (1 + 1 + 1 + … + 1) [(2014 – 671) times]
= 671 ¥ 4 + 1 ¥ 1343
= 2684 + 1343
= 4027
26. We have,
1
x + = 1
x
fi x 2 + 1 = x
fi x 2 – x + 1 = 0
fi x = –w, –w 2
When x = –w,
x 10 + x 20 + x 30 + … + x 100
27. Let
fi
fi
= (–w) 10 + (–w) 20 + (–w) 30 + (–w) 40 + … + (–w) 100
= (w + w 2 + 1) + (w + w 2 + 1) + (w + w 2 + 1) + w
= w
1 1 1 2
+ + =
a + x b+ x c+
x x
( b+ x)( c+ x) + ( a+ x)( c+ x) + ( a+ x)( b+
x) 2
=
( a+ x)( b+ x)( c+
x)
x
2
3x + 2( a + b+ c) x+ ( ab+ bc+
ca) 2
=
( a + x)( b+ x)( c+
x)
x
fi 3x 3 + 2(a + b + c)x 2 + (ab + bc + ca)x
= 2(a + x)(b + x)(c + x)
= 2(x 3 + (a + b + c)x 2 + (ab + bc + ca)x + abc)
fi x 3 – (ab + bc + ca)x – 2abc = 0
which is a cubic in x, whose roots are w, w 2 .
Let y is the third root.
Therefore, w + w 2 + y = 0 fi y = 1
1 1 1 2
Thus,
2
a + 1 + b+ 1 + c+
1 = 1
=
28. We have,
1 + 1 +
1
2
x –1 xw
- 1 xw
- 1
ˆ
˜
¯
ˆ
˜
¯
2
2