1.Algebra Booster

4.84 Algebra Booster
fi
fi
4
Ê3z
- 1ˆ Á = e
Ë z - 2 ˜
¯
Ê3z
- 1ˆ Á = e
Ë z - 2 ˜
¯
i(2n+
1) p
(2n
+ 1) p
i
4
(2n
+ 1) p
i
4
(2n
+ 1) p
i
4
fi
1-
2e
z =
, where n = 0, 1, 2, 3.
3 - e
Put n = 0, we get
fi
Ê 1 i ˆ
i p 1- 2 +
1-
2e
4 Á
Ë
˜
2 2¯
z = =
i p
3 - e
4 Ê 1 i ˆ
3 - Á +
Ë
˜
2 2¯
2 - 2(1 + i) ( 2 - 2) - 2i
= =
3 2 - (1 + i) (3 2 -1)-i
(( 2 - 2) - 2 i)(3 2 - 1) + i)
=
2
(3 2 - 1) + 1
10 - 7 2 (-5 2)
= + i = a + ib
20 - 6 2 20 - 6 2
Similarly for n = 1, 2, 3, we get three other roots of the
given equation.
41. We have,
arg(–z) – arg(z)
Ê-z
ˆ
= argÁ
Ë
˜
z ¯
= arg(–1)
= p
42. Given,
|z 1
| = 1
fi |z 1
| 2 = 1
fi zz 1 1 = 1
fi
1
z 1 =
z1
1
Similarly, z 2 = ,
z2
and
1
z 2 = ,
z
Now,
2
1 1 1
+ + = 1
z z z
1 2 3
fi z1 z2 z3
| + + | = 1
| z + z + z | = 1
fi 1 2 3
fi |z 1
+ z 2
+ z 3
| = 1
43. Let z = (1) 1/n = (cos(2rp) + i sin(2rp)) 1/n
Ê 2r
2r
i 2rp
Ê pˆ Ê pˆˆ
fi z = cos isin
e
n
Á + =
Ë
Á
Ë
˜ Á ˜
n ¯ Ë n ¯˜
¯
where r = 0, 1, 2, 3,....., (n-1)
i 2kp
Let z1= 1and z
n
2=
e
It is given that,
( z - 0) = ( z -0) e i p
2
2 1
i 2kp
i p
n 2
fi e = e
fi 2 kp
p
=
n 2
fi n = 4k
Hence, the result.
44. We have
Ê z1- z2ˆ 1-
i 3
Á =
Ëz2-
z ˜
3¯
2
Ê 1 3ˆ
=-Á- + i
Ë
˜
2 2 ¯
fi
Ê z2-
z1ˆ Ê 1 3ˆ
= Á - + i ˜
Ë
Á z - z ¯
˜ Ë 2 2 ¯
2 3
i 2
e p
3
=
So, z 1
, z 2
, z 3
are the vertices of an equilateral triangle.
45. Given
1 1 1
1
2
-1-w
2
w
1
2
w
4
w
1 1 1
= 0
2
–2 – w
2
w -1
0
2
w -1 w -1
2 2
-2-w
w -1
=
2
w -1 w-1
- 1+ w w -1
=
2
w -1 w -1
= (w – 1) 2 – (w 2 – 1) 2
= (w 2 – 2w + 1) – (w 4 – 2w 2 + 1)
= (w 2 – 2w + 1) – (w – 2w 2 + 1)
= 3w 2 – 3w
= 3w(w – 1)
46. Hence, the minimum value of |z 1
– z 2
|
= PQ
= OQ – OP
= 12 – 10
= 2
2