1.Algebra Booster

4.86 Algebra Booster
53.
fi
2 2
2 ( a - k b) ( a - bk
)
|| z - z - z
2 2
(1 - k ) (1 - k )
2 2
(| a| - k | b| )
+ = 0
2
(1 - k )
Thus, the centre of a circle is
and its radius =
2
( a - k b) 2 .
(1 - k )
2
2
2 2
( a - k b) Ê| a|
- k bbˆ
-
2 Á 2 ˜
(1 - k ) Ë (1 - k ) ¯
Ê
2 2 2 2
( a - k b) ˆÊ( a - k b) ˆ Ê| a|
- k bbˆ
Á 2 ˜Á 2 ˜ Á 2 ˜
= -
Ë (1 - k ) ¯Ë (1 - k ) ¯ Ë (1 - k ) ¯
k( a - b)
=
2
1 - k
Qz ( 2)
( z 0 )
O(1, 0)
Pz ( 1)
Rz ( 3)
Sz ( )
Since the centre of a square coincides with the centre of
z1+
z3 the circle, so = 1
2
fi z 1
+ z 3
= 2
z = 2– z = 2 -(2+ i 3) = -i
3
fi 3 1
p
Here, – zzz 1 0 2=
2
By the rotation theorem,
Êz - z ˆ Êz - z ˆ ip/2
Á
= e = i
Ë z z ˜
¯
Á
Ë z z ˜
¯
2 0 2 0
1- 0 1-
0
fi (z 2
– z 0
) = i(z 1
– z 0
)
fi z 2
= z 0
+ i(z 1
– z 0
)
fi z2 = 1 + i(2+ i 3 - 1) = 1+ i - 3
= (1 - 3) + i
Also, z4= 2- z2= 2–(1- 3) + i
fi z4 = (1 + 3) -i
54. The shaded region is outside the circle
|z + 1| = 2
Thus, |z + 1| > 2
4
Also, – QPR = p as DQPR is a right triangle.
2
p
Therefore, Arg ( z + 1) <
2
55. Let z = |a + bw + cw 2 |
|z| 2 = |a + bw + cw 2 | 2
2 2
= ( a + bw + cw )(1 + bw + cw
)
= a 2 + b 2 + c 2 – ab – bc – ca
2 2 2
=
1 [( ) ( ) ( ) ]
2 a - b + b- c + c -a
It is minimum only when a = b
and (b – c) 2 = 1 = (c – a) 2
Thus minimum of |z| 2 is 1
Therefore minimum of |z| is 1.
Êw-
wzˆ
56. Let z1
= Á
Ë 1 - z ˜
¯
It will be pure real if z1=
z1
fi
Êw–
wzˆ Êw-
wzˆ
Á =
Ë 1– z ˜
¯
Á
Ë 1-
z ˜
¯
fi ( w- wz)(1 - z) = ( w- wz)(1 - z)
fi ( w- wz – wz + wz◊ z) = ( w- wz – wz + wz◊z)
fi
fi
fi
( w- w) + ( w- w)| z| = 0
( w- w)(1 - |z| ) = 0
2
2
2
(1 - | z| ) = 0
( w π w)
fi |z| = 1 and z π 1
i
e p
57. Let OA = 3, so that the complex number A is 3
4.
X¢
Pz ()
O
Y
Y¢
p/4
A
(3 e ip/4 )
Let P be the complex number z.
Then by the rotation theorem, we have
Ê
ip/4
ˆ
-ip/2
e
ip/4
= = -
z –3e 4 4i
Á ˜
Ë0–3e
¯ 3 3
fi 3(z – 3e ipp/ ) = –4i(–3e ip/4 )
= 12ie ip/4
fi 3z – 9e ip/4 = 12ie ip/4
fi z – 3e ip/4 = 4ie ip/4
fi z = (3 + 4i)e ip/4
X