1.Algebra Booster

Complex Numbers 4.39
fi 8 = |z + i| + |z – i|
fi 8 = |z + i| + |z – i| ≥ |z + i + z – i|
fi |z + i + z – i| £ 8
fi |2z| £ 8
fi |z| £ 4
Hence, the maximum value of |z| is 4.
41. (i) Let z = 1 + i = (1, 1).
-1 1 -1
p
We have a= tan = tan (1) =
1 4
Since, the given complex number lies in the first
p
first quadrant, so Arg ( z)
= q = a =
4
(ii) Let z = 1 – i = (1, –1).
-1 -1
-1
p
We have a= tan = tan (1) =
1 4
Since the complex number z lies in the fourth
quadrant, so
p
Arg ( z)
= q = - a = -
4
(iii) Let z = –1 + i = (–1, 1)
-1 1 -1
p
We have a = tan = tan (1) =
-1 4
Since the complex number lies in the second quadrant,
so
p 3p
Arg ( z) = q = p - a = p - = .
4 4
(iv) Let z = –1 – i = (–1, –1)
-1 -1
-1
p
We have a = tan = tan (1) =
-1 4
Since the complex number z lies in the third quadrant,
so
Arg(z) = q = –(p – a)
3 i Ê 3 1ˆ
42. Now - z =- - = Á- , -
2 2 Ë
˜
2 2¯
-1 -1/2 -1Ê 1 ˆ p
We have a = tan = tan Á ˜ =
- 3/2 Ë 3¯
6
Since the complex number –z lies in the third quadrant,
so
Arg(–z) = q = –(p – a)
Ê pˆ
5p
=-Áp
- =-
Ë
˜
6¯
6
Ê pˆ
3p
=-Áp
- =-
Ë
˜
4¯
4
Similarly, you can solve the other parts easily.
43. Since Arg(z) < 0, so z lies in the third quadrant and (–z)
lies in the first quadrant
Thus,
Arg(z) – Arg(–z) = –(p – a) – q = –p
44. Given |z + 1| = |z – 1|
fi |x + iy + 1| = |x + iy – 1|
fi
|(x + 1) + iy| = |(x – 1) + iy|
fi
2 2 2 2
( x+ 1) + y = ( x- 1) + y
fi (x + 1) 2 + y 2 = (x – 1) 2 + y 2
fi 4x = 0 fi x = 0
Ê z - 1ˆ p
Also, AmpÁ
=
Ë z + 1˜
¯ 4
p
fi Amp ( z -1) - Amp ( z + 1) =
4
fi
-1Ê y ˆ -1Ê y ˆ p
tan Á - tan =
Ë x- 1˜ ¯
Á
Ë x+
1˜
¯ 4
fi
Ê y y ˆ
-
Á
-1 x- 1 x+
1 ˜ p
tan Á
y y
˜ =
Á1
+ ◊ ˜ 4
Ë x- 1 x+
1¯
fi
- 1 Ê xy + y - xy + yˆ p
tan
Á 2 2 =
Ë x + y -1
˜
¯ 4
fi
-1
Ê 2y
ˆ p
tan
Á 2 2 =
Ë x + y -1˜
¯ 4
fi
2y
= 1
2 2
x + y -1
fi x 2 + y 2 – 1 = 2y
fi x 2 + y 2 – 2y – 1 = 0
fi y 2 – 2y – 1 = 0, since x = 0
2±
8
fi y = = (1 ± 2)
2
Thus, the complex number,
z = x + iy = i(1±
2)
45. Given,
|z 1
+ z 2
| = |z 1
– z 2
|
fi |z 1
+ z 2
| 2 = |z 1
– z 2
| 2
fi |z 1
| 2 + |z 1
| 2 + 2Re(z 1
z –) = |z 2 1 |2 + |z 1
| 2 + 2Re(z 1
z –)
2
fi 4Re(z 1
z –) = 0 2
i( fi 1 2)
e q + q =
Re( ) 0
fi Re(cos(q 1
– q 2
) – i(q 1
– q 2
)) = 0
fi cos(q 1
– q 2
) = 0
fi
p
q1- q2
=
2
Ê z1
ˆ p
Thus, Amp Á ( q1- q2)
=
Ë z ˜
¯ 2
46. We have,
Êcos
q+
isin
qˆ
z = Á
Ëcos
q-
isin
q ˜
¯
2
= (cos 2q + i sin 2q)
p p
Also, it is given that < q < 4 2