1.Algebra Booster

4.38 Algebra Booster
p q
fi = = l(say), lŒI
-{0}
7 3
Thus, |z| 2 will be minimum only when l = 1.
Therefore, p = 7 and q = 3
Hence, the minimum value of |z| 2 = 58(49 + 9)
= 58 ¥ 58
= 3364
33. Given,
|b| = 1
fi |b| 2 = 1
fi b◊ b = 1
fi
1
b =
b
Now,
b-a b-a
=
1 - ab 1
1 -a ◊ b
bb ( - a)
=
b - a
| bb || - a|
=
| b - a|
| bb || - a|
= = 1
|( b-
a)|
34. We have |z + 1| = z + 2(1 + i).
Let z = x + iy.
Then |x + iy + 1| = (x + iy) + 2(1 + i)
fi |(x + 1) + iy| = (x + 2) + i(y + 2)
2 2
fi ( x+ 1) + y = ( x+ 2) + i( y + 2)
Comparing the real and imaginary parts, we get
2 2
( x+ 1) + y = ( x+ 2), ( y + 2) = 0
when y = –2, x = 1/2
Hence, the complex number is
z = x + iy = 1 - 2i
2
35. We have
|z – 1| 2 + |z + 1| 2 = 5
fi |(x + iy) – 1| 2 + |(x + iy) + 1| 2 = 5
fi |(x – 1) – iy| 2 + |(x + 1) + iy| 2 = 5
fi (x – 1) 2 + y 2 + (x + 1) 2 + y 2 = 8
fi 2(x 2 + y 2 ) + 2 = 8
fi 2(x 2 + y 2 ) = 6
fi (x 2 + y 2 ) = 3
fi |z| 2 = 3
36. Given,
|z – 2| = 2|z – 1|
fi |z – 2| 2 = 4|z – 1| 2
fi (z – 2)(z – – 2) = 4(z – 1)(z– – 1)
1 1
fi (z◊z – – 2z – 2z – + 4) = 4(z◊z – – z – z – + 1
fi (|z| 2 – 2z – 2z – + 4) = 4(|z| 2 – z – z – + 1)
fi
fi
3|z| 2 – 2z –2z – = 0
3|z| 2 – 2(z –2z – ) = 0
fi 3|z| 2 – 2(z –2z – ) = 2.2Re(z) = 4Re(z)
2 4
fi || z = Re() z
3
Hence, the result.
37. Given,
|z + 6| = |3z + 2|
fi |z + 6| 2 = |3z + 2| 2
fi (z + 6)(z – + 6) = (3z + 2)(3z – + 2)
fi (z.z – + 6(z + z – ) + 36) = (9z.z – + 6(z + z – ) + 4)
fi (z.z – + 36) = (9z.z – + 4)
fi (|z| 2 + 36) = (9|z| 2 + 4)
fi 8|z| 2 = 32
fi |z| 2 = 4
fi |z| = 2
Hence, the value of |z| is 2.
38. Given,
|z + 6| = |2z + 3|
fi |x + iy + 6| = |2(x + iy) + 3|
fi |(x + 6) + iy| = |(2x + 3) + i.2y|
fi |(x + 6) + iy| 2 = |(2x + 3) + i.2y| 2
fi (x + 6) 2 + y 2 = (2x + 3) 2 + 4y 2
fi x 2 + 12x + 36 + y 2 = 4x 2 + 12x + 9 + 4y 2
fi 3x 2 + 3y 2 = 27
fi x 2 + y 2 = 9
Hence, the locus of z is x 2 + y 2 = 9.
39. Given,
|z| = 1
fi |z| 2 = 1
fi z.z – = 1
1
fi z =
z
Now, 2Re(w) = (w + w – )
Ê z -1ˆ Ê z -1ˆ
= Á +
Ë z + 1˜ ¯
Á
Ë z + 1˜
¯
Ê z -1ˆ Ê z -1ˆ
= Á +
Ë z + 1˜ ¯
Á
Ë z + 1˜
¯
Ê z -1 ˆ Ê(1/ z) -1ˆ
= Á +
Ë z + 1 ˜
¯
Á
Ë(1/ z) + 1˜
¯
Ê z -1ˆ Ê1-
zˆ
= Á +
Ë z + 1˜ ¯
Á
Ë1+
z˜
¯
Ê z - 1+
1– zˆ
= Á
Ë z + 1 ˜
¯
= 0
fi Re(w) = 0
40. We have,
|z + i| + |z – i| = 8