1.Algebra Booster

Complex Numbers 4.35
10. We have z = (n + i) 4
For n = 0, z = i 4 = 1, an integer
For n = 1, z = (1 + i) 4 = 1 – 6 + 1 = –4, an integer
For n = –1, z = (1 + i) 4 = 1 – 6 + 1 = –4, an integer
For n = R – {–1, 0, 1}, z, an imaginary number.
Thus, the number of integral values of n is 3.
11. We have,
z 2 = (1 + i) 2
= 1 + i 2 + 2i
= 2i
Hence, the multiplicative inverse of z 2 is = 1 i
.
2i =- 2
1+ 2 i (1+ 2)(3 i + 4) i
12 We have z = =
3 -4 i (3 - 4 i)(3 + 4 i)
3+ 4i + 6i -8 - 5+
10i
z = =
25 25
Hence, the multiplicative inverse of z is
25
- 5+
10i
= 25( -5 -10 i)
( - 5 + 10 i)( - 5 - 10 i)
25( -5 -10 i)
=
125
(-5 -10 i)
= =-1-2i
5
13. The given relation a + ib > c + id holds good only when
if b = 0, d = 0.
Thus, b + d + 2016 = 0 + 0 + 2016 = 2016.
14. We have,
n
Ê1+ iˆ 2Ê -1 -1Ê1ˆˆ
Á = sin x + sec Á ˜
Ë1
- i˜ ¯ p
Á
Ë
Ë x¯˜
¯
fi
fi
n
1+
i 2 (sin
-1 cos
-1
x x )
Ê ˆ
Á = +
Ë1
- i˜
¯ p
n
Ê1+
iˆ 2 p
Á = ¥ = 1
Ë1-
i˜
¯ p 2
n
fi
Ê1
+ iˆ
Á
Ë1
- i˜
¯
= 1
fi i n = 1
fi i n = i 4
fi n = 4
Thus, the positive integer is 4.
15. We have
(1 + i) x-2 i (2 - 3 i) y + i
+ = i
(3 + i) (3 -i)
fi
1 + ix ( – 2) 2+ i(1-3 y)
+ = i
(3 + i) (3 -i)
fi
(4 + 2i)x + (9 – 7i)y – 3i – 3 = 10i
Equating the real and imaginary parts, we get,
2x – 7y = 13 and 4x + 9y = 3.
Hence, x = 3, y = –1.
16. We have
1008 2016
2 (1 + i)
x+ iy = +
2016 1008
(1 + i) 2
1008 2
1008
Ê 2 ˆ Ê(1+
i) ˆ
Á 2 ˜ Ë ¯
= +Á ˜
Ë(1 + i)
¯ 2
1008 1008
Ê 2ˆ Ê2iˆ
= Á +
Ë
˜
2i¯ Á
Ë
˜
2¯
1008
Ê1ˆ
1008
= Á + () i
Ë
˜
i¯
= (–i) 1008 + (i) 1008
= 1 + 1
= 2
Thus, x = 2 and y = 0.
17. We have
z 1/3 = a + b
fi (x + i y) 1/3 = a + i b
fi (x + i y) = (a + i b) 3
= a 3 + 3a 2 (i b) + 3a (i b) 2 + (i b) 3
fi (x + iy) = a 3 + i 3 a 2 b – 3ab 2 – ib 3
fi (x + iy) = (a 3 – 3ab 2 ) + i(3a 2 b – b 3 )
fi x = (a 3 – 3ab 2 ) and y = (3a 2 b – b 3 )
x 2 2 y 2 2
fi = a - 3b and = 3a -b
a
b
x y 2 2 2 2
fi + = ( a - 3 b ) + (3 a -b
)
a b
Adding, we get
x y
fi
2 2
+ = (4a
-4 b )
a b
x y
fi
2 2
+ = 4( a -b
)
a b
18. Given
x = 3 + 2i
fi (x – 3) 2 = –4
fi x 2 – 6x + 9 = –4
fi x 2 – 6x + 13 = 0
We have x 4 – 8x 3 + 4x 2 + 4x + 39
= x 2 (x 2 – 6x + 13) – 2x 3 – 9x 2 + 4x + 39
= –2x 3 – 9x 2 + 4x + 39
= –2x(x 2 – 6x + 13) – 21x 2 + 30x + 39
= –21x 2 + 30x + 39
= –21(x 2 – 6x + 13) – 96x + 312
= –96x + 312
= –96(3 + 2i) + 312
= –288 – 192i + 312
= 24 – 192i
Thus, a = 24 and b = –192.