1.Algebra Booster

Complex Numbers 4.61
2
2
Ê
n
1 - z ˆ
= z Á 2 ˜
Ë 1 - z ¯
Ê
2n+
1
z - z ˆ
= z Á 2 ˜
Ë 1 - z ¯
Ê z - 1 ˆ
= z Á 2
Ë1
- z ˜
¯
Ê 1 ˆ
= Á
Ë1
+ z ˜
¯
z + 1
Now, a+ b= - = -1
z + 1
z 1
and a◊ b= =
2
( z + 1) 1
z + + 2
Again,
z
1
z + = (cos q + isin q) + (cos q -isin q)
z
2p
= 2 cos q, where q = Ê ˆ
Á
Ë2n
+ 1 ˜
¯
1
2 Êq
ˆ
fiz
+ + 2 = 2 cos q + 2 = 2(1 + cos q) = 4 cos Á
z Ë
˜
2¯
2 Ê p ˆ
= 4cos Á
Ë2n
+ 1˜
¯
Hence, the required equation is
2
1
x + x + = 0.
2 Ê p ˆ
4cos Á
Ë2n
+ 1˜
¯
45. Let z = (1) 1/n = (cos(2rp) + i sin (2rp)) 1/n
È 2 2
2
Ê rpˆ Ê rpˆ˘
i
= ÍcosÁ + isin
= e
Ë
˜ Á ˜
n ¯ Ë n ¯˙
Î
˚
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.
46. Given w 5 = 1
we have log 2
|1 + w + w 2 + w 3 – w –1 |
2 3 1
= log2
1 + w + w + w -
w
2 3 4
w + w + w + w -1
= log2
w
2 3 4
1+ w + w + w + w - 2
= log2
w
rp
n
= log
= log
2
2
Ê
5
1 - w ˆ
Á - 2
Ë 1 - w
˜
¯
w
0-
2
w
Ê| - 2| ˆ
= log2
Á ˜
Ë | w | ¯
= log 2
|2| – log 2
|w|
= log 2
2 – log 2
1
= 1 – 0
= 1
47. Given b n = 1
Let S n
= 1 + 3b + 5b 2 + … + (2n – 1)b n–1
fi S n
b = b + 3b 2 + … + (2n – 3)b n–1 + (2n – 1)b n
Substracting we get,
fi (1 – b)S n
= 1 + 2b + 2b 2 + … + 2b n–1 – (2n – 1)b n
= 1 + 2b + 2b 2 + … + 2b n–1 – (2n – 1)
= 2b + 2b 2 + … + 2b n–1 – 2n
Ê
= 2 1 n
- b ˆ
Á - 2n
Ë 1 - b
˜
¯
fi
= – 2n ( b n = 1)
2n
Sn
=
b - 1
Hence, the result.
48. Given w 5 = 3
We have,
x = w + w 2
= (w + w 2 ) 5
= w 5 + 5w.w 2 + 10w 3 .w 4 + 10w 2 .w 6 + 5w.w 8 + 5w 10
= 3 + 15w + 30w 2 + 30w 3 + 15w 4 + 9
= 12 + 15w 2 + 30w 2 + 15w 4 + (15w 2 + 15w)
= 12 + 15(w 2 + 2w 2 + w 4 ) + 15(w 2 + w)
= 12 + 15(w + w 2 ) 2 + 15(w + w 2 )
= 12 + 15x 2 + 15x
fi x 5 – 15x 2 – 15x = 12
fi x 5 – 15x 2 – 15x + 18 = 12 + 18 = 30
49. We have
(x 2 + 2ix) – (3x 2 + iy) = (3 – 5i) + (3x 2 + iy)
fi –2x 2 + i(2x – y) = 3(x 2 – 1) + i(y – 5)
Comparing the real and imaginary parts, we get,
3(x 2 – 1) = –2x 2 and (2x – y) = (y – 5)
fi
2 5
5x = 3and( x- y)
=-
2
2 3 5
fi x = and ( x- y)
=-
5 2
fi
3 5
x=± and y = x+
5 2