1.Algebra Booster

4.34 Algebra Booster
HINTS AND SOLUTIONS
LEVEL I
1. We have,
i n + i n+1 + i n+2 + i n+3
= i n (1 + i + i 2 + i 3 )
= 0
2. We have,
i 2010 + i 2011 + i 2012 + i 2013
= i 2010 (1 + i + i 2 + i 3 )
= 0
3. We have,
n
Ê1
+ iˆ Á = 1
Ë1-
i˜
¯
fi
fi
Ê1+ i 1+
iˆ
Á ¥ = 1
Ë1- i 1+
i˜
¯
2
n
Ê (1 + i)
ˆ
Á
1
2 2˜
=
Ë1 - (-i)
¯
fi
Ê
2
1+ i + 2iˆ
Á
Ë 1+
1
˜
¯
= 1
fi (i) n = 1
fi n = …, –8, –4, 0, 4, 8, 12, …
Clearly, n is not defined
n
n
Notes
1. The smallest positive integer n for which
n
Ê1
+ iˆ Á = 1
Ë1
- i˜
, then n = 4
¯
2. The smallest non negative integer n for which
n
Ê1
+ iˆ Á = 1
Ë1
- i˜
, then n = 0
¯
3. The smallest positive integer n for which
n
Ê1
+ iˆ
Á
Ë1-
i˜
is real, then n = 2.
¯
4. We have,
2013
1
( n n
i + i + )
Â
n=
1
Ê
ˆ
2 3 4 2012 2013
= Ái + i + i + i + … + i + i ˜
Á
Ë
˜
= 0
¯
Ê
+ Ái + i + i + i +º+ i + i
Á
Ë
= 0
= i 2013 + i 2014
= i + i 2
= i – 1
2 3 4 5 2013 2014
ˆ
˜
˜
¯
5. We have,
i P + i Q + i R + i S
= i P (1 + i + i 2 + i 3 )
= i P ¥ 0
= 0
6. We have,
i 2015 + i 2016 + i 2017 + i 2018
= i 2015 (1 + i + i 2 + i 3 )
= i 2015 ¥ 0
= 0
7. We have,
fi
2016 2018
k p
Â
Â
i + i = x+
iy
k= 0 p=
0
2016 2018
k p
Â
Â
x+ iy = i + i
k= 0 p=
0
fi x + iy = i 2016 + i 2016 + i 2017 + i 2018
fi x + iy = 1 + 1 + i + i 2 = 1 + i
Thus, x = 1, y = 1
Hence, the value of
x + y + 2
= 1 + 1 +2
= 4
8. We have,
(1 + i) 2n = (1 – i) 2n
2n
(1 + i)
fi
= 1
2n
(1 - i)
fi
fi
fi
fi
Ê1
+ iˆ
Á
Ë1
- i˜
¯
2n
= 1
2n
Ê1+ i 1+
iˆ
Á ¥ = 1
Ë1- i 1+
i˜
¯
Ê
2
(1 + i) ˆ
Á 2 2˜
Ë(1) - ( i)
¯
Ê
2
1+ i + 2iˆ
Á
Ë 1+
1
˜
¯
2n
2n
= 1
= 1
fi (i) 2n = 1
fi n = 2
Hence, the positive integer n is 2.
9. We have
(1 + i) 5 + (1 + i 3 ) 5 + (1 + i 5 ) 7 + (1 + i 7 ) 7
= (1 + i) 5 + (1 – i) 5 } + {(1 + i) 7 + (1 – i) 7
= 21 ± 5 C 1
i + 5 C 1
i 2 ± 5 C 1
i 3 + 5 C 1
i 4 ± 5 C 1
i 5 }
2(1 ± 7 C 1
i + 7 C 2
i 2 ± 7 C 3
i 3 + 7 C 4
i 4 ± 7 C 5
i 5 + 7 C 6
i 6 ± 7 C 7
i 7 )
= (2 – 20 + 10) + (1 – 21 + 35 – 7)
= –8 + 8
= 0