1.Algebra Booster

4.10 Algebra Booster
In an equilateral triangle, the circumcentre and the centroid
are the same point.
Ê z1+ z2+
z3ˆ
Therefore, z0
= Á
Ë
˜
3 ¯
2 2
fi ( z + z + z ) = 9z
fi
1 2 3 0
2 2 2 2
1 + 2 + 3 + 2( 1 2+ 2 3+ 3 1) = 9 0
z z z z z z z z z z
fi (z 1
z 2
+ z 2
z 3
+ z 3
z 1
)
Ê
2 2 2 2
9z0 - z1 - z2 - z ˆ
3
= Á
Ë
˜
2 ¯
since z 1
, z 2
, z 3
are the vertices of an equilateral triangle,
then
2 2 2
fi z + z + z = z z + z z + z z
1 2 3 1 2 2 3 3 1
2 2 2
2 2 2 1 2 3
9 - z - z - z
Thus, z1 + z2 + z3
=
2
2 2 2 2
fi 3( z + z + z ) = 9z
fi
1 2 3 0
2 2 2 2
1 + 2 + 3 = 0
( z z z ) 3z
which is the required condition.
11. Condition of Parallelism of Two Lines
Az ( 1)
Bz ( 2)
Cz ( 3)
Dz ( 2)
Ê 1 2
Let arg z - z ˆ
Á = q
z3 z ˜ .
Ë - 4¯
If AB||CD, then q = 0 or ±p
Ê z1-
z2ˆ fi argÁ
= 0 or ± p
Ë z3-
z ˜
4¯
fi
Ê z1-
z2ˆ
Á z3 z ˜ is real
Ë - 4¯ 12. Condition of Perpendicularity
Az ( 1)
q
Dz ( 4)
Bz ( 2)
Cz ( 3)
p
If AB ^ CD, then q =±
2
Ê z1-
z2ˆ p
arg Á =±
Ë z3-
z ˜
4¯
2
Ê z1-
z2ˆ
Á z3 z ˜ is purely imaginary number.
Ë - 4¯ (z 1
– z 2
) = ±k(z 3
– z 4
),
where k is purely imaginary number.
13. Condition of a parallelogram
Az ( 1)
Dz ( 4)
Bz ( 2)
Cz ( 3)
As we know that the diagonals of a parallelogram bisect
each other.
Thus, mid point of AC = Mid-point of BD
1 3 2 4
fi z + z z +
=
z
2 2
fi (z 1
+ z 3
) = (z 2
+ z 4
)
14. Regular polygon of n sides
(i) Given z 1
and z 0
, then find z 2
and z 3
in terms of z 1
and z 0 z5
z 4
z 0
z n z 3
2 p/n
z 1
Centre at Z 0
is at an angle 2p/n.
Ê z2- z0ˆ z2-
z0
Á = ¥ e
Ë z1- z ˜
0¯
z1-
z0
Ê z2- z0ˆ | z2-
z0|
fi Á = ¥ e
Ë z1- z ˜
0¯
| z2-
z0|
z
2
2 z0
i p
Ê - ˆ
fi
e
n
Á =
Ë z1-
z ˜
0¯
i 2
fi ( z - z ) = ( z - z ) ¥ e p
n
fi
z 2
2 0 1 0
i 2
z2= z0+ ( z1- z0) ¥ e p
n
i 2p
n
i 2p
n
similarly, we can easily show that,
z = z + ( z - z ) ¥ e
3 0 1 0
z = z + ( z - z ) ¥ e
o
4 0 1 0
i 4p
n
i 6p
n
2( n –1) p
i
n
zn
= z0+ ( z1- z0)
¥ e
(ii) Given z 1
, z 2
, find z 3
, z 4
, z 5
and so on.
z 5
z 4
z 0
z n z 3
z 1
( n – 2) p
n
z 2