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Illustration 1: The minimum value of |a+bω+cω2|, where a, b and c are all not equal integers and ω≠1 is a cube root of unity, is (2005)
1. √3
2. 1/2
3. 1
4. 0
Solution: We are required to compute the value of | a+bω+cω2|
So, let us assume z =| a+bω+cω2|
Then, z2 = | a+bω+cω2|2
               = (a2 + b+ c2 – ab– bc – ca)
Or z2 = 1/2 {(a-b)2 + (b-c)2 + (c-a)2}   ….. (1)
Since, a, b and c are all integers but not all simultaneously equal.
Hence, if a = b, then a ≠ c and b ≠ c.
Since, the difference of integers is an integer hence, (b-c)2 ≥ 1{as minimum difference of two consecutive integers is ±1}
Also (c-a)2≥1.
And we have taken a =b so (a-b)2 = 0.
From equation (1), z2≥1/2 (0+1+1)
Hence, z2 ≥ 1
Hence, minimum value of |z| is 1.

Illustration 2: If a and b are real numbers between 0 and 1 such that the points z1 = a + i, z2 = 1+bi, z3 = 0 form an equilateral triangle then what are the values of a and b? (1990)
Solution: Since, z1, z2, zform an equilateral triangle
z12 + z22+ z3= z1z2 + z2z3+ z3z1
Hence, (a+i)2 + (1+bi)2 + (0)2 = (a+i)(1+bi) + 0 + 0
a2 – 1 + 2ia + 1 – b2 + 2ib = a + i(ab + 1) – b
Hence, (a2– b2) + 2i(a+b) = (a-b) + i(ab + 1)
So, a– b= a-b
And 2(a+b) = ab +1
Hence, a = b or a+b = 1
And 2 (a+b) = ab +1
If a=b, 2(2a) = a+1
Hence, a2 – 4a +1 = 0
So, a = 2±√3
If a+b = 1,
2 = a(1-a) + 1
Hence, a2 – a + 1 = 0
so, a = (1±√1-4)/2
Since a and b both belong to R, so only solution exists when a = b
So a = b = 2±√3.

Illustration 3: If iz+ z2 – z + i = 0, then show that |z| = 1. (1995)
Solution: Given that iz3 + z2 – z + i = 0
Hence, iz3–i2z– z + i = 0
Hence, iz2 (z-i) – 1(z-i) = 0
Hence, (iz– 1)(z-i) = 0
Hence, either (iz– 1) = 0 or (z-i) = 0
So, z = i or z= 1/i = -i
If z = i, then |z| = |i| = 1
If z2 = -I, then |z2| = |-i| = 1
Hence, |z| = 1.

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