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**Illustration 1: The minimum value of |a+bÏ‰+cÏ‰ ^{2}|, where a, b and c are all not equal integers and Ï‰â‰ 1 is a cube root of unity, is (2005)**

So, let us assume z =| a+bÏ‰+cÏ‰

Then, z

= (a

Or z

Since, a, b and c are all integers but not all simultaneously equal.

Hence, if a = b, then a â‰ c and b â‰ c.

Since, the difference of integers is an integer hence, (b-c)

Also (c-a)

And we have taken a =b so (a-b)

From equation (1), z

Hence, z

Hence, minimum value of |z| is 1.

**Illustration 2: If a and b are real numbers between 0 and 1 such that the points z _{1} = a + i, z_{2} = 1+bi, z_{3} = 0 form an equilateral triangle then what are the values of a and b? (1990)**

z

Hence, (a+i)

a

Hence, (a

So, a

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

Hence, a

So, a = 2Â±âˆš3

If a+b = 1,

2 = a(1-a) + 1

Hence, a

so, a = (1Â±âˆš1-4)/2

Since a and b both belong to R, so only solution exists when a = b

So a = b = 2Â±âˆš3.

**Illustration 3: If iz ^{3 }+ z^{2} â€“ z + i = 0, then show that |z| = 1. (1995)**

Hence, iz

Hence, iz

Hence, (iz

Hence, either (iz

So, z = i or z

If z = i, then |z| = |i| = 1

If z

Hence, |z| = 1.

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