 # Solved Examples on Complex Numbers

Illustration 1: The minimum value of |a+bω+cω2|, where a, b and c are all not equal integers andω≠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 + b2 + 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 ≠ cand 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, z3 form an equilateral triangle

z12 + z22+ z32 = 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, a2 – b2 = 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) = a2 +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 iz3 + z2 – z + i = 0, then show that |z| = 1. (1995)

Solution: Given that iz3 + z2 – z + i = 0

Hence, iz3–i2z2 – z + i = 0

Hence, iz2 (z-i) – 1(z-i) = 0

Hence, (iz2 – 1)(z-i) = 0

Hence, either (iz2 – 1) = 0 or (z-i) = 0

So, z = i or z2 = 1/i = -i

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

If z2 = -I, then |z2| = |-i| = 1

Hence, |z| = 1.

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