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.