Question

Match the statements of column A and B.

Column A	Column B
(a) The value of 1 + i2 + i4 + i6 + ... i20 is	(i) purely imaginary complex number
(b) The value of `i^(-1097)` is	(ii) purely real complex number
(c) Conjugate of 1 + i lies in	(iii) second quadrant
(d) `(1 + 2i)/(1 - i)` lies in	(iv) Fourth quadrant
(e) If a, b, c ∈ R and b2 – 4ac < 0, then the roots of the equation ax2 + bx + c = 0 are non real (complex) and	(v) may not occur in conjugate pairs
(f) If a, b, c ∈ R and b2 – 4ac > 0, and b2 – 4ac is a perfect square, then the roots of the equation ax2 + bx + c = 0	(vi) may occur in conjugate pairs

Answer 1

Column A	Answers
(a) The value of 1+ i2 + i4 + i6 + ... i20 is	(ii) purely real complex number
(b) The value of `i^(-1097)` is	(i) purely imaginary complex number
(c) Conjugate of 1 + i lies in	(iv) Fourth quadrant
(d) `(1 + 2i)/(1 - i)` lies in	(iii) second quadrant
(e) If a, b, c ∈ R and b2 – 4ac < 0, then the roots of the equation ax2 + bx + c = 0 are non real (complex) and	(vi) may occur in conjugate pairs
(f) If a, b, c ∈ R and b2 – 4ac > 0, and b2 – 4ac is a perfect square, then the roots of the equation ax2 + bx + c = 0	(v) may not occur in conjugate pairs

Explanation:

(a) Because 1 + i² + i⁴ + i⁶ + ... i²⁰

=  1 – 1 + 1 – 1 + ... + 1 = 1 ......(Which is purely a real complex number.)

(b) Because `i^(-1097)` =  `1/((i)^1097)`

= `1/(i^(4 xx 274 + 1)`

= `1/((i^4)^274i)`

= `1/i`

= `i/i^2`

= –i

Which is purely imaginary complex number.

(c) Conjugate of 1 + i is 1 – i which is represented by the point (1, –1) in the fourth quadrant.

(d) Because `(1 + 2i)/(1 - i) = (1 + 2i)/(1 - i) xx (1 + i)/(1 + i)`

= `(-1 + 3i)/2`

= `-1/2 + 3/2 i`

Which is represented by the point `(- 1/2, 3/2)` in the second quadrant.

(e) If b² – 4ac < 0 = D < 0 i.e., square root of D is a imaginary number.

Therefore, roots are x = `(-b +- "Imaginary Number")/(2a)`

i.e., roots are in conjugate pairs.

(f) Consider the equation `x^2 - (5 + sqrt(2)) x + 5 sqrt(2)` = 0, Where a = 1, b = `-(5 + sqrt(2))`, c = `5 sqrt(2)`, Clearly a, b, c ∈ R.

Now D = b² – 4ac = `{- (5 + sqrt(2))}^2 - 4.1.5 sqrt(2) = (5 - sqrt(2))^2`.

Therefore x = `(5 + sqrt(2) +- 5 - sqrt(2))/2` = `5sqrt(2)` which do not form a conjugate pair.