# 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 - Mathematics

Match the Columns

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

#### Solution

 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.

Concept: Concept of Complex Numbers
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#### APPEARS IN

NCERT Mathematics Exemplar Class 11
Chapter 5 Complex Numbers and Quadratic Equations
Solved Examples | Q 18 | Page 86

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