Complex Numbers and Quadratic Equations

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New question posted•a year ago

Complex Numbers and Quadratic Equations Maths NCERT Exemplar Solutions Class 11th Chapter Five

(i) For any two complex numbers <math><mrow><msub><mrow><mi>z</mi></mrow><mrow><mn>1</mn></mrow></msub><mo>,</mo><msub><mrow><mi>z</mi></mrow><mrow><mn>2</mn></mrow></msub></mrow></math> and any real numbers <math><mrow><mi>a</mi><mo>,</mo><mi>b</mi></mrow></math> , <math> <mrow> <mo>?</mo> <msup> <mrow> <mi>a</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <msub> <mrow> <mi>z</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> <mo>+</mo> <msup> <mrow> <mi>b</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <msub> <mrow> <mi>z</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> <mo>?</mo> <mo>−</mo> <mo>?</mo> <msup> <mrow> <mi>b</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <msub> <mrow> <mi>z</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> <mo>+</mo> <msup> <mrow> <mi>a</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <msub> <mrow> <mi>z</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> <mo>?</mo> <mo>=</mo> <mo>.</mo> <mo>.</mo> <mo>.</mo> <mo>.</mo> <mo>.</mo> </mrow> </math> (ii) The value of <math><mrow><mroot><mrow><mo>−</mo><mn>2</mn><mn>5</mn></mrow><mrow></mrow></mroot><mo>×</mo><mroot><mrow><mo>−</mo><mn>9</mn></mrow><mrow></mrow></mroot></mrow></math> is . (iii) The number <math><mrow><mfrac><mrow><msup><mrow><mrow><mo>(</mo><mrow><mn>1</mn><mo>−</mo><mi>i</mi></mrow><mo>)</mo></mrow></mrow><mrow><mn>3</mn></mrow></msup></mrow><mrow><mn>1</mn><mo>−</mo><mi>i</mi></mrow></mfrac></mrow></math> is equal to . (iv) The sum of the series <math><mrow><mi>i</mi><mo>+</mo><msup><mrow><mi>i</mi></mrow><mrow><mn>2</mn></mrow></msup><mo>+</mo><msup><mrow><mi>i</mi></mrow><mrow><mn>3</mn></mrow></msup><mo>+</mo><mo>…</mo></mrow></math> up to 1000 terms is . (v) Multiplicative inverse of <math><mrow><mn>1</mn><mo>+</mo><mi>i</mi></mrow></math> is . (vi) If <math><mrow><msub><mrow><mi>z</mi></mrow><mrow><mn>1</mn></mrow></msub></mrow></math> and <math><mrow><msub><mrow><mi>z</mi></mrow><mrow><mn>2</mn></mrow></msub></mrow></math> are complex numbers such that <math><mrow><msub><mrow><mi>z</mi></mrow><mrow><mn>1</mn></mrow></msub><mo>+</mo><msub><mrow><mi>z</mi></mrow><mrow><mn>2</mn></mrow></msub></mrow></math> is a real number, then <math><mrow><msub><mrow><mi>z</mi></mrow><mrow><mn>2</mn></mrow></msub><mo>=</mo><mo>.</mo><mo>.</mo><mo>.</mo><mo>.</mo></mrow></math> (vii)  <math> <mrow>arg <mrow><mo>(</mo><mrow><mi>z</mi></mrow><mo>)</mo></mrow><mo>+</mo>arg <mi>z</mi><mrow><mo>(</mo><mrow><mi>z</mi><mo>≠</mo><mn>0</mn></mrow><mo>)</mo></mrow></mrow> </math> is . (viii) If <math><mrow><mo>?</mo><mi>z</mi><mo>+</mo><mn>4</mn><mo>?</mo><mo>≤</mo><mn>3</mn></mrow></math> , then the greatest and least values of <math><mrow><mo>?</mo><mi>z</mi><mo>+</mo><mn>1</mn><mo>?</mo></mrow></math> are . and . (ix) If <math><mrow><mfrac><mrow><mi>z</mi><mo>−</mo><mn>2</mn></mrow><mrow><mi>z</mi><mo>+</mo><mn>6</mn></mrow></mfrac><mo>=</mo><mfrac><mrow><mi>π</mi></mrow><mrow><mn>6</mn></mrow></mfrac><mo>,</mo></mrow></math> then the locus of <math><mrow><mi>z</mi></mrow></math> is . (x) If <math><mrow><mo>?</mo><mi>z</mi><mo>?</mo><mo>=</mo><mn>4</mn></mrow></math> and <math><mrow><mrow><mo>(</mo><mrow><mi>z</mi></mrow><mo>)</mo></mrow><mo>=</mo><mfrac><mrow><mn>5</mn><mi>π</mi></mrow><mrow><mn>6</mn></mrow></mfrac></mrow></math> , then <math><mrow><mi>z</mi><mo>=</mo><mo>…</mo><mo>…</mo><mo>…</mo><mo>…</mo></mrow></math>

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New answer posted•a year ago

Complex Numbers and Quadratic Equations Class 11th

42. If ${(\frac{1 + i}{1 - i})}^{m} = 1$ , then find the least positive integral value of m.

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Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

42. We have,

${(\frac{1 + i}{1 - i})}^{m}$ = 1

=> ${(\frac{1 + i}{1 - i} \times \frac{1 +}{1 +})}^{m}$ = 1 [multiply denominator and numerator of LHS by (1 + i)]

=> ${(\frac{1 + i + i + i^{2}}{1^{2} - i^{2}})}^{m}$ = 1 [since, (a – b) (a + b) = a² – b²]

=> ${(\frac{1 + 2 i - 1}{1 + 1})}^{m}$ = 1 [since, i² = –1]

=> ${(\frac{2 i}{2})}^{m}$ = 1

=>i^m = 1

=>i^m = i^4k [since, i^4k = 1]

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New answer posted•a year ago

Complex Numbers and Quadratic Equations Class 11th

41. If (a + ib) (c + id) (e + if) (g + ih) = A + iB, then show that(a²+ b²) (c²+ d²) (e²+ f²) (g²+ h²) = A²+ B²

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Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

41. Given,

(a + ib) (c + id) (e + if) (g + ih) = A + iB

We know that,

Hence proved.

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New answer posted•a year ago

Complex Numbers and Quadratic Equations Class 11th

40. Find the number of non-zero integral solutions of the equation

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Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

40.

So, the only solution of the given equation is 0.

Hence, there is no non – zero integral solution of the given equation.

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New question posted•a year ago

Complex Numbers and Quadratic Equations Class 11th

40. Find the number of non-zero integral solutions of the equation<img src="https://images.shiksha.com/mediadata/images/articles/1742367821phpk1Qd5W.jpeg" alt="" width="64" height="28">.

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Asked by Payal Gupta

New answer posted•a year ago

Complex Numbers and Quadratic Equations Class 11th

39. If (x + iy)3 = u + iv, then show that $\frac{u}{x} + \frac{v}{y} = 4 (x^{2} - y^{2})$ .

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Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

39. (x + iy)³ = u + iv

=>x³ + (iy)³ + 3.x.iy (x + iy) = u + iv [since, (a + b)³ = a³ + b³ + 3ab (a + b)]

=>x³ – iy³ + 3x²yi + 3xy²i² = u + iv

=>x³ – iy³ + 3x²yi – 3xy² = u + iv [since, i² = -1]

=> (x³ –

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New answer posted•a year ago

Complex Numbers and Quadratic Equations Class 11th

38. Find the modulus of $\frac{1 + i}{1 - i} - \frac{1 - i}{1 + i}$

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Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

38. $\frac{1 + i}{1 - i}$ – $\frac{1 - i}{1 + i}$

= $\frac{{(1 + i)}^{2} - {(1 - i)}^{2}}{(1 - i) (1 + i)}$

= $\frac{1^{2} + i^{2} + 2.1 . i - (1^{2} + i^{2} - 2.1 . i)}{1^{2} - i^{2}}$ [Since, (a + b)² = a² + b² + 2ab

(a – b)² = a² + b² – 2ab

a² – b² = (a + b) (a – b)]

= $\frac{1 - 1 + 2 i - 1 + 1 + 2 i}{1 + 1}$ [Since, i² = –1]

= $\frac{4 i}{2}$

= 2i

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New answer posted•a year ago

Complex Numbers and Quadratic Equations Class 11th

37. Find the real numbers x and y if (x – iy) (3 + 5i) is the conjugate of –6 – 24i.

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Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

37. Let z = (x – iy) (3 + 5i)

= 3x + 5xi – 3yi – 5yi²

= (3x + 5y) + (5x – 3y)i

Given, $\bar{z}$ = –6 – 24i

=> (3x + 5y) – (5x – 3y)i = –6 – 24i

Equating real and imaginary part,

3x + 5y = –6 - (1)

5x – 3y = 24 - (2)

Multiplying (1) by 3 and (2) by 5 and adding them, we get

9x + 15y + 25x – 15y = –18 + 120

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New answer posted•a year ago

Complex Numbers and Quadratic Equations Class 11th

36. Let z₁= 2 – i, z₂ = –2 + i. Find (i) $R E (\frac{z_{1} z_{2}}{z_{1}})$ (ii) $I M (\frac{1}{z_{1} z_{1}})$

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Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

36. Z₁= 2 – i, z₂ = –2 + i

Z₁z₂= (2 – i)(–2 + i)

= –4 + 2i + 2i – i2

= –4 + 4i + 1[since, i² = –1]

= –3 + 4i

$\bar{z_{1}}$ = 2 + i

i. $\frac{z_{1} z_{2}}{\bar{z_{1}}}$ = $\frac{- 3 + 4 i}{2 + i}$

= $\frac{- 3 + 4 i}{2 + i}$ × $\frac{2 - i}{2 - i}$ [multiply denominator and numerator by (2 – i)]

= $\frac{- 6 + 3 i + 8 i - 4 i^{2}}{2^{2} - i^{2}}$

= $\frac{- 6 + 11 i + 4}{4 - (- 1)}$ [since, i2 = –1]

= $\frac{- 6 + 4 + 11 i}{4 + 1}$

= $\frac{- 2 + 11 i}{5}$

= $\frac{- 2}{5}$ + $\frac{11}{5} i$

So, Re(&nbs

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New answer posted•a year ago

Complex Numbers and Quadratic Equations Class 11th

35. If a + ib $= \frac{{(x + i)}^{2}}{2 x^{2} + 1}$ , prove that a² + b² $= \frac{{(x^{2} + 1)}^{2}}{{(2 x^{2} + 1)}^{2}} .$

13 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

35. Let, z = a + ib

= $\frac{{(x + i)}^{2}}{2 x^{2} + 1}$

= $\frac{x^{2} + i^{2} + 2 x i}{2 x^{2} + 1}$ [since, (a + b)² = a² + b² + 2ab]

= $\frac{x^{2} - 1 + 2 x i}{2 x^{2} + 1}$ [since, i2 = –1]

= $\frac{x^{2} - 1}{2 x^{2} + 1}$ + $\frac{i 2 x}{2 x^{2} + 1}$

So, |z|²= a² + b²

= $\frac{{(x^{2} - 1)}^{2}}{{(2 x^{2} + 1)}^{2}}$ + $\frac{{(2 x)}^{2}}{{(2 x^{2} + 1)}^{2}}$

= $\frac{{(x^{2})}^{2} + 1^{2} - 2 . x^{2} . 1 + 4 x^{2}}{{(2 x^{2} + 1)}^{2}}$ [since, (a + b)²= a²+ b² + 2ab]

= $\frac{x^{4} + 1 - 2 x^{2} + 4 x^{2}}{{(2 x^{2} + 1)}^{2}}$

= $\frac{x^{4} + 2 x^{2} + 1}{{(2 x^{2} + 1)}^{2}}$

= $\frac{{(x^{2} + 1)}^{2}}{{(2 x^{2} + 1)}^{2}}$ [as, (a + b)² = a² + b² + 2ab]

Hence proved.

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