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Complex Numbers and Quadratic Equations

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New answer posted•a year ago
Complex Numbers and Quadratic EquationsMaths NCERT Exemplar Solutions Class 11th Chapter Five

If ?z1?=?z2?=1 , is it necessary that z1=z2 ?

3 Views|Posted a year ago
Asked by Payal Gupta
S
a year ago

This is Other Questions as classified in NCERT Exemplar

L e t     z 1 = x 1 + y 1 i     a n d     z 2 = x 2 + y 2 i ∴                     | x 1 + y 1 i | = | x 2 + y 2 i | ⇒           x 1 2 + y 1 2 = x 2 2 + y 2 2 ⇒                   x 1 2 + y 1 2 = x 2 2 + y 2 2           ⇒ x 1 2 = x 2 2     a n d     y 1 2 = y 2 2 ⇒                   x 1 = ± x 2     a n d     y 1 = ± y 2 S o ,               z 1 = x 1 + y 1 i     a n d     z 2 = ± x 2 ± y 2 i ∴                       z 1 ≠ z 2 H e n c e ,     i t     i s     n o t     n e c e s s a r y     t h a t     z 1 = z 2 .

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New answer posted•a year ago
Complex Numbers and Quadratic EquationsMaths NCERT Exemplar Solutions Class 11th Chapter Five

What is the conjugate of 2−i(1−2i) z

2 Views|Posted a year ago
Asked by Payal Gupta
S
a year ago

This is Other Questions as classified in NCERT Exemplar

G i v e n     t h a t         z = 2 − i ( 1 − 2 i ) 2 = 2 − i 1 + 4 i 2 − 4 i = 2 − i 1 − 4 − 4 i                                                             = 2 − i − 3 − 4 i = 2 − i − 3 − 4 i × − 3 + 4 i − 3 + 4 i                                                             = − 6 + 8 i + 3 i − 4 i 2 ( − 3 ) 2 − ( 4 i ) 2 = − 6 + 1 1 i + 4 9 − 1 6 i 2                                                             = − 2 + 1 1 i 9 + 1 6 = − 2 2 5 + 1 1 2 5 i ∴                                               z ¯ = − 2 2 5 − 1 1 2 5 i H e n c e ,                       z ¯ = − 2 2 5 − 1 1 2 5 i

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New answer posted•a year ago
Complex Numbers and Quadratic EquationsExemplar Solution

 Find z if ?z?=4 and (z)=5π6 .

2 Views|Posted a year ago
Asked by Payal Gupta
S
a year ago

This is Other Questions as classified in NCERT Exemplar

G i v e n     t h a t : | z | = 4     a n d     a r g ( z ) = 5 π 6     ⇒ θ = 5 π 6         | z | = 4     ⇒ r = 4 S o     P o l a r     f o r m     o f     z = r [ c o s θ + i s i n θ ]                                                                                           = 4 [ c o s 5 π 6 + i s i n 5 π 6 ]                                                                                           = 4 [ c o s ( π − π 6 ) + i s i n ( π − π 6 ) ]                                                                                           = 4 [ − c o s π 6 + i s i n π 6 ] = 4 [ 3 2 + i . 1 2 ]                                                                                             = − 2 3 + 2 i H e n c e ,     z = − 2 3 + 2 i .

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New answer posted•a year ago
Complex Numbers and Quadratic EquationsMaths NCERT Exemplar Solutions Class 12th Chapter Five

If (a2+1)22a−i=x+iy,  what is the value of x2+y2 ?

4 Views|Posted a year ago
Asked by Payal Gupta
S
a year ago

This is Other Questions as classified in NCERT Exemplar

G i v e n     t h a t : ( a 2 + 1 ) 2 2 a − i = x + y i                                                                   … ( i ) T a k i n g     c o n j u g a t e     o n     b o t h     s i d e s                                                   ( a 2 + 1 ) 2 2 a + i = x − y i                                                                   … ( i i ) M u l t i p l y i n g     e q n . ( i )     a n d     ( i i )     w e     h a v e             ( a 2 + 1 ) 2 ( a 2 + 1 ) 2 ( 2 a − i ) ( 2 a + i ) = x 2 + y 2 ⇒                                         ( a 2 + 1 ) 4 4 a 2 − i 2 = x 2 + y 2 ⇒                                         ( a 2 + 1 ) 4 4 a 2 + 1 = x 2 + y 2 H e n c e ,     t h e     v a l u e     o f     x 2 + y 2 = ( a 2 + 1 ) 4 4 a 2 + 1 .

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New answer posted•a year ago
Complex Numbers and Quadratic EquationsMaths NCERT Exemplar Solutions Class 12th Chapter Five

If ?z1?=?z2?=1 , is it necessary that z1=z2 ?

3 Views|Posted a year ago
Asked by Payal Gupta
S
a year ago

This is Other Questions as classified in NCERT Exemplar

L e t     z 1 = x 1 + y 1 i     a n d     z 2 = x 2 + y 2 i ∴                     | x 1 + y 1 i | = | x 2 + y 2 i | ⇒           x 1 2 + y 1 2 = x 2 2 + y 2 2 ⇒                   x 1 2 + y 1 2 = x 2 2 + y 2 2           ⇒ x 1 2 = x 2 2     a n d     y 1 2 = y 2 2 ⇒                   x 1 = ± x 2     a n d     y 1 = ± y 2 S o ,               z 1 = x 1 + y 1 i     a n d     z 2 = ± x 2 ± y 2 i ∴                       z 1 ≠ z 2 H e n c e ,     i t     i s     n o t     n e c e s s a r y     t h a t     z 1 = z 2 .

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New answer posted•a year ago
Complex Numbers and Quadratic EquationsMaths NCERT Exemplar Solutions Class 11th Chapter Five

What is the conjugate of 2−i(1−2i) z

4 Views|Posted a year ago
Asked by Payal Gupta
S
a year ago

This is Other Questions as classified in NCERT Exemplar

G i v e n     t h a t         z = 2 − i ( 1 − 2 i ) 2 = 2 − i 1 + 4 i 2 − 4 i = 2 − i 1 − 4 − 4 i                                                             = 2 − i − 3 − 4 i = 2 − i − 3 − 4 i × − 3 + 4 i − 3 + 4 i                                                             = − 6 + 8 i + 3 i − 4 i 2 ( − 3 ) 2 − ( 4 i ) 2 = − 6 + 1 1 i + 4 9 − 1 6 i 2                                                             = − 2 + 1 1 i 9 + 1 6 = − 2 2 5 + 1 1 2 5 i ∴                                               z ¯ = − 2 2 5 − 1 1 2 5 i H e n c e ,                       z ¯ = − 2 2 5 − 1 1 2 5 i

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New answer posted•a year ago
Complex Numbers and Quadratic EquationsMaths NCERT Exemplar Solutions Class 11th Chapter Five

(i) The order relation is defined on the set of complex numbers.

(ii) Multiplication of a non-zero complex number by −i rotates the point about the origin through a right angle in the anti-clockwise direction.

(iii) For any complex number z , the minimum value of ?z+1z? is 1.

(iv) The locus represented by ?z−1?=?z−i? is a line perpendicular to the join of (1,0) and (0,1) .

(v) If z is a complex number such that z≠0 and Re(z)=0 , then Im(z2)=0 .

(vi) The inequality ?z−4?<?z−2? represents the region given by x>3 .

(vii) Let z1 and z2 be two complex numbers such that ?z1+z2?=?z1?+?z2? , then arg(z1−z2)=0 .

(viii) 2 is not a complex number.

6 Views|Posted a year ago
Asked by Payal Gupta
S
a year ago

This is a True or False Type Questions as classified in NCERT Exemplar

( i ) C o m p a r i s o n     o f     t w o     p u r e l y     i m a g i n a r y     c o m p l e x     n u m b e r s     i s     n o t     p o s s i b l e .     H o w e v e r ,     t h e     t w o               p u r e l y     i m a g i n a r y     r e a l     c o m p l e x     n u m b e r s     c a n     b e     c o m p a r e d .               S o ,     i t     i s     ' F a l s e ' . ( i i ) L e t     z = x + y i                 z . i = ( x + y i ) i = x i − y     w h i c h     r o t a t e s     a t     a n g l e     o f     1 8 0 0               S o ,     i t     i s     ' F a l s e ' . ( i i i ) L e t     z = x + y i                 ∴ | z | + | z + 1 | = x 2 + y 2 + ( x − 1 ) 2 + y 2                 T h e     v a l u e     o f     | z | + | z + 1 |     i s     m i n i m u m     w h e n     x = 0 ,     y = 0     i . e . ,     1 .               H e n c e ,     i t     i s     ' T r u e ' . ( i v ) L e t     z = x + y i                 G i v e n     t h a t :                       | z − 1 | = | z − i |                 t h e n                                     | x + y i − 1 | = | x + y i − i | ⇒                                                     | ( x − 1 ) + y i | = | x − ( 1 − y ) i | ⇒                                           ( x − 1 ) 2 + y 2 = x 2 + ( 1 − y ) 2 ⇒                                           ( x − 1 ) 2 + y 2 = x 2 + ( 1 − y ) 2 ⇒                             x 2 − 2 x + 1 + y 2 = x 2 + 1 + y 2 − 2 y ⇒                                                     − 2 x + 2 y = 0 ⇒                                                                       x − y = 0     w h i c h     i s     a     s t r a i g h t     l i n e .                 S l o p e = 1         Now  equation  of  a  line  through  the  point  (1,0)  and  (0,1)                                                                                 y − 0 = 1 − 0 0 − 1 ( x − 1 ) ⇒                                                                                     y = − x + 1     w h o s e     s l o p e = − 1 .                 N o w     t h e     m u l t i p l i c a t i o n     o f     t h e     s l o p e s     o f     t w o     l i n e s = − 1 × 1 = − 1 ,                 S o     t h e y     a r e     p e r p e n d i c u l a r .               H e n c e ,     i t     i s     ' T r u e ' .

( v ) L e t     z = x + y i ,     z ≠ 0     a n d     R e ( z ) = 0         Since  real  part  is  0  ⇒x=0               ∴ z = 0 + y i = y i             ∴ I m ( z 2 ) = y 2 i 2 = − y 2     w h i c h     i s     r e a l .               H e n c e ,     i t     i s     ' F a l s e ' . ( v i ) G i v e n     t h a t :         | z − 4 | < | z − 2 |                 L e t     z = x + y i ⇒                                                     | x + y i − 4 | < | x + y i − 2 | ⇒                                             | ( x − 4 ) + y i | < | ( x − 2 ) + y i | ⇒                                   ( x − 4 ) 2 + y 2 < ( x − 2 ) 2 + y 2 ⇒                                           ( x − 4 ) 2 + y 2 < ( x − 2 ) 2 + y 2 ⇒                                                               ( x − 4 ) 2 < ( x − 2 ) 2 ⇒                                               x 2 − 8 x + 1 6 < x 2 − 4 x + 4 ⇒                                                         − 8 x + 4 x < − 1 6 + 4 ⇒                                                                             − 4 x < − 1 2 ⇒                                                                                             x > 3               H e n c e ,     i t     i s     ' T r u e ' . ( v i i ) L e t     z 1 = x 1 + y 1 i     a n d       z 2 = x 2 + y 2 i ⇒                                                     | z 1 + z 2 | = | z 1 | + | z

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New answer posted•a year ago
Complex Numbers and Quadratic EquationsMaths NCERT Exemplar Solutions Class 12th Chapter Five

If a and b are real numbers such that (2+α)4=a+bα , where α=-1+i32 , then a+b is equal to:

8 Views|Posted a year ago
Asked by Alok Kumar Singh
S
a year ago

x2-3x+p=0

α, β, γ, δ in G.P.

α+αr=3

x2-6x+q=0

αr2+αr3=6

(2)÷ (1)⇒r2=2

So,  2q+p2q-p=2r5+r2r5-r=2r4+12r4-1=97

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New answer posted•a year ago
Complex Numbers and Quadratic EquationsMaths NCERT Exemplar Solutions Class 12th Chapter Five

Let α , β , be the roots of the equation x2−4λx+5=0  and  α,γ be the roots of the equation x2−(32+23)x+7+3λ3=0, If β+γ=32, then (α+2β+γ)2 is equal to.

8 Views|Posted a year ago
Asked by Payal Gupta
S
a year ago

α , β are roots of x2 - 4 λ x + 5 = 0  

∴ α + β = 4 λ   a n d     α β = 5               

Also α , γ  are roots of

x 2 − ( 3 2 + 2 3 ) x + 7 + 3 3 λ = 0 , λ > 0

∴ α + γ = 3 2 + 2 3 , α γ = 7 + 3 2 λ               

? α is common root

∴ α 2 − 4 λ α + 5 = 0     …….(i)

And

α 2 − ( 3 2 + 2 3 ) α + 7 + 3 3 λ = 0      ….(ii)

From (i) – (ii) ; we get

α = 2 + 3 3 λ 3 2 + 2 3 − 4 λ     &nb

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New question posted•a year ago
Complex Numbers and Quadratic EquationsMaths NCERT Exemplar Solutions Class 11th Chapter Five

<p><strong>(i) For any two complex numbers&nbsp;<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>&nbsp;and any real numbers&nbsp;<math><mrow><mi>a</mi><mo>,</mo><mi>b</mi></mrow></math> , <span class="mathml" contenteditable="false"> <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> </span></strong></p> <p><strong>(ii) The value of&nbsp;<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>&nbsp;is .</strong></p> <p><strong>(iii) The number&nbsp;<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>&nbsp;is equal to .</strong></p> <p><strong>(iv) The sum of the series&nbsp;<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>&nbsp;up to 1000 terms is .</strong></p> <p><strong>(v) Multiplicative inverse of&nbsp;<math><mrow><mn>1</mn><mo>+</mo><mi>i</mi></mrow></math>&nbsp;is .</strong></p> <p><strong>(vi) If&nbsp;<math><mrow><msub><mrow><mi>z</mi></mrow><mrow><mn>1</mn></mrow></msub></mrow></math>&nbsp;and&nbsp;<math><mrow><msub><mrow><mi>z</mi></mrow><mrow><mn>2</mn></mrow></msub></mrow></math>&nbsp;are complex numbers such that&nbsp;<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>&nbsp;is a real number, then&nbsp;<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><br></strong></p> <p><strong>(vii)&nbsp;<span contenteditable="false">&nbsp;<math><mrow><mrow><mo>(</mo><mrow><mi>z</mi></mrow><mo>)</mo></mrow><mo>+</mo><mi>z</mi><mrow><mo>(</mo><mrow><mi>z</mi><mo>≠</mo><mn>0</mn></mrow><mo>)</mo></mrow></mrow></math>&nbsp;</span>is .</strong></p> <p><strong>(viii) If&nbsp;<span class="mathml" contenteditable="false"> <math> <mrow> <mo>?</mo> <mi>z</mi> <mo>+</mo> <mn>4</mn> <mo>?</mo> <mo>≤</mo> <mn>3</mn> </mrow> </math> </span>, then the greatest and least values of <span class="mathml" contenteditable="false"> <math> <mrow> <mo>?</mo> <mi>z</mi> <mo>+</mo> <mn>1</mn> <mo>?</mo> </mrow> </math> </span>are . and .</strong></p> <p><strong>(ix) If&nbsp;<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>&nbsp;then the locus of&nbsp;<math><mrow><mi>z</mi></mrow></math>&nbsp;is .</strong></p> <p><strong>(x) If <span class="mathml" contenteditable="false"> <math> <mrow> <mo>?</mo> <mi>z</mi> <mo>?</mo> <mo>=</mo> <mn>4</mn> </mrow> </math> </span></strong><strong>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>&nbsp;, then&nbsp;<math><mrow><mi>z</mi><mo>=</mo><mo>…</mo><mo>…</mo><mo>…</mo><mo>…</mo></mrow></math></strong></p>

4 Views|Posted a year ago
Asked by Payal Gupta