chapter-5 Complex Number And Quadratic Equations Miscellaneous Exercise

chapter-5 Complex Number And Quadratic Equations Miscellaneous Exercise

NCERT Solutions for Class-11 Maths Chapter-5 Complex Number and Quadratic Equations

NCERT Solutions For Class-11 Maths Chapter-5 Complex Number and Quadratic Equations Miscellaneous Exercise prepared by the expert of Physics Wallah score more with Physics Wallah NCERT Class-11 maths solutions. You can download solution of all chapters from Physics Wallah NCERT solutions of class 11.

NCERT Solutions for Class-11 Maths Chapter-5 Complex Number and Quadratic Equations Miscellaneous Exercise

Question 1. Evaluate: [i ¹⁸ + (1 / i) ²⁵ ] ³

Solution :
Given: [i ¹⁸ + (1 / i) ²⁵ ] ³


Question 2. For any two complex numbers z ₁ and z ₂ , prove that
Re (z ₁ z ₂ )  =  Re z ₁ Re z ₂ –  Im z ₁ Im z ₂

Solution :
Let z ₁ = x ₁ +  iy ₁ and z ₂ = x ₂ + iy ₂
∴z ₁ z ₂ = (x ₁ +iy ₁ ) (x ₂ + iy ₂ )
=x1(x ₂ + iy ₂ )  +  iy ₁ (x ₂ + ty ₂ )
=x ₁ x ₂ +  ix ₁ y ₂ +  iy ₁ x ₂ +  i2y ₁ y ₂
=x ₁ x ₂ + ix ₁ y ₂ + iy ₁ x ₂ − y ₁ y ₂
=(x ₁ x ₂ −  y ₁ y ₂ ) +  i(x ₂ y ₂ +  y ₁ x ₂ )
⇒Re(z ₁ z ₂ )=  x ₁ x ₂ −  y ₁ y ₂
⇒Re (z ₁ z ₂ )  =  Rez ₁ Rez ₂ −  lmz ₁ Imz ₂
Hence, proved.
Question 3. Reduce to the standard form.

Solution :


Question 4. If prove that

Solution :



Question 5. Convert the following in the polar form:
(i)
(ii)

Solution :


This is the required polar form.

(ii) Here

Solve each of the equations in exercises 6 to 9:
Question 6. 3x ² -4x + 20/3 = 0

Solution :

Question 7. x ² -2x + 3/2 = 0

Solution . Given:

Question 8. Solve the equation 27x ² – 10x + 1 = 0

Solution :
Given:

Question 9. Solve the equation 21x ² – 28x + 10 = 0

Solution :
Given:

Question 10. If find

Solution :



Question 11. If prove that

Solution :



Question 12. Let z ₁ =2−i,z ₂ =−2+i. Find
(i) Re (z ₁ z ₂ /z ₁ ),
(ii) Im (1z ₁ / z ₁ )

Solution :



Question 13. Find the modulus and argument of the complex number

Solution :



Question 14. Find the real numbers x and y if (x – iy) (3 + 5i) is the conjugate of –6 – 24i.

Solution :
Let = (x−ty) (3+5i)z
=3x + 5xi − 3yi − 5yi ²
=3x + 5xi − 3yi + 5y
=(3x+5y) + i(5x−3y)
∴z =(3x+5y) −i (5x−3y)
It is given that,z=−6−24i
∴(3x+5y) −i(5x−3y) =−6−24i
Equating real and imaginary parts, we obtain
3x+5y=−6
5x−3y=24
Multiplying equation (i) by 3 and equation (ii) by 5 and then adding them, we obtain
Putting the value ofxin equation (i),
we obtain 3(3)+ 5y=−6⇒ 5y=−6−15 ⇒y=−3
Thus, the values of x and y are 3 and − 3 respectlvely.
Question 15. Find the modulus of

Solution :


Question 16. If then show that
Ans. Given:

Question 17. If  α and  β are different complex numbers with | β | = 1  then find

Solution :


Question 18. Find the number of non-zero integral solutions of the equation | 1−i| ^x = 2 ^x

Solution :


Question 19. If (a + ib) (c + id) (e + if) (g + ih) = A + iB, then show that
(a ² + b ² ) (c ² + d ² ) (e ² + f ² ) (g ² + h ² ) = A ² + B ² .

Solution :


Question 20. If then find the least positive integral value of m.

Solution :