In this chapter, we provide NCERT Exemplar Problems Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations for English medium students, Which will very helpful for every student in their exams. Students can download the latest NCERT Exemplar Problems Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations pdf, free NCERT Exemplar Problems Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations book pdf download. Now you will get step by step solution to each question.
|Chapter Name||Complex Numbers and Quadratic Equations|
NCERT Exemplar Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations
Short Answer Type Questions
Q6. If a = cos θ + i sin θ, then find the value of (1+a/1-a)
Sol: a = cos θ + i sin θ
Q10. Show that the complex number z, satisfying the condition arg lies on arg (z-1/z+1) = π/4 lies on a circle.
Sol: Let z = x + iy
Q11. Solve the equation |z| = z + 1 + 2i.
Sol: We have |z| = z + 1 + 2i
Putting z = x + iy, we get
|x + iy| = x + iy + 1+2i
Long Answer Type Questions
Q12. If |z + 1| = z + 2( 1 + i), then find the value of z.
Sol: We have |z + 11 = z + 2(1+ i)
Putting z = x + iy, we get
Then, |x + iy + 11 = x + iy + 2(1 + i)
⟹|x + iy + l|=x + iy + 2(1 +i)
Q13. If arg (z – 1) = arg (z + 3i), then find (x – 1) : y, where z = x + iy.
Sol: We have arg (z – 1) = arg (z + 3i), where z = x + iy
=> arg (x + iy – 1) = arg (x + iy + 3i)
=> arg (x – 1 + iy) = arg [x + i(y + 3)]
Q14. Show that | z-2/z-3| = 2 represents a circle . Find its center and radius .
Sol: We have | z-2/z-3| = 2
Puttingz=x + iy, we get
Q15. If z-1/z+1 is a purely imaginary number (z ≠1), then find the value of |z|.
Sol: Let z = x + iy
Q17. If |z1 | = 1 (z1≠ -1) and z2 = z1 – 1/ z1 + 1 , then show that real part of z2 is zero .
Q18. If Z1, Z2 and Z3, Z4 are two pairs of conjugate complex numbers, then find arg (Z1/ Z4) + arg (Z2/ Z3)
Sol. It is given that z1 and z2 are conjugate complex numbers.
Q20. If for complex number z1 and z2, arg (z1) – arg (z2) = 0, then show that |z1 – z2| = | z1|- |z2 |
Q21. Solve the system of equations Re (z2) = 0, |z| = 2.
Sol: Given that, Re(z2) = 0, |z| = 2
Q22. Find the complex number satisfying the equation z + √2 |(z + 1)| + i = 0.
Fill in the blanks
True/False Type Questions
Q26. State true or false for the following.
(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 origin through a right angle in the anti-clockwise direction.
(iii) For any complex number z, the minimum value of |z| + |z – 11 is 1.
(iv) The locus represented by |z — 11= |z — i| is a line perpendicular to the join of the points (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 |z, + z2| = |z1 j + |z2|, then arg (z1 – z2) = 0.
(viii) 2 is not a complex number.
We can compare two complex numbers when they are purely real. Otherwise comparison of complex numbers is not possible or has no meaning.
Let z = x + iy, where x, y > 0
i.e., z or point A(x, y) lies in first quadrant. Now, —iz = -i(x + iy)
= -ix – i2y = y – ix
Now, point B(y, – x) lies in fourth quadrant. Also, ∠AOB = 90°
Thus, B is obtained by rotating A in clockwise direction about origin.
Matching Column Type Questions
Q24. Match the statements of Column A and Column B.
|Column A||Column B|
|(a)||The polar form of i + √3 is||(i)||Perpendicular bisector of segment joining (-2, 0) and (2,0)|
|(b)||The amplitude of- 1 + √-3 is||(ii)||On or outside the circle having centre at (0, -4) and radius 3.|
|(c)||It |z + 2| = |z – 2|, then locus of z is||(iii)||2/3|
|(d)||It |z + 2i| = |z – 2i|, then locus of z is||(iv)||Perpendicular bisector of segment joining (0, -2) and (0,2)|
|(e)||Region represented by |z + 4i| ≥ 3 is||(v)||2(cos /6 +I sin /6)|
|(0||Region represented by |z + 4| ≤ 3 is||(Vi)||On or inside the circle having centre (-4,0) and radius 3 units.|
|(g)||Conjugate of 1+2i/1-I lies in||(vii)||First quadrant|
|(h)||Reciprocal of 1 – i lies in||(viii)||Third quadrant|
Q28. What is the conjugate of 2-i / (1 – 2i)2
Q29. If |Z1| = |Z2|, is it necessary that Z1 = Z2?
Sol: If |Z1| = |Z2| then z1 and z2 are at the same distance from origin.
But if arg(Z1) ≠arg(z2), then z1 and z2 are different.
So, if (z1| = |z2|, then it is not necessary that z1 = z2.
Consider Z1 = 3 + 4i and Z2 = 4 + 3i
Q30.If (a2+1)2 / 2a –i = x + iy, then what is the value of x2 + y2?
Sol: (a2+1)2 / 2a –i = x + iy
Q31. Find the value of z, if |z| = 4 and arg (z) = 5π/6
Q34. Where does z lies, if | z – 5i / z + 5i | = 1?
Sol: We have | z – 5i / z + 5i |
Instruction for Exercises 35-40: Choose the correct answer from the given four options indicated against each of the Exercises.
Q35. sin x + i cos 2x and cos x – i sin 2x are conjugate to each other for
Q41. Which of the following is correct for any two complex numbers z1 and z2?
All Chapter NCERT Exemplar Problems Solutions For Class 11 Maths
All Subject NCERT Exemplar Problems Solutions For Class 11
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