# complex numbers multiple choice questions with answers pdf

Question 5.If ω is an imaginary cube root of unity, then (1 + ω – ω²)7 equals(a) 128 ω(b) -128 ω(c) 128 ω²(d) -128 ω², Answer: (d) -128 ω²Hint:Given ω is an imaginary cube root of unity.So 1 + ω + ω² = 0 and ω³ = 1Now, (1 + ω – ω²)7 = (-ω² – ω²)7⇒ (1 + ω – ω2)7 = (-2ω2)7⇒ (1 + ω – ω2)7 = -128 ω14⇒ (1 + ω – ω2)7 = -128 ω12 × ω2⇒ (1 + ω – ω2)7 = -128 (ω3)4 ω2⇒ (1 + ω – ω2)7 = -128 ω2, Question 6.The least value of n for which {(1 + i)/(1 – i)}n is real, is(a) 1(b) 2(c) 3(d) 4, Answer: (b) 2Hint:Given, {(1 + i)/(1 – i)}n= [{(1 + i) × (1 + i)}/{(1 – i) × (1 + i)}]n= [{(1 + i)²}/{(1 – i²)}]n= [(1 + i² + 2i)/{1 – (-1)}]n= [(1 – 1 + 2i)/{1 + 1}]n= [2i/2]n= inNow, in is real when n = 2 {since i2 = -1 }So, the least value of n is 2, Question 7.Let z be a complex number such that |z| = 4 and arg(z) = 5π/6, then z =(a) -2√3 + 2i(b) 2√3 + 2i(c) 2√3 – 2i(d) -√3 + i, Answer: (a) -2√3 + 2iHint:Let z = r(cos θ + i × sin θ)Then r = 4 and θ = 5π/6So, z = 4(cos 5π/6 + i × sin 5π/6)⇒ z = 4(-√3/2 + i/2)⇒ z = -2√3 + 2i, Question 8:The value of i-999 is(a) 1(b) -1(c) i(d) -i, Answer: (c) iHint:Given, i-999= 1/i999= 1/(i996 × i³)= 1/{(i4)249 × i3}= 1/{1249 × i3} {since i4 = 1}= 1/i3= i4/i3 {since i4 = 1}= iSo, i-999 = i, Question 9.Let z1 and z2 be two roots of the equation z² + az + b = 0, z being complex. Where can you find best quality multiple choice questions? ... Then multiply the number by its complex conjugate. a) Find b and c b) Write down the second root and check it. More.. Download MCQs for JEE Mathematics Complex Numbers, Get MCQs for Complex Numbers Mathematics for important topics for all chapters based on 2021 syllabus and pattern. Start below. शिक्षण कौशल (Teaching Skill) h &� CJ UVaJ j h &� h &� EH��UjQ@yP � � : : N. � � � : : � : � : . You will see that, in general, you proceed as in real numbers, but using i 2 =−1 where appropriate. Check the below NCERT MCQ Questions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations with Answers Pdf free download. The probability that the number on the card taken out is an even number, is. Choose your answers to the questions and click 'Next' to see the next set of questions. Register for online coaching for IIT JEE (Mains & Advanced), NEET, Engineering and Medical entrance exams. &. bjbjLULU J .? EMBED Equation.3 45 27 9 + 36i -27 - 36i 5. . Questions on Complex Numbers . Check the below NCERT MCQ Questions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations with Answers Pdf free download. If you send me questions using the forms in MCQ Sets d. Get the most authentic answers for your subjective questions 31. Question 12.The value of x and y if (3y – 2) + i(7 – 2x) = 0(a) x = 7/2, y = 2/3(b) x = 2/7, y = 2/3(c) x = 7/2, y = 3/2(d) x = 2/7, y = 3/2, Answer: (a) x = 7/2, y = 2/3Hint:Given, (3y – 2) + i(7 – 2x) = 0Compare real and imaginary part, we get3y – 2 = 0⇒ y = 2/3and 7 – 2x = 0⇒ x = 7/2So, the value of x = 7/2 and y = 2/3, Question 13.Find real θ such that (3 + 2i × sin θ)/(1 – 2i × sin θ) is imaginary(a) θ = nπ ± π/2 where n is an integer(b) θ = nπ ± π/3 where n is an integer(c) θ = nπ ± π/4 where n is an integer(d) None of these, Answer: (b) θ = nπ ± π/3 where n is an integerHint:Given,(3 + 2i × sin θ)/(1 – 2i × sin θ) = {(3 + 2i × sin θ)×(1 – 2i × sin θ)}/(1 – 4i² × sin² θ)(3 + 2i × sin θ)/(1 – 2i × sin θ) = {(3 – 4sin² θ) + 8i × sin θ}/(1 + 4sin² θ) …………. . Then(a) a² = b(b) a² = 2b(c) a² = 3b(d) a² = 4b, Answer: (c) a² = 3bHint:Given, z1 and z2 be two roots of the equation z² + az + b = 0Now, z1 + z2 = -a and z1 × z2 = bSince z1 and z2 and z3 from an equilateral triangle.⇒ z12 + z22 + z32 = z1 × z2 + z2 × z3 + z1 × z3⇒ z12+ z22 = z1 × z2 {since z3 = 0}⇒ (z1 + z2)² – 2z1 × z2 = z1 × z2⇒ (z1 + z2)² = 2z1 × z2 + z1 × z2⇒ (z1 + z2)² = 3z1 × z2⇒ (-a)² = 3b⇒ a² = 3b, Question 10:The complex numbers sin x + i cos 2x are conjugate to each other for(a) x = nπ(b) x = 0(c) x = (n + 1/2) π(d) no value of x, Answer: (d) no value of xHint:Given complex number = sin x + i cos 2xConjugate of this number = sin x – i cos 2xNow, sin x + i cos 2x = sin x – i cos 2x⇒ sin x = cos x and sin 2x = cos 2x {comparing real and imaginary part}⇒ tan x = 1 and tan 2x = 1Now both of them are not possible for the same value of x.So, there exist no value of x, Question 11.The curve represented by Im(z²) = k, where k is a non-zero real number, is(a) a pair of striaght line(b) an ellipse(c) a parabola(d) a hyperbola. 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