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51.

Convert -1-i into polar form.(a) \(\sqrt{2}\), 5π/4(b) \(\sqrt{2}\), 3π/4(c) \(\sqrt{2}\), -3π/4(d) \(\sqrt{2}\), π/4This question was posed to me in an online interview.This interesting question is from Argand Plane and Polar Representation topic in division Complex Numbers and Quadratic Equations of Mathematics – Class 11

Answer»

Right answer is (C) \(\SQRT{2}\), -3π/4

The BEST I can explain: R=\(\sqrt{x^2+y^2} = \sqrt{(-1)^2+1^2} = \sqrt{1+1} = \sqrt{2}\).

r COS θ = -1 and r sin θ = -1 => θ is in 3^rd quadrant since sin and cos both negative.

tan θ = 1 => θ= -3π/4.
Discussion
52.

Find mirror image of point representing x+i y on imaginary axis.(a) (x, y)(b) (-x, -y)(c) (-x, y)(d) (x, -y)This question was posed to me in an international level competition.I need to ask this question from Argand Plane and Polar Representation topic in section Complex Numbers and Quadratic Equations of Mathematics – Class 11

Answer» RIGHT choice is (C) (-x, y)

Explanation: MIRROR image of POINT (x, y) on IMAGINARY axis is (-x, y).

Since imaginary axis is acting as mirror y-coordinate remains same whereas x-coordinate gets inverted. So, (-x, y) is mirror image of (x, y) on imaginary axis.
Discussion
53.

(2-i)^3 =________________(a) 2-3i(b) 8-i(c) 2-11i(d) 2+11iThis question was posed to me during an online interview.This is a very interesting question from Complex Numbers-2 topic in section Complex Numbers and Quadratic Equations of Mathematics – Class 11

Answer» RIGHT CHOICE is (c) 2-11i

Easy explanation: We KNOW, (a-b)^3 = a^3-b^3-3ab(a-b)

So, (2-i)^3 = 2^3-(i)^3-3(2)(i) (2-i)

= 8-(-i)-6i(2-i)

= 8+i-12i-6

= 2-11i.
Discussion
54.

If z1 = 2+3i and z2 = 5+2i, then find sum of two complex numbers.(a) 4+8i(b) 3-i(c) 7+5i(d) 7-5iThe question was posed to me in a national level competition.This interesting question is from Complex Numbers-1 in section Complex Numbers and Quadratic Equations of Mathematics – Class 11

Answer» CORRECT answer is (c) 7+5i

For EXPLANATION I would say: In ADDITION of two COMPLEX numbers, corresponding parts of two complex numbers are added i.e. real parts of both are added and IMAGINARY parts of both are added.

So, sum = (2+5) + (3+2) i = 7+5i.
Discussion
55.

1+0i is _________________ for complex number z.(a) additive inverse(b) additive identity element(c) multiplicative identity element(d) multiplicative inverseI got this question during an internship interview.I'm obligated to ask this question of Complex Numbers-1 topic in portion Complex Numbers and Quadratic Equations of Mathematics – Class 11

Answer»

The correct choice is (c) multiplicative identity ELEMENT

The explanation is: On MULTIPLYING one (1+0i) to a COMPLEX NUMBER, we GET same complex number so 1+0i is multiplicative identity element for complex number z i.e. z*1=z.
Discussion
56.

(x+3) + i(y-2) = 5+i2, find the values of x and y.(a) x=8 and y=4(b) x=2 and y=4(c) x=2 and y=0(d) x=8 and y=0This question was posed to me in an interview for job.I'd like to ask this question from Complex Numbers-1 in division Complex Numbers and Quadratic Equations of Mathematics – Class 11

Answer»

The correct answer is (b) x=2 and y=4

To elaborate: If two COMPLEX numbers are equal, then corresponding parts are equal i.e. REAL parts of both are equal and imaginary parts of both are equal.

x+3 = 5 and y-2 = 2

x = 5-3 and y = 2+2

x=2 and y=4.
Discussion
57.

Convert -1+i into polar form.(a) \(\sqrt{2}\), 5π/4(b) \(\sqrt{2}\), 3π/4(c) –\(\sqrt{2}\), π/4(d) \(\sqrt{2}\), π/4The question was posed to me in an interview.Enquiry is from Argand Plane and Polar Representation in division Complex Numbers and Quadratic Equations of Mathematics – Class 11

Answer»

Right choice is (b) \(\sqrt{2}\), 3π/4

To elaborate: r=\(\sqrt{x^2+y^2}=\sqrt{(-1)^2+1^2}=\sqrt{1+1}=\sqrt{2}\).

r cos θ = -1 and r sin θ = 1 So, θ is in 2^nd quadrant SINCE sin is POSITIVE and cos is negative.

tan θ = -1 => tan θ = -tan π/4

=> tan θ = tan (π-π/4) = tan 3π/4

=> θ=3π/4.
Discussion
58.

If acosθ + bsinθ = c have roots α and β. Then, what will be the value of sinα * sinβ ?(a) 2bc/(a^2 + b^2)(b) 0(c) 1(d) (c^2 + a^2)/(a^2 + b^2)The question was posed to me in exam.Enquiry is from Quadratic Equations in division Complex Numbers and Quadratic Equations of Mathematics – Class 11

Answer»

The correct OPTION is (d) (c^2 + a^2)/(a^2 + b^2)

For explanation I would say: Given, acosθ + bsinθ = c

So, this implies acosθ = c – bsinθ

Now squaring both the sides we get,

(acosθ)^2 = (c – bsinθ)^2

a^2 cos^2 θ = c^2 + b^2 sin^2 θ – 2b c sinθ

a^2 (1- sin^2 θ) = c^2 + b^2 sin^2 θ – 2b c sinθ

a^2 – a^2 sin^2 θ = c^2 + b^2 sin^2 θ – 2b c sinθ

Now rearranging the elements,

(a^2 + b^2) sin^2 θ – 2b c sinθ+( c^2 – a^2) = θ

So, as sum of the roots are in the formc/a if there is a QUADRATIC equation ax^2 + BX + c = 0

Now , we can CONCLUDE that

sinα + sinβ = (c^2 + a^2)/(a^2 + b^2).
Discussion
59.

6i is point on ____________________(a) x-axis(b) y-axis(c) z-axis(d) any axisI had been asked this question in an online quiz.Enquiry is from Argand Plane and Polar Representation topic in chapter Complex Numbers and Quadratic Equations of Mathematics – Class 11

Answer» CORRECT option is (b) y-axis

For explanation I WOULD say: Since real part of complex number is zero. So, it is PLOTTED on IMAGINARY axis i.e. y-axis.

6i is point on y-axis.
Discussion
60.

i^-35 =___________________(a) 1(b) -1(c) i(d) -iThis question was addressed to me by my school teacher while I was bunking the class.I need to ask this question from Complex Numbers-2 topic in portion Complex Numbers and Quadratic Equations of Mathematics – Class 11

Answer» RIGHT option is (c) i

The EXPLANATION: We KNOW, i^-35= 1/i^35 = i/i^36

= i/(i^4)^9 = i/1 = i.
Discussion
61.

i^7 =______________(a) 1(b) -1(c) i(d) -iThis question was addressed to me in an online interview.My question is based upon Complex Numbers-2 in division Complex Numbers and Quadratic Equations of Mathematics – Class 11

Answer» CORRECT choice is (d) -i

Easiest explanation: We know, i = \(\sqrt{-1}\)

 => i^2 = -1 => i^4 = 1.

So, i^7 = i^4.i^3 = 1*i^2*i = (-1)*i = -i.
Discussion
62.

In polar representation of a complex number (r, π/2) lies on ____________(a) x-axis(b) y-axis(c) z-axis(d) any axisI got this question by my college professor while I was bunking the class.My question is from Argand Plane and Polar Representation topic in section Complex Numbers and Quadratic Equations of Mathematics – Class 11

Answer»

Correct choice is (b) y-AXIS

To explain I would say: To convert POLAR representation (r, θ) into argand plane (X, y), substitute x=r cos θ and y=r sin θ.

x=r cos π/2 = 0 and y=r sin π/2 = r.

Argand plane representation is (0, r). Since real part is zero, so it LIES on IMAGINARY axis i.e. y-axis.
Discussion
63.

0+0i is ______________________for complex number z.(a) additive inverse(b) additive identity element(c) multiplicative identity element(d) multiplicative inverseThe question was asked in a national level competition.This intriguing question originated from Complex Numbers-1 in section Complex Numbers and Quadratic Equations of Mathematics – Class 11

Answer» RIGHT OPTION is (b) additive identity ELEMENT

For EXPLANATION I would say: On adding zero (0+0i) to a complex number, we GET same complex number so 0+0i is additive identity element for complex number z i.e. z+0 = z.
Discussion
64.

Square roots of -7 are____________(a) 7i and -7i(b) \(\sqrt{7}\) i(c) –\(\sqrt{7}\) i(d) \(\sqrt{7}\) i and –\(\sqrt{7}\) iI got this question by my college director while I was bunking the class.This question is from Complex Numbers-2 in portion Complex Numbers and Quadratic Equations of Mathematics – Class 11

Answer» CORRECT choice is (d) \(\sqrt{7}\) i and –\(\sqrt{7}\) i

For explanation: We know, i^2 = -1.

-7 = 7(i^2)

Square ROOT of i^2 is ±i so, square root of -7 are \(\sqrt{7}\)i and –\(\sqrt{7}\)i.
Discussion
65.

Solve \(\sqrt{3}x^2 – \sqrt{2} x + 3\sqrt{3}\) = 0(a) \(\frac{-1±i\sqrt{17}}{2}\)(b) \(\frac{1±i\sqrt{17}}{2}\)(c) \(\frac{1±\sqrt{17}}{2}\)(d) \(\frac{-1±\sqrt{17}}{2}\)I got this question during an interview for a job.The question is from Quadratic Equations topic in section Complex Numbers and Quadratic Equations of Mathematics – Class 11

Answer»

Right choice is (B) \(\FRAC{1±i\sqrt{17}}{2}\)

The EXPLANATION is: \(\sqrt{3}x^2 – \sqrt{2} x + 3\sqrt{3}\) = 0

=>3x^2 – \(\sqrt{6}\)x + 9 = 0

=>D=(-\(\sqrt{6}\))^2 – 4.3.9 = 6-108 = -102.

Since D ≤ 0, imaginary roots are there.

=>x = \(\frac{\sqrt{6}±i\sqrt{102}}{2.3} = \frac{1±i\sqrt{17}}{2}\).
Discussion
66.

Find multiplicative inverse of 3+5i.(a) 87+145i(b) 87-145i(c) 145-87i(d) 145+87iI have been asked this question in homework.My enquiry is from Complex Numbers-2 in division Complex Numbers and Quadratic Equations of Mathematics – Class 11

Answer»

The correct ANSWER is (B) 87-145i

The explanation is: We KNOW, Z*\(\bar{z}\) = |z|^2.

(1/z) = \(\bar{z}\)|z|^2

z^-1=(3-5i) (3^2+5^2) = (3-5i) (29) = 87-145i.
Discussion
67.

z1=1+2i and z2=2+3i. Find z1z2.(a) 2+6i(b) -4+0i(c) -4+7i(d) 8+7iThe question was posed to me by my college professor while I was bunking the class.My question is from Complex Numbers-2 in division Complex Numbers and Quadratic Equations of Mathematics – Class 11

Answer»

The CORRECT answer is (c) -4+7i

Best explanation: Z1Z2 = (1+2i) (2+3I)

= 2 + 3i + 4I -6

= -4 + 7i.
Discussion