Mathematics Questions and Answers – Argand Plane and Polar Representation

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This set of Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Argand Plane and Polar Representation”.

1. Which axis is known as real axis in argand plane?
a) x-axis
b) y-axis
c) z-axis
d) any axis
View Answer

Answer: a
Explanation: The plane having a complex number assigned to each of its point is called the
complex plane or the Argand plane. When (x + y i) is plotted in argand plane, x-axis is real axis.
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2. Which axis is known as imaginary axis in argand plane?
a) x-axis
b) y-axis
c) z-axis
d) any axis
View Answer

Answer: b
Explanation: The plane having a complex number assigned to each of its point is called the
complex plane or the Argand plane. When (x + y i) is plotted in argand plane, y-axis is imaginary axis.

3. 2+i0 is point on ______________
a) x-axis
b) y-axis
c) z-axis
d) any axis
View Answer

Answer: a
Explanation: Since imaginary part of complex number is zero. So, it is plotted on real axis i.e. x-axis.
2+i0 is point on x-axis.
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4. 6i is point on ____________________
a) x-axis
b) y-axis
c) z-axis
d) any axis
View Answer

Answer: b
Explanation: 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.

5. \(\sqrt{x^2+y^2}\) distance of point representing complex number x+y i from origin.
a) True
b) False
View Answer

Answer: a
Explanation: Since complex number x + y i is represented by (x, y) on argand plane, distance of point (x, y) from origin is \(\sqrt{x^2+y^2}\).
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6. Find mirror image of point representing x+i y on real axis.
a) (x, y)
b) (-x, -y)
c) (-x, y)
d) (x, -y)
View Answer

Answer: d
Explanation: Mirror image of point (x, y) on real axis is (x, -y).
Since real axis is acting as mirror x-coordinate remains same whereas y-coordinate gets inverted.
So, (x, -y) is mirror image of (x, y) on real axis.

7. Find mirror image of point representing x+i y on imaginary axis.
a) (x, y)
b) (-x, -y)
c) (-x, y)
d) (x, -y)
View Answer

Answer: c
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.
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8. If P and Q are conjugate complex numbers then their points on argand plane are mirror image on __________________
a) x-axis
b) y-axis
c) z-axis
d) any axis
View Answer

Answer: a
Explanation: Conjugate complex numbers means their real part is same and imaginary part is inverted i.e. same x part and opposite imaginary part. So, they are mirror image on real axis i.e. x-axis.

9. In polar representation of a complex number (r, 2π) lies on ____________
a) x-axis
b) y-axis
c) z-axis
d) any axis
View Answer

Answer: a
Explanation: To convert polar representation (r, θ) into argand plane (x, y), substitute x = r cos θ and y = r sin θ.
x = r cos 2π = r and y = r sin 2π = 0.
Argand plane representation is (r, 0). Since imaginary part is zero, so it lies on real axis i.e. x-axis.
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10. In polar representation of a complex number (r, π/2) lies on ____________
a) x-axis
b) y-axis
c) z-axis
d) any axis
View Answer

Answer: b
Explanation: 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.

11. Convert (8, 2π/3) into Argand plane representation.
a) (-4, 4\(\sqrt{3}\))
b) (4, 4\(\sqrt{3}\))
c) (4\(\sqrt{3}\), 4)
d) (-4\(\sqrt{3}\), 4)
View Answer

Answer: a
Explanation: To convert polar representation (r, θ) into argand plane (x, y), substitute x=r cos θ and y=r sin θ.
x=8cos 2π/3 = 8cos(π-π/3) = 8(-1/2) = -4.
y=8sin 2π/3 = 8sin(π-π/3) = 8(\(\sqrt{3}\)/2) = 4\(\sqrt{3}\).

12. 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/4
View Answer

Answer: b
Explanation: 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 2nd 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.

13. 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/4
View Answer

Answer: c
Explanation: 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 3rd quadrant since sin and cos both negative.
tan θ = 1 => θ= -3π/4.

Sanfoundry Global Education & Learning Series – Mathematics – Class 11.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter