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

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.

2. Which axis is known as imaginary axis in argand plane?
a) x-axis
b) y-axis
c) z-axis
d) any axis

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

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

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

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)

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)

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.

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

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

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.

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

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)

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

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

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.

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