Complex Analysis Questions and Answers – Regions in the Complex Plane

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This set of Complex Analysis Multiple Choice Questions & Answers (MCQs) focuses on “Regions in the Complex Plane”.

1. What is the shape of the region formed by the set of complex numbers z satisfying |z-ω|≤ α?
a) circle of radius ω
b) circle with center ω
c) disk of radius α
d) disk with center α
View Answer

Answer: d
Explanation: The equation |z-ω|≤ α implies that the distance of z from ω is less than or equal to α. This means that a disk is formed with center ω and radius α.
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2. The complex number given by [(√3/2)+i/2]5+[(√3/2)-i/2]5 lies, on which of the following regions?
a) imaginary axis
b) real axis
c) first quadrant
d) fourth quadrant
View Answer

Answer: b
Explanation: Let z=[(√3/2)+i/2]5+[(√3/2)-i/2]5 and ω=[(√3/2)+i/2]5 ⇒ \(\overline{\omega}\)=[(√3/2)-i/2]5
⇒ z=ω+\(\overline{\omega}\) ⇒ z is real (sum of conjugates is real) ⇒ z lies on real axis.

3. Find the area of the region given by 11≤|z| ≤ 19.
a) 120π sq. units
b) 180π sq. units
c) 240π sq. units
d) 320π sq. units
View Answer

Answer: c
Explanation: The region formed is an annulus of inner radius 11 units and outer radius 19 units. Therefore, the required area=π(192–112)=240π.

4. Find the largest angle of the triangle formed by thevertices z1=8(1-i), z2=8(i-1) and
Z3=10+2√7i.

a) π/3 radians
b) 2π/3 radians
c) π/2 radians
d) 3π/4 radians
View Answer

Answer: c
Explanation: Note that z1 and z2 are the opposite ends of a diameter of a circle of radius 8√2 units, centered at the origin. Also note that z3 lies on this circle (distance of z3 from origin = 8√2). Hence, angle corresponding to z3=π/2 radians.

5. Find the equation of the circle passing through the origin and having intercepts a and b on real and imaginary axes, respectively, on the arg and plane.

a) zz̅=a(Im z)–b(Re z)
b) zz̅=a(Im z)+b(Re z)
c) zz̅=a(Re z)–b(Im z)
d) zz̅=a(Re z)+b(Im z)
View Answer

Answer: d
Explanation: Consider a point z on the circle. Therefore, arg [(z-a)/(z-ib)]=±π/2
⇒ (z-a)/(z-ib)+(z̅-a)/(z̅+ib)=0 ⇒ zz̅-a(z+z̅)/2–b(z-z̅)/2i=0
⇒ zz̅=a(Re z)+b(Im z).
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6. On the arg and plane, the complex numbers z1, z2, z3, z4 are the vertices of a parallelogram. Evaluate (z4–z1+z2)/z3 .
a) 1
b) 2
c) 3
d) 4
View Answer

Answer: a
Explanation: For a parallelogram, the diagonals bisect each other. Hence, their midpoints coincide.
This implies that (z1+z3)/2=(z2+z4)/2 ⇒ z1+z3=z2+ z4 ⇒ ( z4– z1+ z2)/ z3=1.

7. Consider the shape formed by the set of points z=ω-1/ω, where |ω|=2. Which of the following is incorrect?
a) eccentricity=4/5
b) |z|≤3
c) shape is an ellipse
d) major axis is of length=5/2
View Answer

Answer: d
Explanation: | ω|=2 ⇒ ω=2(cosθ+isinθ) ⇒ z=x+iy=2(cosθ+isinθ)-1/2(cosθ-isinθ)(⇒ |z|≤3)
=3/2cosθ+i5/2sinθ⇒x2/(3/2)2+y2/(5/2)2=1 ⇒ ellipse ⇒ e2=1-(9/4)/(25/4)=16/25
⇒ e=4/5.

8. Find the area enclosed by the curve formed by iz3+z2–z+i=0.
a) π/2
b) π
c) 3π/4
d) 2π
View Answer

Answer: b
Explanation: Dividing the equation by i on both sides, z3-iz2+iz+1=0
⇒ z2(z-i)+i(z-i)=0 ⇒ (z-i)(z2+i)=0 ⇒ z=i or z2=-i ⇒ |z|=|i|=1 or |z2|=|z|2=|-i|=1
⇒ |z|=1 ⇒ circle of radius 1 is formed. Hence, area=π(12)=π.

9. Given a vertex of the square circumscribing the circle |z-1|=√2 as 2+√3i, which of the following is not a vertex of this square ?
a) (1-√3)+i
b) –i√3
c) (√3+i)-i
d) i√3
View Answer

Answer: d
Explanation: The given circle has z0=1 as its center and √2 as radius. Let z1=2+i√3. Now, obtain z2 by rotating z1 anticlockwise by 900 about z0 ⇒ z2=(1-√3)+i. Now, z0 is midpoint of z1 and z3 and z2 and z4.
؞(z1+z3)/2 ⇒ (2+i√3+z3)/2=1 ⇒ z3=-i√3 and(z2+z4)/2=z0 ⇒ z4=(√3+i)-i.
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10. Find the area of the region bounded by arg|z|≤π/4 and |z-1|<|z-3|.
a) 1 sq. units
b) 2 sq. units
c) 3 sq. units
d) 4 sq. units
View Answer

Answer: d
Explanation: |z-1|<|z-3| ⇒ (x-1)2+y2<(x-3)2+y2 ⇒ x<2.
Therefore, a triangle is formed with base length 4 and height 2 (along x-axis). Hence, the required area=1/2×4×2=4.

11. Find the locus of z/(1-z2), where z lies on the circle of radius 1 centered at origin and z≠±1.
a) line not passing through origin
b) |z|=√2
c) real axis
d) imaginary axis
View Answer

Answer: d
Explanation: Given |z|=1 and z≠±1, write ω=z/(1-z2)=z/(zz̅-z2)=1/(z̅-z).
Hence ω is a purely imaginary number and lies on imaginary axis.

12. Describe the region given by |z-i|z||-|z+i|z||=0.
a) real axis
b) imaginary axis
c) circle centered at origin
d) quadrant 2
View Answer

Answer: a
Explanation: |z/|z|-i|=|z/|z|+i|, z≠0 ⇒ z/|z| is unimodular complex number and lies on the perpendicular bisector of i and –i ⇒ z/|z|=±1 ⇒ z=±|z| ⇒ z is real.

13. The area of the region enclosed by the curve zz̅+a(z̅+z)+a=0 is 2π. If a2–7a+10=0, find the area of the region enclosed by the curve zz̅+2a(z̅+z)+a=0.
a) 4π sq. units
b) 10π sq. units
c) 14π sq. units
d) 22π sq. units
View Answer

Answer: c
Explanation: The curve represents a circle with center–a and radius (a2–a)1/2. Therefore, Area=π(a2–a)=2π ⇒ a=-1,2. Also, a2–7a+10=0 ⇒ a=2,5. Hence, a=2.
Hence required area=π(4a2–a)=14π.
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14. Find the area of the region common to the sets S1={z∈C: |z|<4}, S2={z∈C:Im[(z-1+√3i)/(1-√3i)]>0} and S3={z∈C: Re z>0}.
a) 10π/3
b) 20π/3
c) 16π/3
d) 32π/3
View Answer

Answer: b
Explanation: S1: |z|<4, z lies inside the circle of radius 4; S2: √3x+y>0, z lies above the line √3x+y=0; S3: Re(z)>0, z lies to the right of the imaginary circle. Therefore, required area=π×42/4+π×42/6=20π/3.

15. In the triangle shown, if the angle corresponding to z3 is said to be π/2,

Find a possible value of z3 in terms of z1, z2 and z4.
a) z4+3/5(z2– z1)eiπ/2
b) z4-3/5(z2– z1)eiπ/2
c) z4+3/5(z2– z1)e-iπ/2
d) no such z3 is possible
View Answer

Answer: d
Explanation: If we draw a circle with z2-z1 as a diameter, we see that z3 would be outside the circle, since the radius of the circle is 5(<6). But we know that the points on the circle only would result in an angle of π/2. Hence, no such z3 is possible.

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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