Complex Analysis Questions and Answers – Functions of a Complex Variable

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This set of Complex Analysis Multiple Choice Questions & Answers (MCQs) focuses on “Functions of a Complex Variable”.

1. Find the domain of the function defined by f(z)=z/(z+z̅).
a) Im(z)≠0
b) Re(z)≠0
c) Im(z)=0
d) Re(z)=0
View Answer

Answer: b
Explanation: Write z=x+iy ⇒ f(x+iy)=(x+iy)/(x+iy+x-iy)=(x+iy)/2x
=1/2+iy/2x ⇒ x≠0 ⇒ Re(z)≠0 .
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2. Let f(z)=z+1/z. What will be the definition of this function in polar form?
a) (r+1/r)cosθ+i(r-1/r)sinθ
b) (r-1/r)cosθ+i(r+1/r)sinθ
c) (r+1/r)sinθ+i(r-1/r)cosθ
d) (r+1/r)sinθ+i(r-1/r)cosθ
View Answer

Answer: a
Explanation: Write z=r(cosθ+isinθ), therefore, f(z)=z+1/z=r(cosθ+isinθ)+1/[r(cosθ+isinθ)]
=re+(1/r)e-iθ=r(cosθ+isinθ)+1/r(cosθ-isinθ)=(r+1/r)cosθ+i(r-1/r)sinθ.

3. For the function f(z)=zi, what is the value of |f(ω)|+Arg f(ω), ω being the cube root of unity with Im(ω)>0?
a) e-2π/3
b) e2π/3
c) e-2π/3+2π/3
d) e-2π/3-2π/3
View Answer

Answer: a
Explanation: Let y=zi⇒ ln y=iln z=i(ln |z|+iarg z)=iln |z|-arg z
⇒ y=eiln |z|/earg z ⇒ |y|=earg z and Arg y=ln |z| ⇒ |f(ω)|+Arg f(ω)=e-2π/3+0=e-2π/3.

4. Let f(z)=(z2–z–1)7. If α2+α+1=0 and Im(α)>0, then find f(α).
a) 128α
b) -128α
c) 128α2
d) -128α2
View Answer

Answer: c
Explanation: Note that α=ω. Therefore, f(α)=f(ω)=(ω2–ω–1)7
=(ω22)7=(2ω2)7=27ω14=128ω2=128α2.

5. For all complex numbers z satisfying Im(z)≠0, if f(z)=z2+z+1 is a real valued function, then find its range.
a) (-∞, -1]
b) (-∞, 1/3)
c) (-∞, 1/2]
d) (-∞, 3/4)
View Answer

Answer: d
Explanation: Let y=f(z). then z2+z+1=y has imaginary roots (∵Im(z)≠0)
⇒ D<0 ⇒ 1–4(1–y)<0 ⇒ 4y<3 ⇒ y<3/4 . Also, putting Re z=-1/2 and Im z=∞, we get, f(z)=-∞.
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6. Let x, y, z be integers, not all simultaneously equal. If ω is a cube root of unity with Im(ω)≠1, and if f(z)=az2+bz+c, then find the range of |f(ω)|.
a) (0, ∞)
b) [1, ∞)
c) (√3/2, ∞)
d) [1/2, ∞)
View Answer

Answer: b
Explanation: ω=-1/2+i√3/2. Therefore, |f(ω)|=|a+b(-1/2+i√3/2)+c(-1/2-i√3/2)|
=|(2a-b-c)/2+i(b√3-c√3)/2|=1/2[(2a-b-c)2+3(b-c)2]1/2={1/2[(a-b)2+(b-c)2+(c-a)2]}1/2. Putting b=c=0 and a=1 gives us the minimum value=1, while, a=∞ gives us the maximum value=∞.

7. Let f(z)=arg 1/(1 – z), then find the range of f(z) for |z|=1, z≠1.
a) (-∞, π/2)
b) (-π/2, π/2)
c) (-∞, ∞)
d) [0, π/2)
View Answer

Answer: b
Explanation: Let y=1/(1-z) ⇒ z=1-1/y
|z|=1 ⇒ |1-1/y|=1 ⇒ |y-1|=|y| ⇒ locus of y is the perpendicular bisector of line segment joining 0 and 1 ⇒ arg y ∈(-π/2, π/2).

8. Define f(z)=z2+bz−1=0 and g(z)=z2+z+b=0. If there exists α satisfying f(α)=g(α)=0, which of the following cannot be a value of b?
a) √3i
b) -√3i
c) 0
d) √3i/2
View Answer

Answer: d
Explanation: α2+bα−1=0 and α2+α+b=0 ⇒ (b−1)α−1−b=0 ⇒ α=(b+1)/(b-1)
⇒ (b+1)2/(b-1)2+(b+1)/(b-1)+b=0 ⇒ b=√3i, -√3i, 0.

9. Let f(z)=2(z+z̅)+3i(z-z̅) and g(z)=|z|. f(z)=2 divides the region g(z)≤6 into two parts. If Q={(2+3i/4), (5/2+3i/4), (1/4-i/4), (1/8+i/4)}, then find the number of elements of Q lying inside the smaller part.

a) 1
b) 2
c) 3
d) 4
View Answer

Answer: b
Explanation: Writing z=x+iy, we get L as 2x–3y–1 and S as x2+y2–6, a point z1 lies in the smaller region if L1>0 and S1<0. ∴ (2+3i/4) and (1/4-i/4) lie in the smaller region.
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10. Find the range of the function defined by f(z)=Re[2iz/(1-z2)].
a) (−∞, 0) ⋃ (0, ∞)
b) [2, ∞)
c) (−∞, −1] ⋃ [1, ∞)
d) (−∞, 0] ⋃ [2, ∞)
View Answer

Answer: c
Explanation: z=2i(x+iy)/(1-(x+iy)2)=2i(x+iy)/(1-(x2-y2+2ixy))
Using 1-x2=y2, z=(2ix-2y)/(2y2-2ixy)=-1/y
∵ –1≤y≤1 ⇒ –1/y≤-1 or -1/y≥1.

11. Let f(z)=|z|2+Re z(2(z+z̅)+3(z-z̅)/2i, the find the maximum value of |z|2/f(z).
a) 1
b) 2
c) 3
d) 4
View Answer

Answer: b
Explanation: Write z=|z|(cosθ+isinθ) ⇒ |z|2/f(z)=1/(1+4cos2θ+3sinθcosθ)
=1/(1+4cos2θ+3/2sin2θ)=1/[2(1+cos2θ)+1+3/2sin2θ].
Now, 2(1+cos2θ)+1+3/2sin2θ=3+2cos2θ+3/2sin2θ≥3-(4+9/4)1/2=1/2.
Hence, maximum value is 2.

12. Consider a function f(z) of degree two, having real coefficients. If z1 and z2 satisfying f(z1)=f(z2)=0 are such that Re z1=Re z2=0 and if z3 satisfies f(f(z3))=0, then select the correct statement.
a) Re z3=0
b) Im z3=0
c) Re z3×Imz3≠0
d) Re z3=0 and Im z3=0
View Answer

Answer: c
Explanation: f(z)=az2+b, with a, b of same sign ⇒ f(f(z))=a(az2+b)2+b
If z∈R or iz∈R ⇒ z2∈R ⇒ f(z)∈R ⇒ f(f(z))≠0 ⇒ Hence real or purely imaginary number cannot satisfy f(f(z))=0.

13. Let f(z)=|1–z|, if zk=cos(2kπ/10)+isin(2kπ/10), then find the value of f(z1)×f(z2)×…×f(z9).
a) 10
b) 15
c) 20
d) 30
View Answer

Answer: a
Explanation: z10–1=(z-1)(z-z1)…(z-z9) ⇒ (z-z1)(z-z2) …(z-z9)=1+z+z2+…+z9.
Now, putting z=1, we get, (z-z1)(z-z2)…(z-z9)=f(z1)×f(z2)×…×f(z9)=10.
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14. For a∈R, let f(z)=z5-5z+a. Select the correct statement for α satisfying f(α)=0.
a) α has exactly three possible real values for a>4
b) α has exactly one possible real value for a>4
c) α has exactly three possible real values for a<-4
d) α has exactly one possible real value for -4<a<4
View Answer

Answer: b
Explanation: z5-5z+a=0 ⇒ z5-5z=-a ⇒ z(z-51/4)(z+51/4)(z2+51/2)=-a f'(z)=5z4–5=0 ⇒ (z2+1)(z2-1)=0 ⇒ (z-1)(z+1)(z2+1)=0 ⇒ α has exactly one possible real value for a>4 and exactly three possible real values for -4<a<4.

15. Let f(z)=z4+a1z3+a2z2+a3z+a4=0; a1, a2, a3, a4 being real and non-zero. If f has a purely imaginary root, then what is the value of the expression a3/(a1a2)+ a1a4/(a2a3) ?
a) 0
b) 1
c) -2
d) 2
View Answer

Answer: b
Explanation: For real x(≠0), let ix be the root⇒x4-a1x3i- a2x2+a3xi+a4=0⇒x4-a2x2+a4=0 and a1x3-a3x=0
a1x3-a3x=0 ⇒ a1x2-a3=0 ⇒x2=a3/a1, putting this value in the equation, a3/(a1a2)+a1a4/(a2a3)=1.

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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