Check whether limz→0zˉzlimz→0zzˉ exists in complex sense
A Equals 11
B Equals −1−1
C Does not exist
D Equals 00
Write z=reiθz=reiθ. Then zˉ/reiθ=re−iθ/reiθ=e−2iθzˉ/reiθ=re−iθ/reiθ=e−2iθ, which depends on θθ. Different approach directions give different values, so no limit.
Determine limz→0∣z∣zlimz→0z∣z∣ along different paths
A Does not exist
B Equals 11
C Equals 00
D Equals −1−1
Take z=rz=r (positive real): ∣z∣/z=1∣z∣/z=1. Take z=reiπ/2=irz=reiπ/2=ir: ∣z∣/z=r/(ir)=−i∣z∣/z=r/(ir)=−i. Different path values imply the complex limit does not exist.
Find limz→0zzˉzlimz→0zzzˉ if it exists
A 11
B 00
C Does not exist
D ∞∞
For z≠0z=0, zzˉz=zˉzzzˉ=zˉ. As z→0z→0, zˉ→0zˉ→0. Conjugation is continuous, so the limit exists and equals 00.
Evaluate limz→0sin(zˉ)zlimz→0zsin(zˉ) using series and paths
A Equals 11
B Equals 00
C Equals −1−1
D Does not exist
For small zz, sin(zˉ)∼zˉsin(zˉ)∼zˉ. So expression behaves like zˉ/z=e−2iθzˉ/z=e−2iθ, which depends on approach angle θθ. Hence the limit is path dependent.
Decide analyticity of f(z)=ℜ(z)f(z)=ℜ(z) on CC
A Nowhere analytic
B Entire function
C Analytic at 0
D Analytic on real
ℜ(z)=xℜ(z)=x gives u=x, v=0u=x, v=0. Then ux=1ux=1 but vy=0vy=0, so CR fails everywhere. Hence ff is not complex differentiable anywhere.
Decide analyticity of f(z)=ℑ(z)f(z)=ℑ(z) on CC
A Entire function
B Analytic at 0
C Nowhere analytic
D Analytic on axis
ℑ(z)=yℑ(z)=y corresponds to u=y, v=0u=y, v=0. Then ux=0ux=0 while vy=0vy=0 matches, but uy=1uy=1 and −vx=0−vx=0 fails. So CR fails everywhere.
Which function is analytic on C∖{0}C∖{0} but not entire
A 1/z1/z
B ezez
C sinzsinz
D z3z3
1/z1/z is a rational function, analytic wherever defined. It fails at z=0z=0 due to a pole, so it cannot be entire. Others listed are entire on all CC.
For f(z)=zˉ2f(z)=zˉ2, choose the correct statement
Determine whether f(z)=∣z∣2f(z)=∣z∣2 is analytic at any point
A Everywhere analytic
B Nowhere analytic
C Only at 00
D Only on unit circle
f=x2+y2f=x2+y2 gives u=x2+y2, v=0u=x2+y2, v=0. CR requires 2x=02x=0 and 2y=02y=0, so only at (0,0)(0,0). Hence differentiable only at 0.
If f(z)=∣z∣2f(z)=∣z∣2, then f′(0)f′(0) equals
A 00
B 11
C Does not exist
D ∞∞
f′(0)=limz→0∣z∣2−0z=limz→0zzˉz=limz→0zˉ=0f′(0)=limz→0z∣z∣2−0=limz→0zzzˉ=limz→0zˉ=0. The limit exists, so derivative at 0 is 0.
Compute limz→∞(z+1z−1)limz→∞(z−1z+1) in complex sense
A 00
B 11
C −1−1
D ∞∞
Divide numerator and denominator by zz: 1+1/z1−1/z→1+01−0=11−1/z1+1/z→1−01+0=1 as ∣z∣→∞∣z∣→∞. Limit exists in extended complex plane.
Evaluate limz→∞(z−z2+1)limz→∞(z−z2+1) using asymptotic idea
A −12z→0−2z1→0
B −∞−∞
C 00
D 11
Multiply by conjugate: z−z2+1=z2−(z2+1)z+z2+1=−1z+z2+1z−z2+1=z+z2+1z2−(z2+1)=z+z2+1−1. Denominator grows like 2z2z, so limit is 0.
Which mapping sends a horizontal strip 0<ℑz<2π0<ℑz<2π to
A Slit plane, no zero
B Punctured plane only
C Unit disk only
D Single circle only
In the strip, y∈(0,2π)y∈(0,2π) covers all arguments except 00 (positive real axis). Also ez≠0ez=0. Hence the image is C∖[0,∞)C∖[0,∞) (and 00 excluded automatically
Under w=ezw=ez, what is the image of the line ℑz=πℑz=π
A Positive real axis
B Unit circle
C Imaginary axis
D Negative real axis
If z=x+iπz=x+iπ, then w=ex(cosπ+isinπ)=ex(−1+0i)=−exw=ex(cosπ+isinπ)=ex(−1+0i)=−ex, which is the negative real axis (excluding 0), traced as xx varies.
For Möbius map w=1zw=z1, the line ℜz=1ℜz=1 maps to a
A Line through origin
B Circle through origin
C Circle not through
D Half-plane region
Möbius maps send lines/circles to lines/circles. A line not through the origin maps under inversion to a circle passing through the origin. Thus ℜz=1ℜz=1 becomes a circle through 0.
If ff is analytic and nonconstant on a region, then ∣f∣∣f∣ can have a local maximum only if
A Pole exists
B CR fails
C Constant function
D Branch cut
By the maximum modulus principle, a nonconstant analytic function cannot attain a strict local maximum of ∣f(z)∣∣f(z)∣ inside the region. If such a maximum occurs, the function must be constant.
If ff is analytic, zeros cannot accumulate unless
A f≡0f≡0
B ff is polynomial
C ff is bounded
D ff is periodic
Identity theorem: if an analytic function has zeros with a limit point inside its domain, it must be identically zero. Therefore nonzero analytic functions have isolated zeros only.
Check differentiability of f(z)=zˉf(z)=zˉ at z=0z=0 using quotient
A Equals 11
B Equals −1−1
C Does not exist
D Equals 00
Compute limz→0zˉ−0z=limz→0zˉ/z=e−2iθlimz→0zzˉ−0=limz→0zˉ/z=e−2iθ, dependent on approach angle. Since the limit depends on direction, derivative at 0 does not exist.
Using the principal branch where z2+1∼zz2+1∼z as ∣z∣→∞∣z∣→∞, evaluate limz→∞(z−z2+1)limz→∞(z−z2+1)
A 11
B −1−1
C ∞∞
D 00
Rationalize: z−z2+1=−1z+z2+1z−z2+1=z+z2+1−1. With the chosen branch, z2+1∼zz2+1∼z, so denominator ∼2z→∞∼2z→∞, hence the limit is 00.
If uu is harmonic on a simply connected region, then a harmonic conjugate vv exists
A Always
B Never
C Only at boundary
D Only at poles
In a simply connected domain, every harmonic function uu has a harmonic conjugate vv (up to a constant) such that u+ivu+iv becomes analytic. Simple connectivity prevents multi-valued issues.
Choose the correct polar form CR equations
A ur=rvθur=rvθ
B ur=1rvθur=r1vθ
C uθ=vruθ=vr
D ur=vrur=vr
In polar coordinates, CR becomes ur=1rvθur=r1vθ and vr=−1ruθvr=−r1uθ. These are used when u,vu,v are easier to express in r,θr,θ.
If f(z)=z3f(z)=z3, then f′(z)f′(z) in polar form changes argument by
A Adds 3θ3θ
B Halves θθ
C Adds 2θ2θ
D Keeps θθ
f′(z)=3z2f′(z)=3z2. If z=reiθz=reiθ, then z2=r2ei2θz2=r2ei2θ. Multiplying by 3 changes only modulus, not angle. Thus derivative direction involves angle 2θ2θ.
Evaluate limz→0z2zˉlimz→0zˉz2 in complex sense
A 00
B Does not exist
C 11
D ∞∞
Let z=reiθz=reiθ, zˉ=re−iθzˉ=re−iθ. Then z2zˉ=r2ei2θre−iθ=rei3θ→0zˉz2=re−iθr2ei2θ=rei3θ→0 as r→0r→0, independent of θθ.
Evaluate limz→0zˉ2zlimz→0zzˉ2 in complex sense
A Does not exist
B 00
C 11
D ∞∞
With z=reiθz=reiθ, zˉ2=r2e−i2θzˉ2=r2e−i2θ. Then zˉ2z=r2e−i2θreiθ=re−i3θ→0zzˉ2=reiθr2e−i2θ=re−i3θ→0 as r→0r→0, path independent.
Evaluate limz→0z∣z∣limz→0∣z∣z for complex approach
A Equals 11
B Equals 00
C Equals −1−1
D Does not exist
Write z=reiθz=reiθ. Then z/∣z∣=reiθ/r=eiθz/∣z∣=reiθ/r=eiθ, which depends on direction θθ. Different approach angles give different values, so no limit.
If f(z)=z−zˉzf(z)=zz−zˉ, the limit as z→0z→0 is
A 00
B 22
C Does not exist
D −2−2
z−zˉ=2iyz−zˉ=2iy. For z=reiθz=reiθ, expression becomes 1−zˉ/z=1−e−2iθ1−zˉ/z=1−e−2iθ, which depends on θθ. Hence limit varies with path and fails to exist.
Which statement about \Logz\Logz is correct for analyticity
A Analytic on slit
B Entire on CC
C Analytic at 0
D Single-valued always
A chosen principal branch of \Logz\Logz is analytic on a slit plane (commonly C∖(−∞,0]C∖(−∞,0]). It cannot be entire due to the branch point at 0 and argument multivaluedness.
For f(z)=\Logzf(z)=\Logz (principal branch), f′(z)f′(z) equals
A \Logz\Logz
B 1/z1/z
C zz
D ezez
On any branch where \Logz\Logz is analytic, its derivative is 1/z1/z. This follows from differentiating \Logz=ln∣z∣+i\Argz\Logz=ln∣z∣+i\Argz within the branch domain.
Find the residue of f(z)=1(z−a)2f(z)=(z−a)21 at z=az=a
A 11
B −1−1
C 00
D aa
Residue is the coefficient of (z−a)−1(z−a)−1 in Laurent expansion. Here f(z)=(z−a)−2f(z)=(z−a)−2 has no (z−a)−1(z−a)−1 term, so the residue is 0.
Residue of f(z)=1z−af(z)=z−a1 at aa equals
A 11
B 00
C −1−1
D aa
The Laurent expansion about aa is exactly (z−a)−1(z−a)−1. The coefficient of (z−a)−1(z−a)−1 is 1, so residue at z=az=a is 1. This is the simplest residue.
Residue of f(z)=z+1z−2f(z)=z−2z+1 at z=2z=2 equals
A 11
B 22
C −3−3
D 33
For a simple pole at z=2z=2, residue is limz→2(z−2)z+1z−2=z+1∣z=2=3limz→2(z−2)z−2z+1=z+1z=2=3. This is a standard simple-pole shortcut.
Find coefficient of (z−a)−1(z−a)−1 in 1z−a+2(z−a)2z−a1+(z−a)22
A 11
B 22
C 00
D aa
The Laurent form already shows terms. The coefficient of (z−a)−1(z−a)−1 is 1, while (z−a)−2(z−a)−2 does not contribute to residue. Hence residue is 1.
If a Laurent series has infinitely many negative powers, the singularity is
A Removable
B Simple pole
C Essential
D Double pole
Infinite negative-power terms in the principal part mean the function behaves extremely irregularly near the point. This exactly defines an essential singularity, unlike poles which have only finitely many negative terms.
Which function has an essential singularity at z=0z=0
A 1/z31/z3
B e1/ze1/z
C z2z2
D (z−1)−1(z−1)−1
Expanding e1/z=∑n=0∞1n!z−ne1/z=∑n=0∞n!1z−n gives infinitely many negative powers, so z=0z=0 is an essential singularity. 1/z31/z3 is a pole of order 3.
For f(z)=sinzzf(z)=zsinz, the singularity at z=0z=0 is
A Simple pole
B Essential
C Branch point
D Removable
sinz=z−z33!+⋯sinz=z−3!z3+⋯. Then sinzz=1−z23!+⋯zsinz=1−3!z2+⋯ has a finite limit 1 at 0. Defining f(0)=1f(0)=1 removes the hole.
For f(z)=1−coszz2f(z)=z21−cosz, the removable value at z=0z=0 is
A 1/21/2
B 11
C 00
D 22
Using cosz=1−z22+z424−⋯cosz=1−2z2+24z4−⋯, we get 1−cosz=z22−⋯1−cosz=2z2−⋯. Dividing by z2z2 gives limit 1/21/2. Define value to remove singularity.
If ff is analytic in a simply connected region, then ff has an antiderivative
A No, never
B Only if bounded
C Yes, always
D Only if periodic
In a simply connected domain, analyticity implies path-independent integrals. Hence there exists a primitive FF with F′=fF′=f. This is a major consequence of Cauchy’s theorem.
A key test for path independence of ∫f(z) dz∫f(z)dz in a region is that ff is
A Only continuous
B Analytic on region
C Only bounded
D Only real-valued
If ff is analytic on a simply connected region, contour integrals over closed loops are zero, implying path independence and existence of antiderivative. Continuity alone is not sufficient.
If f=u+ivf=u+iv is analytic, then level curves of uu and vv meet at
A Right angles
B 60° angles
C 45° angles
D Random angles
CR implies gradients of uu and vv are perpendicular. Therefore curves u=constantu=constant and v=constantv=constant intersect orthogonally wherever the gradients are nonzero. This supports conformal geometry.
Which statement about w=z2w=z2 near z=0z=0 is correct
A Conformal at 0
B Distance preserving
C Not conformal at 0
D Area preserving
A map is conformal where analytic and derivative is nonzero. For w=z2w=z2, derivative w′=2zw′=2z equals 0 at z=0z=0. Hence conformality fails at 0 though it is analytic.
If ff is analytic and nonconstant, then its image of an open set is
A Closed set
B Single point
C Line segment
D Open set
Open mapping theorem: a nonconstant analytic function maps open sets to open sets. This is a deep property that distinguishes complex analysis from real calculus, showing analytic maps cannot “collapse” interiors.
If ff is analytic and one-to-one, then f′f′ in that region is
A Always zero
B Never zero
C Sometimes infinite
D Always real
If ff is analytic and injective (conformal mapping), it cannot have a critical point f′(z)=0f′(z)=0 inside the region, because that would locally fold the map and destroy one-to-one behavior.
Consider f(z)=z2f(z)=z2. At any point z≠0z=0, the map is
A Conformal
B Not analytic
C Multivalued
D Discontinuous
z2z2 is analytic everywhere. For z≠0z=0, f′(z)=2z≠0f′(z)=2z=0, so it preserves angles locally and is conformal there. Only at z=0z=0 derivative vanishes.
If f(z)=1zf(z)=z1, then near z=0z=0 the function has
A Essential point
B Removable hole
C Pole order 1
D Branch point
1/z1/z behaves like (z−0)−1(z−0)−1, so the singularity at 0 is a simple pole. Its Laurent series has a single negative power term, not infinitely many.
If f(z)=e1/zf(z)=e1/z, then residue at z=0z=0 equals
A 11
B 00
C ee
D −1−1
Expand e1/z=∑n=0∞1n!z−ne1/z=∑n=0∞n!1z−n. The coefficient of z−1z−1 occurs at n=1n=1 and equals 1/1!=11/1!=1. Hence residue at 0 is 1.
If a function has a removable singularity at aa, then its Laurent series about aa has
A One negative power
B Two negative powers
C Infinite negative
D No negative powers
Removable means the function can be made analytic at aa. That happens when the Laurent expansion has no principal part (no negative powers). Then it is just a power series.
If limz→af(z)=∞limz→af(z)=∞, then aa is a
A Removable point
B Pole type point
C Regular point
D Zero point
If ∣f(z)∣→∞∣f(z)∣→∞ as z→az→a, the function has a pole-like singularity at aa. More precisely, it indicates a pole if growth is like finite negative powers in Laurent form.
For f(z)=1z2f(z)=z21, residue at z=0z=0 is
A 11
B −1−1
C 00
D 22
1/z21/z2 has Laurent term z−2z−2 only. Residue is coefficient of z−1z−1, which is absent. Therefore residue at 0 is 0, even though 0 is a pole.
A correct statement about \Argz\Argz is that it is
A Discontinuous on cut
B Entire everywhere
C Analytic at 0
D Constant on circles
Any single-valued choice of argument needs a branch cut where the angle jumps by 2π2π. So \Argz\Argz cannot be continuous everywhere on C∖{0}C∖{0} without excluding a cut.
If ff is analytic and ∣f(z)∣∣f(z)∣ is constant on a region, then ff must be
A Linear function
B Quadratic function
C Constant function
D Periodic function
If an analytic function has constant modulus on a connected open region, it must be constant. This follows from the maximum modulus principle and the open mapping theorem: a nonconstant analytic map cannot keep modulus fixed.