Evaluate limz→0ez−1zlimz→0zez−1 using series idea
A 00
B ee
C 11
D Does not exist
Using ez=1+z+z22!+⋯ez=1+z+2!z2+⋯, we get ez−1z=1+z2!+⋯→1zez−1=1+2!z+⋯→1 as z→0z→0. Path does not matter.
Find limz→0sinzzlimz→0zsinz in complex sense
A 11
B 00
C −1−1
D Undefined
The power series sinz=z−z33!+⋯sinz=z−3!z3+⋯ gives sinzz=1−z23!+⋯→1zsinz=1−3!z2+⋯→1 as z→0z→0, same as real case.
Compute limz→01−coszz2limz→0z21−cosz using series
A 11
B 00
C 22
D 1/21/2
Use cosz=1−z22!+z44!−⋯cosz=1−2!z2+4!z4−⋯. Then 1−cosz=z22−⋯1−cosz=2z2−⋯, so 1−coszz2→12z21−cosz→21.
If f(z)=z2+zzf(z)=zz2+z for z≠0z=0, then limz→0f(z)limz→0f(z) equals
A 00
B 11
C ∞∞
D Does not exist
For z≠0z=0, f(z)=z+1f(z)=z+1. Hence as z→0z→0, f(z)→1f(z)→1. This shows the original expression has a removable singularity at z=0z=0.
A limit at infinity of f(z)f(z) is defined as LL when
A f(w)→Lf(w)→L only
B f(zˉ)→Lf(zˉ)→L
C f(1/w)→Lf(1/w)→L
D f(z)→0f(z)→0 always
limz→∞f(z)=Llimz→∞f(z)=L means f(z)f(z) approaches LL as ∣z∣→∞∣z∣→∞. Setting w=1/zw=1/z converts this to limw→0f(1/w)=Llimw→0f(1/w)=L.
Evaluate limz→∞z3+22z3−1limz→∞2z3−1z3+2
A 1/21/2
B 22
C 00
D ∞∞
Divide by z3z3: 1+2/z32−1/z3→122−1/z31+2/z3→21 as ∣z∣→∞∣z∣→∞. Same degree terms dominate, giving ratio of leading coefficients.
Which statement is true about continuity in CC
A Needs CR equations
B Only on real axis
C Only for polynomials
D Uses same limit idea
Continuity at aa in CC means limz→af(z)=f(a)limz→af(z)=f(a). It’s the same concept as real continuity, but the approach to aa can be from any direction.
If ff is analytic in a region, then it is
A Differentiable once only
B Not differentiable
C Infinitely differentiable
D Discontinuous
Analytic (holomorphic) functions are very smooth: once complex differentiable in an open region, they have derivatives of all orders there and can be represented locally by power series.
Which function is analytic on C∖{0}C∖{0}
A zˉzˉ
B 1/z1/z
C ∣z∣∣z∣
D argzargz
1/z1/z is analytic everywhere except at z=0z=0, where it has a pole. The conjugate and modulus are not analytic due to CR failure, and argzargz is not analytic.
For f(z)=1z−af(z)=z−a1, the type of singularity at z=az=a is
A Simple pole
B Removable
C Essential
D Branch point
Near z=az=a, f(z)f(z) behaves like 1/(z−a)1/(z−a) and blows up in magnitude. This is exactly a pole of order 1, called a simple pole.
For f(z)=1(z−a)3f(z)=(z−a)31, the singularity at z=az=a is
A Simple pole
B Removable
C Pole order 3
D Essential
The function has denominator (z−a)3(z−a)3, so it diverges like 1/∣z−a∣31/∣z−a∣3. That makes it a pole of order 3 at z=az=a, a higher-order pole.
A function is conformal at a point if it is analytic there and
A Derivative equals zero
B Modulus equals one
C Argument equals zero
D Derivative nonzero
Conformality means angle preservation locally. An analytic function preserves angles at points where f′(z)≠0f′(z)=0. If the derivative is zero, angles may not be preserved.
Under w=z2w=z2, the right half-plane maps mainly to a
A Plane slit line
B Unit circle only
C Single point only
D Horizontal strip
Squaring doubles arguments. The right half-plane has arguments between −π/2−π/2 and π/2π/2. Doubling gives −π−π to ππ, covering the plane except typically the nonpositive real axis boundary behavior.
Which mapping sends rays from origin to rays from origin
A w=z+aw=z+a
B w=z+1w=z+1
C w=znw=zn
D w=zˉ+1w=zˉ+1
If z=reiθz=reiθ, then zn=rneinθzn=rneinθ. A ray θ=θ= constant maps to a ray with angle multiplied by nn. Translations generally do not keep rays from the origin.
For the mapping w=1/zw=1/z, which set maps to itself
A Real axis only
B Unit circle
C Upper half-plane
D Disk ∣z∣<1∣z∣<1
If ∣z∣=1∣z∣=1, then ∣w∣=∣1/z∣=1∣w∣=∣1/z∣=1. So the unit circle maps onto itself under inversion, though points are reversed in angle (reflection in unit circle sense).
If f(z)=u+ivf(z)=u+iv** is analytic, then CR implies**
A u=vu=v always
B ux=vxux=vx
C uy=vyuy=vy
D ux2+uy2=vx2+vy2ux2+uy2=vx2+vy2
From CR, ux=vyux=vy and uy=−vxuy=−vx. Squaring and adding gives ux2+uy2=vy2+vx2ux2+uy2=vy2+vx2. This reflects equal scaling in orthogonal directions for conformal maps.
Which condition makes CR equations sufficient for analyticity at a point
A Only continuity of ff
B Only boundedness
C Partial derivatives continuous
D Only existence of limit
CR equations alone are necessary, but for a common sufficient test we assume ux,uy,vx,vyux,uy,vx,vy exist and are continuous in a neighborhood. Then CR implies complex differentiability.
If f(z)=u+ivf(z)=u+iv** is analytic, then uu and vv are related by**
A Harmonic conjugates
B Constant multiples
C Always equal
D Always zero
In an analytic function, uu and vv satisfy CR and both are harmonic. Each can be viewed as a harmonic conjugate of the other (locally), meaning together they form u+ivu+iv.
If u=x2−y2u=x2−y2, a harmonic conjugate vv that makes u+ivu+iv analytic is
A x2+y2x2+y2
B 2xy2xy
C −2xy−2xy
D x−yx−y
For u=x2−y2u=x2−y2, ux=2xux=2x and uy=−2yuy=−2y. CR requires vy=ux=2xvy=ux=2x and vx=−uy=2yvx=−uy=2y. Integrating gives v=2xyv=2xy (plus constant).
If f(z)=z2f(z)=z2, then in terms of x,yx,y, the imaginary part is
A x2+y2x2+y2
B 2xy2xy
C x2−y2x2−y2
D −2xy−2xy
z2=(x+iy)2=x2−y2+i(2xy)z2=(x+iy)2=x2−y2+i(2xy). Thus u=x2−y2u=x2−y2 and v=2xyv=2xy. This classic example satisfies CR everywhere, showing analyticity.
Evaluate the derivative ddz(z−2)dzd(z−2) for z≠0z=0
A −2z−3−2z−3
B −z−1−z−1
C 2z−12z−1
D z−3z−3
Treat z−2z−2 like a power: ddzzn=nzn−1dzdzn=nzn−1 for integer nn. With n=−2n=−2, derivative is −2z−3−2z−3, valid for z≠0z=0.
For f(z)=zz2+1f(z)=z2+1z, where is ff analytic
A Only at z=0z=0
B Only on real axis
C Everywhere
D Except z=±iz=±i
A rational function is analytic wherever its denominator is nonzero. Here z2+1=0z2+1=0 gives z=±iz=±i. So ff is analytic on C∖{±i}C∖{±i}.
If f(z)=logzf(z)=logz, then moving once around the origin changes the value by
A πiπi
B 2π2π
C 2πi2πi
D 00
logz=ln∣z∣+i(\Argz+2πk)logz=ln∣z∣+i(\Argz+2πk). A full turn increases the argument by 2π2π, so the logarithm increases by i(2π)=2πii(2π)=2πi. This shows multivaluedness.
Principal value of \Logz\Logz** is analytic on CC excluding the**
A Positive real axis
B Nonpositive real axis
C Imaginary axis
D Unit circle
The usual principal branch cut for \Logz\Logz is taken along the negative real axis (often including 0), preventing argument jumps across the cut. So it’s analytic on the slit plane.
If f(z)=ezf(z)=ez, what is the image of the vertical line ℜ(z)=cℜ(z)=c
A Circle radius ecec
B Ray angle cc
C Horizontal line
D Unit disk
For z=c+iyz=c+iy, ez=ec(cosy+isiny)ez=ec(cosy+isiny). The modulus is constant ecec while yy varies, creating a circle centered at the origin of radius ecec.
For z=x+iyz=x+iy, the mapping w=ezw=ez** has modulus equal to**
A eyey
B x2+y2x2+y2
C exex
D ∣y∣∣y∣
ex+iy=ex(cosy+isiny)ex+iy=ex(cosy+isiny). The modulus is ∣ez∣=ex∣ez∣=ex, since cos2y+sin2y=1cos2y+sin2y=1. So real part controls magnitude.
Which identity is correct for complex exponential
A ez+π=ezez+π=ez
B ez+i=ezez+i=ez
C ez−1=ezez−1=ez
D ez+2πi=ezez+2πi=ez
Since e2πi=1e2πi=1, we have ez+2πi=eze2πi=ezez+2πi=eze2πi=ez. This shows periodicity in the imaginary direction with period 2π2π.
Solve ez=iez=i** in general form**
A 2πk2πk
B i(π/2+2πk)i(π/2+2πk)
C πkπk
D 1+2πik1+2πik
i=ei(π/2+2πk)i=ei(π/2+2πk). For ex+iy=iex+iy=i, we need x=0x=0 and y=π/2+2πky=π/2+2πk. So z=i(π/2+2πk)z=i(π/2+2πk).
Which statement about analytic functions is correct
A Zeros are isolated
B Zeros form area
C Zeros always finite
D Zeros are poles
A nonzero analytic function cannot have a cluster of zeros inside its domain. If zeros accumulate in the region, the function must be identically zero. Hence ordinary zeros occur as isolated points.
If f(z)=znf(z)=zn, then f′(z)f′(z) equals
A zn+1zn+1
B nznz
C nzn−1nzn−1
D zn/nzn/n
The power rule holds for complex polynomials. Using the difference quotient and algebra, ddz(zn)=nzn−1dzd(zn)=nzn−1 for integer n≥1n≥1. It matches real differentiation.
A function is holomorphic on a region if it is
A Only continuous
B Only bounded
C Only real-valued
D Complex differentiable
Holomorphic means complex differentiable at every point of an open region. This is equivalent to being analytic there. It is a stronger condition than mere continuity or differentiability in real sense.
If ff is analytic on a region, then it can be represented locally by
A Power series
B Fourier series only
C Finite polynomial only
D Piecewise lines
Analytic functions are locally equal to a convergent power series around each point in their domain. This powerful property explains why analytic functions are so smooth and rigid.
The harmonic conjugate of uu is unique up to a
A Variable factor
B Square term
C Constant
D Conjugation
If vv is a harmonic conjugate of uu, then v+Cv+C also works because adding a constant does not change derivatives. Thus harmonic conjugates are unique up to an additive constant.
For Möbius maps, a point where cz+d=0cz+d=0 is mapped to
A Zero
B Infinity
C One
D Same point
For w=az+bcz+dw=cz+daz+b, if cz+d=0cz+d=0, then the denominator vanishes and ∣w∣∣w∣ becomes unbounded. In the extended complex plane, that input point maps to infinity.
Which mapping property is true for all Möbius transformations
A Preserve all distances
B Preserve all areas
C Preserve all lengths
D Preserve generalized circles
Möbius transformations map circles and straight lines (together called generalized circles) to circles or lines. This geometric property is a core reason they are widely used in complex mapping.
In a Laurent series, a removable singularity occurs when
A Infinite negative terms
B One negative term
C No negative terms
D Two negative terms
If the Laurent expansion around aa has no negative powers, then it reduces to a regular power series. That means the function can be defined analytically at aa, so the singularity is removable.
In Laurent form, a simple pole occurs when the principal part has
A Only (z−a)−1(z−a)−1
B Only (z−a)−2(z−a)−2
C No negative term
D Infinite negative terms
A simple pole means the Laurent series has exactly one negative-power term: c−1(z−a)−1c−1(z−a)−1. Higher-order poles include additional negative powers like (z−a)−2(z−a)−2, etc.
Residue at aa in Laurent expansion equals
A Constant coefficient
B Coefficient of (z−a)−1(z−a)−1
C Coefficient of (z−a)(z−a)
D Sum of all terms
The residue is defined as the coefficient c−1c−1 of (z−a)−1(z−a)−1 in the Laurent series. It directly determines contour integrals through the residue theorem.
If a function is analytic inside and on a closed contour, then by Cauchy theorem the integral is
A 2πi2πi
B 11
C 00
D ∞∞
Cauchy’s integral theorem states that for an analytic function on and inside a simple closed contour in a simply connected region, the contour integral around that loop is zero.
A Cauchy sequence idea in CC means
A zn→∞zn→∞
B ∣zn∣→1∣zn∣→1
C ∣zn−zm∣→0∣zn−zm∣→0
D argzn→0argzn→0
A sequence (zn)(zn) is Cauchy if for every ε>0ε>0, there exists NN such that ∣zn−zm∣<ε∣zn−zm∣<ε for all m,n>Nm,n>N. It measures convergence behavior.
If ff is analytic and f′(a)=0f′(a)=0, then conformality at aa is
A Not guaranteed
B Always guaranteed
C Always impossible
D Same as translation
Conformal mapping requires analytic plus nonzero derivative. If f′(a)=0f′(a)=0, the mapping can flatten angles or change local geometry, so angle preservation is not ensured at that point.
Which equality holds for any complex numbers z,wz,w
A ∣z+w∣=∣z∣+∣w∣∣z+w∣=∣z∣+∣w∣
B ∣zw∣=∣z∣∣w∣∣zw∣=∣z∣∣w∣
C ∣z∣=z+zˉ∣z∣=z+zˉ
D zˉ=zzˉ=z
Modulus is multiplicative: ∣zw∣=∣z∣∣w∣∣zw∣=∣z∣∣w∣. This follows from zwzw‾=(zzˉ)(wwˉ)zwzw=(zzˉ)(wwˉ). Triangle equality ∣z+w∣=∣z∣+∣w∣∣z+w∣=∣z∣+∣w∣ is not always true.
Triangle inequality in complex numbers states
A ∣z+w∣=∣z∣∣w∣∣z+w∣=∣z∣∣w∣
B ∣z+w∣≤∣z∣+∣w∣∣z+w∣≤∣z∣+∣w∣
C ∣z+w∣≥∣z∣+∣w∣∣z+w∣≥∣z∣+∣w∣
D ∣z+w∣=∣z∣+∣w∣∣z+w∣=∣z∣+∣w∣
The triangle inequality is a key tool in limit proofs. It states the magnitude of a sum is at most the sum of magnitudes, helping estimate ∣f(z)−L∣∣f(z)−L∣ and other expressions.
If ∣z∣<1∣z∣<1, then the mapping w=1/zw=1/z** sends the point to**
A Outside unit circle
B Inside unit circle
C On unit circle
D At origin
If ∣z∣<1∣z∣<1, then ∣w∣=1/∣z∣>1∣w∣=1/∣z∣>1. So points inside the unit circle map outside it under inversion. The unit circle itself remains invariant.
The mapping w=z2w=z2** sends a line through origin into a**
A Circle through origin
B Disk region
C Vertical strip
D Line through origin
A line through the origin corresponds to a fixed argument θθ (a ray in polar form). Under w=z2w=z2, the argument doubles but remains constant, so it stays a line/ray through the origin.
If ff is analytic, then f′f′ is also
A Discontinuous
B Analytic
C Multivalued
D Undefined
In complex analysis, analyticity is preserved under differentiation. If ff is analytic in a region, then its derivative f′f′ is also analytic in that region, and higher derivatives exist too.
For f(z)=z−1z−1f(z)=z−1z−1** for z≠1z=1, choosing f(1)=1f(1)=1 makes ff**
A Discontinuous everywhere
B Multivalued
C Continuous everywhere
D A pole at 1
For z≠1z=1, f(z)=1f(z)=1. The limit as z→1z→1 is 1. Defining f(1)=1f(1)=1 removes the hole and makes the function continuous everywhere.
A function defined by a power series has radius of convergence that forms a
A Line segment
B Disk region
C Half-plane only
D Square region
For a power series ∑an(z−a)n∑an(z−a)n, convergence depends on ∣z−a∣∣z−a∣. Hence it converges inside a circle ∣z−a∣
Which function is analytic except for a branch issue
A \Logz\Logz
B z4z4
C ezez
D coszcosz
\Logz\Logz is analytic on its chosen branch domain (a slit plane) but cannot be analytic on all of CC due to multivalued argument behavior. Others listed are entire.
The main reason complex limits are harder than real limits is that z→az→a can happen through
A Two directions only
B Infinitely many paths
C One direction only
D Closed loops only
In real limits, approach is from left or right. In complex limits, zz can approach aa from infinitely many directions and curves. The limit exists only if all give the same value.