Instructions: click "edit" in the upper right corner and type in your question after the instructions but before all other questions. If you want to write a math formula, look at how other formulas have been typed on this wiki or type it in plain text. I will answer the question as soon as I can

If I have skipped your question below, or if something is unclear, or if I accidentally deleted it, please leave a comment about it (I was answering the questions today in a hurry and concurrently with some other edits)

Q : Give an example of a non-constant analytic function f(z) such that the equation f(z) = 0 doesn’t have any solutions.

A: f(z)=e^z

Q: Find the number of zeroes of z^2+e^z-2 in the disc |z|<2

A: Is the question about z^2 + e^(z-2) ? As written, it is a very hard question.

Q: True or False: There exists an analytic (one-valued) function on the complex plane whose square is equal to e^z-1.

A: False. If f(z)^2 = e^z - 1, then differentiating both sides we get 2 f(z) f ' (z) = e^z. In particular f(0)^2 = 0 and 2 f(0) f ' (0) = 1. These two equalities are incompatible.

A little more conceptual proof: If an analytic function f has a zero of order k at z=z_0, then f^2 has a zero of order 2k at z_0. Thus a square of an analytic function can have only zeroes of even orders. The function e^z - 1 has a zero of order 1 at z=0, so it can't be a square of an analytic function.

Q: Find the number of zeros of z^2050+4z^4-e^z in the disc abs(z)<1

A: First notice that if |z|=1, then Re(z)<=1 and hence |e^z|=e^(Re(z))<=e.

Thus if |z|=1, |4z^4|=4, while |z^2050-e^z|<=1+e < 4. By Rouche theorem there are 4 zeroes of z^2050+4z^4-e^z in |z|<1.

Q:∫Re(z)dz from 1+i to 2+2i (straight line)

A: Parametrize the line as z=1+i + t(1+i), 0=<t<=1. Then dz=(1+i)dt and Re(z)=1+t. Thus
$\int Re(z) dz = \int_{0}^1 (1+t) (1+i) dt = (1+i) (1+1/2) = 3/2+3/2 i$

Q: lim R-->inf of ∫sin(1/z)dz over the semicircle z=Re^it [t=0 to t=pi]

A: sin(1/z) = 1/z - 1/3! 1/z^3 + 1/5! 1/z^5 +...

Thus
$\int \sin(1/z)\,dz = \int dz / z - 1/3! \int dz/ z^3 + ... = \pi i - 1/3! \frac{z^{-2}}{-2}|\limits_{z=R}^{z=-R} + ...$
As R tends to infinity, all terms but the first one tend to zero. Hence the limit as R tends to infinity is pi i.

Q:In last problem of the 8th lecture, you showed that for f(z)=1/(z (z+1) (z+2), to compute the residue at z=-1 one can write f(z)= 1/(z+1) * 1/z(z+2). Then you substitute z=-1 into 1/z(z+2) to find its residue without take its derivative. However, in the notes of lecture(the one you typed it on wiki), you showed the similar result for residue Res( 1/(z+1/2)^2 *[z/4(z+2)^2]; z=-1/2), at this point, you take its derivative and set z=-1/2 to find its residue. Which one is correct?

A: Both are correct. Suppose you want to compute the residue at z= z_0 of a function of the form f(z)/(z-z_0)^k, where f is analytic and f(z_0) is not zero.

You write the Taylor series expansion of f around z_0:
$f(z) = f(z_0) + f ' (z_0) (z-z_0) + ... + f^{(k-1)}(z_0)/(k-1)! (z-z_0)^{k-1}+\ldots$
Then
$f(z) / (z-z_0)^k = \frac{f(z_0)}{(z-z_0)^k} + \frac{f ' (z_0)}{(z-z_0)^{k-1}} + ... + \frac{ f^{(k-1)}(z_0)/(k-1)! }{z-z_0}+\ldots$
And thus the residue is
$f^{(k-1)}(z_0)/(k-1)!$

Q: Show that every function f analytic on the complex plane and satisfying |f(z)| > 1 for all z is constant.

A: Since |f(z)|>1 for all z, in particular f(z) is not zero for any z. Thus g(z)=1/f(z) is analytic and satisfies |g(z)|<1 for all z. By Liouville's theorem g is constant. Thus f=1/g is constant as well.

Q: Find the number of zeroes of the function f(z) = z^2011 + 4z^4 - 2 in the disc |z| < 1.

A: On |z|=1, |4z^4|=4, while |z^2011-2|<=1+2 =3 < 4. Thus by Rouche theorem there are 4 zeroes inside the disc |z|<1

Q: ∫x^4/(1+x^10) dx from x= - infinity to x= infinity.

A: Very similar to the question

∫ x^4/(1+x^8) dx from negative infinity to infinity.

Q: Show that for real a,

∫[e^(acost)][cos(asint)]dt = Pi (definite integral from t=0 to Pi)

HINT (Not implying that you need one :) ): Let gamma be the unit circle transversed once in the positive sense and consider ∫(e^az)/z dz.

A: Let z=e^it. Then e^(az)=e^(acos t + i a sin t) = e^(a cos t) (cos (a sin t)+ i sin (a sin(t)).
Also (dz)/z = i dt.

Thus
$\int_{|z|=1} e^{az} \frac{dz}{z}=i \int_{-\pi}^{\pi} e^{a \cos t} \cos(a\sin t) dt - \int_0^{2\pi} e^{a \cos t} \sin(a\sin t) dt$

Thus
$\int_{-\pi}^{\pi} e^{a \cos t} \cos(a\sin t) dt = Im( \int_{|z|=1} e^{az} \frac{dz}{z} )$

Finally by residue theorem, the integral of e^(az)/z over the unit circle is 2 pi i. Thus

$\int_{-\pi}^{\pi} e^{a \cos t} \cos(a\sin t) dt = 2 \pi$

Finally the function e^{a \cos t} \cos(a\sin t) is even, and hence its integral from 0 to pi is half the integral from -pi to pi

Q: True or false, 'all values of (-1)^i are real numbers'. Explain your reasoning.
A:
$(-1)^i = e^{(\pi i + 2\pi i k)i} = e^{-\pi - 2\pi k}$
All these numbers are indeed real (k is any integer here).

Q: How many zeros does the polynomial z^8 +3z^2 +12z + 5 have inside the circle abs(z)=2? Inside abs(z)=1? Explain.

A: On |z|=2, |z^8|=2^8=256, while |3z^2+12z+5|<= 12+24+5<256. Thus by Rouche theorem there are 8 zeroes inside the circle |z|=2.

On |z|=1, |12z|=12, while |z^8+3z^2+5|<=1+3+5=9 < 12, so by Rouche theorem there is one zero inside |z|=1

Q: Show that the principal branch of the logarithm satisfies log(1/z) = -log(z)

A: If z=re^(it) with -pi

(remark: I am a little confused by what happens for real negative values of z. If log is defined for them, then the required equality can't possibly hold (it would imply that log (1/(-1)) = - log(-1), i.e. log(-1)=0, which is nonsense). Probably the principal branch of log isn't defined for negative reals or, alternatively, the question is wrong. I will comment on this once I take a look at our book)

Q:
a) Let f be an analytic function on the open unit disk (that is, the set of zEC such that abs(z)<1) and assume that the values of f are on the parabola y = x^2 (this means that if you write Re(f) = u and Im(f) = v then v = u^2). Show that f must be constant on the unit disk.

b) If f is defined on an open set different from the unit disk, is f necessarily constant? (Justify the answer).

A:

a) v=u^2, hence v_x=2uu_x, v_y=2uu_y.
Cauchy-Riemann equations tell that v_y=u_x, v_x=-u_y.

Thus v_x= - 4u^2 v_x, v_y = -4u^2 v_y. Hence (1+4u^2)v_x=0, (1+4u^2)v_y =0. Hence v_x=v_y=0. Hence u_x=0,u_y=0. Hence u,v are constants.

b) No. Example: f(x)= 0 if |x|<1 and 1+i if |x-10|<1. This function is defined on an open set (union of two discs), its values lie on the parabola v=u^2, it is analytic and yet it is not constant.

Q:
For what value z does the limit as w tends to 0 of (|2+z+w|^2-|2+z|^2)/w exist?
A: let f(z)=|2+z|^2. Then the limit in question is by definition f ' (z).

Now verify at what points z Cauchy-Riemann equations hold for f:
f(x+iy)=(2+x)^2+y^2, so Cauchy-Riemann equations tell 2(2+x)=0, 2y=0.
Thus f can be differentiable only at z= - 2.

When z= - 2, indeed we have lim(|w|^2)/w=lim (conjugate(w)) = 0

Q:
Find the radius of convergence of the power series expansion of the function sin(1/(z^2 + 9)) in powers of (z-4)
A: 5, since the function has poles only at z=3i, z=-3i, which are located at distance 5 from the point 4.

Q: Do we need to know how to use the Jordan Lemma?
(I (a student) asked Yuri this last week and I'm quite sure he said we don't need to know it. Please correct me if I'm wrong!)

A: You don't need to know it.

Q: Every time we do an integral of a real function over the real line, do we have to show that the semicircular part of the complex integral tends to zero? Or can we use the results from class?

A: It is much better if you actually show it.

If you want to, you can quote instead the exact formulation of the result from the class you are using and explain why the result applies to your problem.

Q: Evaluate the line integral ∫ 1/(z^2+1) dz over the positively oriented circle |z|=1. My question here is what happens when singularities exist on the boundary of a curve that is being integrated over? Do we treat the integral any differently than an integral around, lets say abs(z-i)=1 ?

A: The integral you asked about is not even defined (the integral is an improper integral which doesn't even converge).

Q: Prove that for a polynomial of degree n p(z) = a_n*z^n + a_n-1*z^n-1 + ... + a_1*z + a_0,

0.5*|a_n| <= |p(z)/z^n| <= 2|a_n| for |z| large enough. I do not understand the proof in the textbook :(

A: the reason is that lim p(z)/z^n as |z|-> infinity is a_n. So when |z| is large, p(z)/z^n must be very close to being a_n. In particular its absolute value will lie between 1/2 |a_n| to 2 |a_n|

Q:
Find the number of zeroes of z^140 + 200z^40 + z + 1 in the annulus 1<|z|<2.
A: When |z|=1, we have |200z^40|=200 > 3 >= |z^140+z+1|, so by Rouche theorem there are 40 zeroes in |z|<1.

When |z|=2 we have |z^140|=2^140 > 200 2^40+2+1 >= |200 z^40 + z+1|, so by Rouche theorem there are 140 zeroes in |z|<2.

Thus there are 100 zeroes in the annulus 1<|z|<2.

Q: For the question
Find the value of the integral of z / (e^π z- e^-π z) dz over the circle of radius 2 centered at z=1+4i (oriented counterclockwise),
how did you find the residue? Please elaborate on your previous answer.

A:
At point z=in we have the following Taylor expansions:
$z = in + (z-in)$
$e^{\pi z}-e^{-\pi z}= (\pi e^{\pi i n}+\pi e^{-\pi i n})(z-in)+\ldots$
Hence
$\frac{z}{e^{\pi z}-e^{-\pi z}}=\frac{in+\ldots}{(\pi e^{\pi i n}+\pi e^{-\pi i n})(z-in)+\ldots}=\frac{in}{2\pi(-1)^n}\frac{1}{z-in}+\ldots$
and hence the residue at z=in is (-1)^n in/(2 pi).

Q：Evalute the line integral ∫ sin(z)/z^5 dz over the positively oriented circle |z|=1.

A: sin z/ z^5 = (z-z^3/3!+z^5/5!-...)/z^5 = 1/z^4 - 1/3! 1/z^2 + 1/5! +..., hence the residue at z=0 is zero and hence the integral over |z|=1 is zero.

Q:Find number of zeroes of z^2+e^(z-2) in the disc |z| <2.

A: When |z|=2, we have |z^2|=4 while |e^(z-2)|=e^Re(z-2)<=e^0=1, so by Rouche theorem there are 2 zeroes in the disc |z|<2.

(remark: we've used the facts that |e^(x+iy)|=e^x for real x, y and that if |z|=2, the Re z<= 2 and hence Re(z-2)<= 0).

Q: Determine how many zeroes does 4z^3-12z^2+2z+10 have in the annulus 1/2<|z-1|<2

A: Let w=z-1. Then we are interested in the function is 4(w+1)^3-12(w+1)^2+2(w+1)+10=4w^3-10w+4. When |w|=2, the inequality |4w^3|=32> 24>=|-10w+4| holds, so by Rouche theorem, the function has 3 zeroes in |w|<2. When |w|=1/2, the inequality |10w|=5 > 4.5>=|4w^3+4| holds, so by Rouche theorem there is one zero in |w|<1/2 and (once again since when |w|=1/2, the inequality |10w|=5 > 4.5>=|4w^3+4| holds) there are no zeroes on |w|=1/2. Hence there are 2 zeroes in 1/2<|w|<2

Q:Find a Mobius transformation T such that T maps the real axis onto itself and the imaginary axis onto the circle |w-1/2|=1/2

A: Real and imaginary axis intersect at z=0, z=infinity. The real axis and the circle |w-1/2|=1/2 intersect at points w=0,w=1. Hence the Mobius transformation in question must send the points z=0,z=infinity to points w=0,w=1. To make sure that the real axis gets mapped to itself, we choose another point on the real axis (e.g. z=1) and send it to some point on the real axis (e.g. w=infinity).

Now a Mobius transformation sending 0,infinity and 1 to 0,1,infinity must map the real line to the real line. Moreover, the imaginary axis gets mapped by such transformation to some circle passing through points 0,1.

It seems that we haven't really solved the problem (we haven't shown that the imaginary axis gets mapped to the circle |w-1/2|=1/2: instead we have showed only that it gets mapped to some circle passing through points 0,1). However an additional argument helps in this case: the real and imaginary axes intersect at right angle and Mobius transformations preserve angles. Thus if the image of the real axis is the real axis, the image of the imaginary axis must be a circle orthogonal to the real axis. Since we also know that it passes through points 0,1, it must be the circle |w-1/2|=1/2.

Finally a Mobius transformation sending 0,infinity and 1 to 0,1,infinity is w=z/(z-1).

Remark: the answer to this question is not unique (we made a choice of sending 1 to infinity, which we could have replaced by any other choice of "send a point on the real axis to a point on the real axis").

Q: For question
∫ x^4/(1+x^8) dx from negative infinity to infinity. Can we simply use the formula offered in the textbook, namely Res(P/Q; x_0) = P(x_0)/Q'(x_0) over the reals, since the rest of the integral (top semi-circle) converges to 0, which has been shown in the textbook as well?

A: If you want to use a theorem from the book, you should quote the theorem precisely and explain why is it applicable in your situation.

For instance here to justify that Res( x^4/(1+x^8) ; x=x_0 ) = x_0^4/(8x_0^7) where x_0 is a zero of the polynomial 1+x^8, you can proceed in two ways:

Way 1:write "
x^4/(1+x^8)=(x_0^4+...)/(0 + (8x_0^7)(x-x_0)+...)=x_0^4/(8x_0^7) 1/(x-x_0) + ..., hence Res( x^4/(1+x^8) ; x=x_0 ) = x_0^4/(8x_0^7)
"

(here the answer is justified by a computation of the Laurent expansion of x^4/(1+x^8) in powers of (x-x_0) and finding the coefficient of 1/(x-x_0) in that expansion)
Way 2: write "
there is a theorem (for instance in the book we were using in class) that if P,Q are two analytic functions in a neighborhood of a point x_0, P(x_0) is not zero and Q(x) has a zero of order one at x=x_0, then Res(P/Q; x_0) = P(x_0)/Q'(x_0). If we let P(x)=x^4, Q(x)=1+x^8, and x_0 be a zero of 1+x^8, then, since x_0^4 is not zero (indeed, x_0=0 is not a root of 1+x^8) and the order of the zero x=x_0 of Q(x) is one (indeed, Q'(x_0)=8x_0^7 is not zero), the theorem implies that Res(x^4/(1+x^8); x=x_0) = x_0^4/(8x_0^7).
"

A solution of the kind "by a theorem from the book Res(x^4/(1+x^8); x=x_0) = x_0^4/(8x_0^7)" will be graded as 1 or 2 out of 10.
A solution of the form "by the theorem from the book that says that Res(P/Q; x_0) = P(x_0)/Q'(x_0), Res(x^4/(1+x^8); x=x_0) = x_0^4/(8x_0^7)" will also be graded as 2 to 4 points out of 10 (the reason is that the theorem is not quoted with the conditions in which it applies and the fact that these conditions do actually hold in the question is not verified).

Similar reasoning applies to "since the rest of the integral (top semi-circle) converges to 0, which has been shown in the textbook as well" --- you should quote the exact fact you are using!

Q: ∫(cos2x)dx/(1 - 2a*cosx + a^2) from 0 to 2Pi, where -1<a<1

A: Make a change of variables to z=e^(ix). Then cos x = (z+1/z)/2, cos(2x)=(z^2+1/z^2)/2, dx=dz/(iz). Then compute the resulting integral using residue theorem. Please let me know what step you want me to elaborate on.

Q:When we find a Mobius transformation,are we supposed to choose the three points arbitrarily?(Based on the question below)

A: In the questions of the type "find a Mobius transformation sending a given line/circle to a given line/circle" the answer is indeed not unique: one can choose any three points on the source line/circle and any three points on the target line/circle and then find a Mobius transformation sending the first triple to the second one.

Q:Find a Mobius transformation that carries the circle |z|=1 onto the line Re((1+i)w)=0
A: Choose three points on the circle (e.g. -i,1,i) and find a Mobius transformation that carries them to three points on the line (e.g. w=0, w=infinity, w=1+i).
Such a transformation is given by w=(1+i)(i-1)/(2i) (z+i)/(z-1)

Q: Find the number of zeroes of z^199 + iz^2 - i in the upper half-plane Imz>0.
A: We will apply argument principle for the domain bounded by the line segment from -R to R and a semicircle in the upper half-plane centered at the origin and having radius R.
First as z moves along the semicircle of radius R, we have |z|=R and if R is large enough, terms iz^2-i are much-much smaller than z^199. In particular as z makes half a circle, f(z) makes 199 half-circles (approximately half-circles: they are actually somewhat perturbed by the terms iz^2-i) of radius R^199.
Then z goes from -R to R along the reals: z=x, -R<x<R. Then f(z)=x^199 + i (x^2-1). As x increases from x=-R to x=R, the point f(z) goes from being far-far to the left (a little bit above the x-axis) to far-far to the right (a little bit above the x-axis). The question is whether this path "completes" the 199-th half-turn to a full turn and whether it makes any additional turns.
For this notice that f(z) crosses the x-axis at two points: when x=-1 and when x=1. As z moves from -R to -1, f(z) moves from more or less -R^199 + i R^2 (a little above x-axis) to point -1, staying above the x-axis. Then as z moves from -1 to 1, f(z) moves from -1 to 1, now passing below the x-axis. And finally as z goes from 1 to R, f(z) moves from 1 to approximately R^199+iR^2, staying above the x-axis.
Since indeed the path went around point zero one additional half-turn, the number of turns the image of the boundary of the domain makes around the origin is 100. Hence there are 100 zeroes in the domain.

Q: please add the 'usual arguments about the integration on the semicircle' for the second question below this one.
A: see below.

Q:∫ x^2/(x^4-4x^2+5) dx from negative infinity to infinity.
A: Similar to the question below, but the answer seems to be much worse.

Let R be a large number, let C be the semicircle x=R e^it, 0<t<pi and let I be the interval [-R,R]. First we compute the integral over the closed contour consisting of I and C. By the residue theorem this integral is equal to

2pi i times sum of residues of x^2/(x^4-4x^2+5) at the roots of the denominator. The denominator factors as (x^2-2+i)(x^2-2-i), so the roots are at
$\pm \sqrt{2\pm i}$

Two of these roots lie in the upper half-plane, so the integral is equal to 2pi i times the sum of the residues of x^2/(x^4-4x^2_5) at these two points. To be more precise, let
$\phi=\arctan 1/2 , (0<\phi<\pi/2)$
Then the relevant roots are
$\sqrt[4]{5}e^{i\phi/2}$
and
$-\sqrt[4]{5}e^{-i\phi/2}$

The residue at such a root x_0 is
$x_0^2/(4x_0^3-2x_0)=x_0/(4x_0^2-2)$
i.e. the residues are
$\pm\sqrt[4]{5}e^{\pm i\phi/2}/(4(2\pm i)-2)= \pm\sqrt[4]{5}e^{\pm i\phi/2}/(6\pm 4i)$
Hence the integral over C and I is equal to
$2\pi i( \sqrt[4]{5}e^{i\phi/2}/(6+4i)- \sqrt[4]{5}e^{-i\phi/2}/(6-4i) ) = 2\pi \sqrt[4]{5}(\sin(\frac{\arctan\frac12}2-\arctan\frac23)/\sqrt{13})$

Finally the integral over C is bounded by pi R (R^2/(R^4-4R^2-5)), which tends to 0 as R tends to infinity. Hence the integral over the realsis equal to the limit of the integral over I and C as R tends to infinity, so it is also equal to
$2\pi \sqrt[4]{5}(\sin(\frac{\arctan\frac12}2-\arctan{2/3})/\sqrt{13})$

Q: ∫ x^4/(1+x^8) dx from negative infinity to infinity.
A: Let R be a large number, let C be the semicircle x=R e^it, 0<t<pi and let I be the interval [-R,R]. First we compute the integral over the closed contour consisting of I and C. By the residue theorem this integral is equal to

2pi i times sum of residues of x^4/(1+x^8) at x=e^(pi i/ 8), x=e^(3 pi/8), x=e^(5 pi / 8), x=e^(7 pi i /8)

The residue at x_0=e^(pi i/8 + 2pi k i / 8) is of x_0^4/(8x_0^7)=1/8 x_0^(-3) i.e. it is
1/8 e^(-3 pi i / 8 - 6 pi i/8)

Hence the integral is equal to 2 pi i / 8 (e^(-3 pi i/ 8) + e^(-9 pi i/8) + e^(-15 pi i/8) + e^(-21 pi i/8)), which can be siplified to
2 pi i / 8 (e^(-3 pi i/ 8) - e^(-pi i/8) + e^(pi i/8) - e^(3 pi i/8)) = 2 pi i /8 (2 i sin(pi/8) - 2 i sin(3 pi/8))=pi/2 (sin(3 pi/8)-sin(pi / 8))

Now we relate this integral to what we actually wanted to compute. For this note that when |x|=R, the integrand is bounded by
|x^4/(1+x^8)|<= R^4/(R^8-1)

Hence by triangle inequality the integral over C is bounded by pi R (R^4/(R^8-1)), which tends to 0 as R tends to infinity. Hence as R tends to infinity, the integral over C and I tends to the integral over the real line. Hence the integral over the real line is also equal to pi/2 (sin(3 pi/8)-sin(pi / 8)).

Q:Find the Laurent series about z0=0 for exp(z+1/z) in 0<|z|<infinity

A:
$e^{z+1/z}=e^z e^{1/z}=(\sum\limits_{k=0}^\infty \frac{z^k}{k!})( \sum\limits_{m=0}^\infty \frac{(1/z)^m}{m!} )= \sum\limits_{k=0,m=0}^{\infty,\infty} \frac{x^{k-m}}{k!m!}=\sum\limits_{n=-\infty}^\infty (\sum\limits_{m=0}^\infty \frac{1}{m!(m+n)!})x^n$
Q:Locate each of the isolated singulatiries of the following functions and tell whether it is a removable singularity,a pole,or an essential singularity.
1. (z^2)/(sinz)
2. (2z+1)/(z+2)

A: For the first function the singularities are at the zeroes of the denominator, i.e. at points z where sin z=0. These points are z= n pi for integer n.

At point z=0 we have (z^2)/(sin z)= (z^2)/(z - z^3/6 +...) = z + ..., so the singularity is removable (the power series expansion starts with non-negative term, or, in other words, the limit of z^2/sin z as z goes to 0 exists).

At z=n pi for non-zero n we have (z^2)/(sin z)= ((n pi)^2 + ...)/( pi cos (n pi) (z-n pi) + ...)= plus/minus n^2 pi^2 / (z-n pi) + ...
This shows that z=n pi for non-zero n is a pole of order 1.

2. Point z=-2 is the only singularity. At this point we have (2z+1)/(z+2)=(-3 + 2(z+2))/(z+2)=(-3)/(z+2) + 2, so the point z= -2 is a pole of order 1.

Q: Find the residue of (cosz - e^z)/(tanz - sinz), around z=0.

A: We'll have to find several terms in the Taylor series expansions of the numerator and the denominator around the point z=0.

The numerator is simple: cos z - e^z = (1-z^2/2+z^4/24+...) - (1+z+z^2/2+z^3/6+z^4/24+...)= - z - z^2 - z^3/6 + 0 z^4 +....

For the denominator we'll first have to compute several terms in Taylor series expansion of tan z = sin z / cos z around z=0:

Suppose sin z/ cos z= (z - z^3/6 + z^5 / 120+...)/(1 - z^2/2 + z^4/ 24 + ...) = z + a_3 z^3 + a_5 z^5 + ...
(note that only terms of odd order can appear in the Taylor series expansion of and odd function tan z around the point z=0).

Then we have (z + a_3 z^3 + a_5 z^5 + ...)(1 - z^2/2 + z^4/ 24 + ...) = z - z^3/6 + z^5 / 120+...
or, opening the brackets:

z + (a_3-1/2)z^3 + (a_5 - a_3/2 + 1/24)z^5 + ... = z - z^3/6 + z^5 / 120 + ...

Hence a_3 = 1/2 - 1/6 = 1/3,
a_5 = 1/120 + 1/6 - 1/24 = 2/15

Thus tan z = z + z^3/3 + 2/15 z^5 + ...

It follows that the denominator, tan z - sin z, is equal to (z + z^3 / 3 + 2/15 z^5 + ...) - (z - z^3/6 + z^5/120 + ...) = z^3 /2 + z^5 / 8 + ...

Finally we compute that
(cosz - e^z)/(tanz - sinz) = (- z - z^2 - z^3/6 + 0 z^4 +....)/(z^3 / 2 + z^5 / 8+ ...) = - 2 / z^2 + a_{-1} / z + ...

where the coefficient a_{-1} that interests us the most can be found from the identity

(- 2 / z^2 + a_{-1} / z + ...)( z^3 / 2 + z^5 / 8+ ... ) = - z - z^2 - z^3/6 + 0 z^4 +....

i.e.
- z + a_{-1}/2 z^2 + ... = - z - z^2 + ...

Thus the residue a_{-1} is a_{-2} = - 2

Q：Evaluate the value of the integral of dx/((1+x)x^(2/3)) form 0 to infinity.

A: See examples 9 and 10 in chapter 2.6 in the book for related examples.

We'll compute the integral along the "keyhole" contour consisting of
(1) the segment from r to R
(2) the circle z=R e^it, t goes from 0 to 2pi
(3) the segment from R to r
(4) the circle z=e e^it, t goes from 2pi to 0

Along the contour we interpret z^2/3 as |z|^2/3 e^(2/3 i phi), where phi is from 0 to 2pi.

The integral over the whole contour is equal to 2pi i times the residue of 1/((1+x)x^(2/3)) at x= - 1, which is 1/(-1)^(2/3)=1/e^(2/3 pi i). So the integral is 2 pi i e^(-2/3 pi i).

The integral (3) is equal to
$\int\limits_R^r \frac{dx}{(1+x)(x^{2/3}e^{\frac23 2 \pi i})}=-\frac{ \int\limits_r^R \frac{dx}{(1+x)x^{2/3}} }{ e^{4 \pi i/3} }$

The integral (2) tends to zero as R tends to infinity:
$|\int_{|x|=R}\frac{dx}{(1+x)x^{2/3}}|\leq 2 \pi R\frac{1}{(R-1)R^{2/3}}\to 0$

Similarly
$|\int_{|x|=r}\frac{dx}{(1+x)x^{2/3}}|\leq 2 \pi r\frac{1}{(1-r)r^{2/3}}\to 0$
as r tends to 0.

Hence
$(1-\frac{1}{e^{2/3\pi i}})\int_0^\infty \frac{dx}{(1+x)x^{2/3}}=2\pi i e^{-2/3 \pi i}$
or
$\int_0^\infty \frac{dx}{(1+x)x^{2/3}}=2\pi i \frac{e^{-2/3 \pi i}}{ 1-\frac{1}{e^{4\pi i/3}}} }=\pi\frac{2i}{e^{2\pi i/3}-e^{-2\pi i/3}}=\frac{\pi}{\sin \frac{2\pi}{3}}=\frac{2\pi}{\sqrt{3}} }$
Q：Find the value of the integral of z / (e^π z- e^-π z) dz over the circle of radius 2 centered at z=1+4i (oriented counterclockwise)

A: First we find the singularities of the function:
$e^{\pi z}-e^{-\pi z}=0$
means that z=i n for some integer n.

Of these points only the points 3i, 4i, 5i are inside the circle |z-(1+4i)|=2. So we have to compute the residues at these three points.

At point z=in we have
$\frac{z}{e^{\pi z}-e^{-\pi z}}=\frac{in+\ldots}{(\pi e^{\pi i n}+\pi e^{-\pi i n})(z-in)+\ldots}=\frac{in}{2\pi(-1)^n}\frac{1}{z-in}+\ldots$
so the residue at z=in is (-1)^n in/(2 pi). In particular the residues at points 3i, 4i, 5i are -3/2 i/pi, 2 i/pi, -5/2 i/pi respectively. Their sum is -2i/pi. Hence the integral is 2pi i (-2i/pi)=4.

Q:Find the order of each of the zeros of (z^2-4z+4)^3.

A: (z^2-4z+4)^3=(z-2)^6, so z=2 is a zero of order 6.

Q: Let f(z)=z^2/(z^2 -1). Find the Laurent series representation of f around point z0=1. Also, give the residue at that point.

A: Let w=z-1. Then the function becomes f(w)=(w+1)^2/(w^2+2w)=1/w ((w^2+2w+1)/(w+2))= 1/w (w + 1/(w+2)).

Now 1/(w+2)=1/2 1/(1+w/2)=1/2 (1 - w/2 + w^2/4 - w^3/8+...), Hence w + 1/(w+2)=1/2+3/4 w + w^2/8 - w^3/16 + ... + (-1)^n w^n/2^(n+1)+... and hence 1/w (w + 1/(w+2)) = 1/2 (1/w) + 3/4 + w/8 - w^2/16 + ...

Thus f(z)=1/2 (1/(z-1)) + 3/4 + (z-1)/8 - (z-1)^2/16 + ....

The residue is 1/2

Q: Let f be an analytic function on the punctured disc, 0<abs(z-z_0)<R and f has an essential singularity at the point z=z_0. Let w be any complex number and let g(z)=1/(f(z) -w).
Show that g is not bounded in any punctured disc 0<abs(z-z_0)<r.

A: The idea is that if it is bounded, it has only removable singularities and hence f(z)-w has only poles.
I will write down the details later.

Q: How is it possible to write a Laurent series for f(z)=z/(sinz)^2 around z=0

A: You can only find several first terms in the expansion (as many as you want, but still it's very hard to find a general formula): first you find sin z = z- z^3/6+ z^5/120+.... Then you find (sin z)^2=(z- z^3/6+ z^5/120+...)(z- z^3/6+ z^5/120+...)=z^2-z^4/3+(1/36+2/120)z^6+...
Then you let z/(sin z)^2= 1/z + a_0 + a_1 z + a_2 z^2 + a_3 z^3... and write down the condition that
(1/z + a_0 + a_1 z + a_2 z^2 + a_3 z^3...)(z^2-z^4/3+(1/36+2/120)z^6+...)=z
and compare coefficients in
z+a_0z^2+(-1/3+a_1)z^3+(a_2-a_0/3)z^4+(a_3-a_1/3+1/36+1/60)z^5+...=z
to find a_0=0, a_1=1/3, a_2=0, a_3 = [[tel:1/9-1/36-1/60|1/9-1/36-1/60]], ...

Q: Consider the function f(z)=pi*cot(pi*z). Find all isolated singularities of f. If it is a removable singularity define a new function to "fix" the old function. If it is a pole, find the order of the pole.

A: pi cot(pi z)= pi cos(pi z)/sin(pi z). The denominator vanishes at integer values of z. If z_0=n is integer, then we have the following Laurent expansion for the function
pi cos(pi z)/sin(pi z):
$\pi \frac{cos(\pi z)}{sin(\pi z)}=\pi \frac{cos(\pi n)+\ldots}{\pi \cos(\pi n)(z-n)+\ldots}=\frac{1}{z-n}+\ldots$

So the function has a pole of order 1 at every integer. The residue at all these poles is 1.

Q:∫ (1/(a+bcos t))dt from 0 to 2π, where a>b>0.

A: Let z=e^it. As t goes from 0 to 2pi, the point z=e^it makes a full turn around 0 along the unit circle |z|=1 in the counter-clockwise direction.

Differentiating z=e^it we get dz=ie^it dt, so dt= -i dz / z.

Also if z=e^it, then cos t = (z+1/z)/2.

Thus the integral can be rewritten as
$\int_{|z|=1} \frac{-i dz/z}{a+b(z+1/z)/2}$

Now this integral can be further rewritten as
$-2i \int_{|z|=1} \frac{ dz}{bz^2+2az+b}$

The denominator vanishes at two points:

$z_{1,2}=\frac{-a\pm \sqrt{a^2-b^2}}{b}=-\frac a b \pm \sqrt{(\frac a b)^2 -1}$

Since a>b>0, the expression under the square root is positive. Thus
$-\frac a b - \sqrt{(\frac a b)^2 -1}$
is outside the circle |z|=1

The other root
$-\frac a b + \sqrt{(\frac a b)^2 -1} = \frac{1}{ -\frac a b - \sqrt{(\frac a b)^2 -1} }$
is inside the circle (the denominator is < -1, so the quotient is between -1 and 0).

Thus we can apply Cauchy formula
$\int \frac{f(z)\,dz}{z-z_0}=2\pi i f(z_0)$
with the function
$f(z)=\frac{1}{b(z-( -\frac a b - \sqrt{(\frac a b)^2 -1} ))}$

and
$z_0= -\frac a b + \sqrt{(\frac a b)^2 -1}$

Thus
$-2i \int_{|z|=1} \frac{ dz}{bz^2+2az+b}=\frac{4\pi}{b(( -\frac a b + \sqrt{(\frac a b)^2 -1} )-( -\frac a b - \sqrt{(\frac a b)^2 -1} ))}=\frac{2\pi}{b\sqrt{(\frac a b)^2-1}}=\frac{2\pi}{\sqrt{a^2-b^2}}$
To be completed later

Q: ∫ (1/(3+sinθ+cosθ))dθ from 0 to 2π.

A: Very similar to the question above it. The answer is 2pi/sqrt(7). Please let me know if you need more details.

Q：find the power series expansion about the given point. (z+2)/(z+3) about z_0=-1

A: first let w=z+1, so that we will have to expand everything in powers of w. Now the original function is (w+1)/(w+2), or, equivalently, 1- 1/(w+2).

To expand 1/(w+2) in powers of w, we rewrite it as 1/2 1/(1+w/2)=1/2 (1-w/2+w^2/4-w^3/8+...), where the last expansion is valid for |w/2|<1, or, equivalently, |w|<2.

Finally we gather all this together to get (z+2)/(z+3)=1/2 + (z+1)/4 - (z+1)^2/8 - (z+1)^3/16 + ...

Q：Find the order of zeros of (z^2+z-2)^3.

A: Factor z^2+z-2 as (z-1)(z+2). Thus the original function is (z-1)^3(z+2)^3, which has zeroes of order 3 at z=1 and at z= -2.

Q:Find the largest disc in which the power series expansion for the following function is valid.

A: The function log(1-z) can be made analytic on any domain that doesn't contain a loop around the point 1 (since log w can be chosen to be analytic on any domain which doesn't contain any loop around the origin). In particular it can be made analytic on |z|<1. By a theorem we proved in class this means that the Taylor series of
[log(1-z)]^2 converges on |z|<1. It doesn't converge on a larger disc, since
[log(1-z)]^2 is not even continuous at z=1, which is on the boundary of the disc.

Q:Find all solutions to the differential equation
f ''(z)+β^2*f(z) = 0, f is an entire function.

A: "Entire" function means it is analytic everywhere. In particular it has a Taylor series around any point and this Taylos series converges on the whole plane.

White
$f(z)=\sum_{n=0}^{\infty} a_n x^n$
Then
$f''(z)=\sum_{n=2}^{\infty} n(n-1)a_n x^{n-2}=\sum_{k=0}^\infty (k+2)(k+1)a_{k+2}x^k$
Hence
$f''(z)+\beta^2 f(z)=\sum_{k=0}^\infty [(k+2)(k+1)a_{k+2}+\beta^2a_k]x^k=0$
Equating coefficients we see that for any k
$(k+2)(k+1)a_{k+2}+\beta^2a_k=0$
We solve this recursion equation. If β=0, then
$f(z)=a_0+a_1z$
since all other coefficients are zero. If β is non-zero, then for even k we have
$a_{k+2}=\frac{\beta^2a_k}{(k+2)(k+1)}= \frac{\beta^4a_{k-2}}{(k+2)(k+1)k(k-1)}=\ldots= \frac{\beta^{k+2}a_0}{(k+2)(k+1)\ldtots 2\cdot 1}= \frac{\beta^{k+2}a_0}{(k+2)!}$
And similarly if k is odd, then
$a_{k+2}=\frac{\beta^2a_k}{(k+2)(k+1)}= \frac{\beta^4a_{k-2}}{(k+2)(k+1)k(k-1)}=\ldots= \frac{\beta^{k+1}a_1}{(k+2)(k+1)\ldtots 3\cdot 2}= \frac{\beta^{k+1}a_1}{(k+2)!}$
Plugging these coefficients back into the definition of f, we find that
$f(z)=a_0\sum_{m=0}^\infty \frac{\beta^{2m}z^{2m}}{(2m)!}+a_1 \sum_{m=0}^\infty \frac{\beta^{2m}z^{2m+1}}{(2m+1)!}$
or
$f(z)=a_0\cosh(\beta z)+\frac{a_1}{\beta}\sinh(\beta z)$
Q:Find the radius of convergence of the power series expansion of the function sin(1/(z^2+9)) in powers of (z-4).

A: sin(1/(z^2+9)) is analytic everywhere except at z=3i or z=-3i. The function sin(1/(z^2+9)) doesn't have any limit as z tends to 3i (or to -3i).

By a theorem we proved in class, this means that the Taylor series around any point converges on the largest disc around that point which doesn't contain neither 3i
nor -3i.

Thus the Taylor series around the point 4 converges in the disc |z-4|<5

Q:Evaluate ∫ (log(z+2) + z*(z conjugate)) dz over |z|=1 - answered below

Q:Determine the image of {z: 0< Re(z) < pi / 2 and Im(z) < 0} under f(z) = sinz.

You can find the answer in the textbook (see p.51), but we didn't discuss images and preimages under sin or cos (it is slightly more complicated than what we did).

Q: Explain why the three lines on the top left-hand corner of this page is a contradiction. (yeah that's what I meant, where is the mistake in the argument...)

Comment: it is obvious that it is a contradiction (e^(-2pi) is not 1). The question should be where is the mistake in the argument.

A: The mistake is in going from the second line to the third. Both (e^2\pi i)^i and 1^i have infinitely many values. The equality on the second line means that the set of all values of
(e^2\pi i)^i coincides with the set of all values of 1^i. On the third line we mistakenly conclude "hence any value on the left is equal to any value on the right"

Here is an analogous, but simpler mistake:
$((-1)^{2})^{1/2}=1^{1/2}$
$-1=1$

Q:Suppose that the function f is analytic at all points of the complex plane.Suppose also that the image of f is contained in the unit circle {u + iv | u^2 + v^2 = 1}.Show that f is constant.

A: See answers to the summer's midterm in the "Midterm Summer 2011" section

Q: g(z) = z^2 - |z|^2.Find g'(z) where it exists.

A: g(x+iy)=(x+iy)^2-|x+iy|^2=2ixy-2y^2. Cauchy-Riemann equations with u(x,y)= - 2y^2, v(x,y)=2xy read
0=2x
-4y=-2y
Hence if the complex derivative exists at some point (x,y), then by Cauchy-Riemann equations x=0,y=0, i.e. z=x+iy=0.

At point z=0 we compute
$g'(0)=\lim_{z\to 0} \frac{z^2-|z|^2}{z}=\lim_{z\to 0} z-\bar{z}=0$

Q: Find the power series expansion of g(z) =z^2*sin(3z) around z0=0 and find its radius of convergence.

A: Since
$\sin z = \sum\limits_{n=0}^\infty (-1)^n z^{2n+1}/(2n+1)!$
we have
$\sin(3z) = \sum \limits_{n=0}^\infty (-1)^n (3z)^{2n+1}/(2n+1)!= \sum\limits _{n=0}^\infty (-1)^n 3^nz^{2n+1}/(2n+1)!$
and
$z^2\sin(3z) = \sum\limits_{n=0}^\infty (-1)^n 3^nz^{2n+3}/(2n+1)!$
The radius of convergence is infinity, just like the one for sine.

Q: Show that cos(z) is an unbounded function.

A:
$\cos(in)=\frac{e^n+e^{-n}}{2}\xrightarrow[n\to\infty]{}\infty$

Q: Evaluate the line integrals
∫ |z-1|^2n*(z-1)^m dz over |z-1|=1
where n and m are integers.

A:
On |z-1|=1 the value of |z-1|^2n is 1. Hence the integral can be rewritten as
$\int\limits_{|z-1|=1}(z-1)^m \,dz=\int \limits_{|w|=1} w^m \, dw$
I've made a change of variable w=z-1 for convenience. This last integral is equal to 0 if m is not -1 and is equal to 2 pi i if m= -1.

Q: Show that 1/2pi *∫e^ikw dw from 0,2pi={0 if k non-zero and 1 if k=0}

A: If k=0, the integrand is equal to 1 hence the integral is 2 pi. If k is not 0, then
$\int\limits_0^{2\pi} e^{ikw} dw = \frac{e^{ikw}}{ik}|_{w=0}^{2\pi}=0$

Q: ∫ (z+1/z)^(2n+1) dz over |z|=1

A: Open the brackets:

$(z+1/z)^{2n+1}=z^{2n+1}+{2n+1 \choose 1} z^{2n-1} + \ldots +{2n+1 \choose 2n} z^{-2n+1}+ z^{-2n-1}$

The integral of
$z^{k}$
over |z|=1 is non-zero only when k= -1. For k= -1 the integral is 2 pi i. Hence only the summand
${2n+1 \choose n+1}z^{-1}$
contributes to the answer and the answer is
$\int (z+1/z)^{2n+1} \,dz={2n+1 \choose n+1} 2 \pi i$

Q: lim n(1-cos(θ/n)-isin(θ/n)) as n--->∞,where θ is some fixed real number.
A: Computing separately lim n(1-cos(θ/n)) = 0 and lim n sin(θ/n) = θ we find that the original limit is - i θ (both limits can be computed using l'Hopital's rule: letting x=1/n the limits can be rewritten as lim ( 1-cos(θx) )/x and lim sin(θx) / x as x-->0)

Q: Integral 1/[z^3(z-2)^2]dz over |z|=3

A: By Cauchy theorem this integral is equal to the integral of the same expression over |z|=R for any R>3.

For R large enough the absolute value of the integrand is at most 1/[R^3(R-2)^2], hence by triangle inequality the absolute value of the integral is at most 2pi R/[R^3(R-2)^2]=2pi/[R^2(R-2)^2], which tends to zero as R tends to infinity. Since this non-negative number (the absolute value of the integral) is smaller than all the values of a function that tends to zero, it must be equal to zero.

Q:Find the power series centered at 0 for 1/(1-z^3)
A: 1+z^3+z^6+z^9+....

Q: Let u(x,y)=g(xy)where g:R--R is a real polynomial. Suppose u(x,y)is harmonic on R-2.Show that g(t)=at+b with real a and b

A:
$u_{xx}=(yg'(xy))_x=y(g'(xy))_x=y^2g''(xy)$
Similarly
$u_{yy}=(xg'(xy))_y=x(g'(xy))_y=x^2g''(xy)$
Since u is harmonic, it follows that
$(x^2+y^2)g''(xy)=0$
Hence for any (x,y) except (0,0) the second derivative at the point g''(xy) vanishes. Thus g'' vanishes identically and hence g is a linear function.

Q: Evaluate integral[log(z+2)+z*z-bar]dz

Over|z|=1

A: The function log(z+2) is analytic inside the circle |z|=1 (point z=-2 is outside), so its integral is zero by Cauchy theorem. The function z*z-bar is equal to |z|^2, i.e. it is equal to 1 on the unit circle. Now the integral of 1 dz is once again zero, since 1 is an analytic function on the unit disc.

Q:Find the value of (i∫z-bar dz) where the integration is along the closed path consisting of the segments from -1-i to 1-i, from 1-i to 2i and from 2i to -1-i (i.e. along the triangle with vertices -1-i,1-i,2i going in counterclockwise direction)

A: One way is to parametrize the sides of the triangle and compute the sum of the integrals. It is a long computation however. For instance the integral over the lower side (from -1-i to 1-i) can be computed as follows: the path is z=-i+t, -1<t<1; dz=dt; hence the integral is
$\int_{-1}^1 (i+t) dt=2i$
The integral over the other two sides can be found in a similar way, but with more tricky computations.

Much simpler approach is to use Green's theorem: write z=x+i y, z-bar = x - i y, dz = dx + i dy

Hence
$i \int \bar z dz = i \int (x-iy)(dx+idy)= i \int (xdx+ydy) + \int (ydx-xdy)$

finally by Green's theorem this expression can be rewritten as

$i \int \int 0+0 dx dy + \int \int -1-1 dx dy = -2 \int \int dx dy = - 2 \cdot \mbox{area of triangle}= - 2 \cdot \mbox{half base times height}= - 6$

Q: Determine whether the given infinite series converges or diverges.
$\sum_{i=1}^\infty{ \frac{1}{n}(\frac{1+i}{\sqrt 2})^n }\!$

A: The first thing to try is to determine the radius of convergence for the series
$\sum_{i=1}^\infty \frac{1}{n} z^n$
The radius of convergence in this case is one. Since
$\left| \frac{1+i}{\sqrt 2} \right|=1$
this doesn't tell us directly whether the series converges or diverges.

In fact if we ask the question whether the series converges absolutely, then the answer is no:
$\sum_{i=1}^\infty{ \left| \frac{1}{n}(\frac{1+i}{\sqrt 2})^n \right| }\!= \sum_{i=1}^\infty{ \frac{1}{n} }\! = \infty$

However the series itself does converge. To see this notice that the powers of (1+i)/sqrt(2) repeat themselves with period 8:
$(\frac{1+i}{\sqrt 2})^1=\frac{1+i}{\sqrt 2}$
$(\frac{1+i}{\sqrt 2})^2=i$
$(\frac{1+i}{\sqrt 2})^3=\frac{-1+i}{\sqrt 2}$
$(\frac{1+i}{\sqrt 2})^4=-1$
$(\frac{1+i}{\sqrt 2})^5=\frac{-1-i}{\sqrt 2}$
$(\frac{1+i}{\sqrt 2})^6=-i$
$(\frac{1+i}{\sqrt 2})^7=\frac{1-i}{\sqrt 2}$
$(\frac{1+i}{\sqrt 2})^8=1$
and then the powers repeat themselves.

Using this information we can split the power series to the real part and to the imaginary part. Namely the real part is equal to
$\frac{1}{\sqrt 2}-\frac 1 3 \frac{1}{\sqrt 2}-\frac 1 4 - \frac 1 5 \frac{1}{\sqrt 2} + \frac 1 7 \frac{1}{\sqrt 2} + \frac 1 8 + \frac 1 9 \frac{1}{\sqrt 2} - \frac 1{10} \frac 1 {\sqrt 2} - \ldots$
In this series there are always three positive terms followed by three negative ones, then three positive ones and so on. Moreover the general term in the series tends to 0. By a result on convergence of alternating series this series converges.

Similar considerations show that the imaginary part of the series is also a convergent series.

Hence the series itself is convergent.

Q: Fisher, section 2.2, Question 22 parts a and b.

Please do type the question itself, not just a reference to it.

A: a) The crucial observation is that quite generally the coefficients of the power series can be reconstructed from the function defined by it. Namely we have the following:
$a_0=f(z_0); a_1=f'(z_0); a_2=f''(z_0)/2; \ldots a_n=f^(n)(z_0)/n!$
(this follows from the power series expansion of f', f'' etc). In particular if the function vanishes everywhere around the point z_0, then all its derivatives vanish at z_0, and hence all a_i's also vanish.

b) If we apply the result of a) to the power series
$0=F(z)-G(z)=\sum_{n=0}^\infty (a_n-b_n) (z-z_0)^n$
we get the desired conclusion a_n=b_n

Q: For what values of z does the series
$\sum_{n=0}^\infty (z^n-z^{n+1})$
converge?

A: If you compute the partial sum
$\sum_{n=0}^N (z^n-z^{n+1})$
you get
$1-z^{N+1}$
since all the terms between the first one and the very last one cancel each other. As N tends to infinity this expression has a limit only if |z|<1 or z=1. If |z|<1 the limit is 1; if z=1, the limit is 0.