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1

It's a mistake that can be corrected by letting $0 < \rho < 1$ and choosing the contour of integration to be $|z| = \rho$, instead of $|z| = 1$. Then the ML inequality gives $|a_k| \le \rho^{-k}$. Letting $\rho \to 1^{-}$, deduce $|a_k| \le 1$.


1

It's better to prove it directly. Let $z$ such that $d(z,\operatorname{fr}(G)) < r/2$. Then there is a $w\in \operatorname{fr}(G) \cap B_{r/2}(z)$. Since $w\in \operatorname{fr}(G)$, by definition of $r$ we have $B_r(w) \cap \{\gamma\} = \varnothing$. Then $B_r(w)$ is a connected set containing $z$ and intersecting $\mathbb{C}\setminus G$, hence $V \cap ...


1

make the substitution $w=z+1$ so $|z+1| \lt 1 $ means $|w| \lt 1$. now note that: $$ \frac1{(1-w)^2} = \sum_{k=0}^\infty \binom{-2}{k} (-w)^k = 1 +2w+3w^2+... $$


1

Your proof is completely correct, and I will continue from : $$f(x)-f(x-1)=i$$ This means that : $$\forall k\in \Bbb N \ \ \ \ \ \ f(k)-f(k-1)=i $$ This implies that: $$\forall k\in \Bbb N \ \ \ \ \ \ f(k)=ik+f(0) $$ so the equation $f(x)-xi-f(0)=0$ has an infinite number of solutions hence: $$\forall x\in \Bbb C \ \ \ \ \ \ f(x)=ix+a $$ with $a\in \Bbb ...


1

If $$f(z)=\frac{1}{z^2}$$ then $$f^{(n)}(z)=(-1)^{n}(n+1)!z^{-{n+2}}$$ so that $$f^{(n)}(z=-1)=(n+1)!$$ Thus, $$\begin{align} f(z)&=\sum_{n=0}^{\infty} \frac{f^{(n)}(-1)}{n!}(z+1)^n\\\\ &=\sum_{n=0}^{\infty} (n+1)\,(z+1)^n \end{align}$$


1

This more like the Taylor Expansion about $z=-1$ of the function $$ f(z)=\frac{1}{z^2}-1 $$ Computing the first derivatives you will see that $$ f^{(0)}(-1)=0; \ \ f^{(1)}(-1)=2; \ \ f^{(2)}(-1)=6=3\cdot2; \ \ f^{(2)}(-1)=14=4\cdot6 $$ So you deduce that $$ f^{(n)}(-1)=(n+1)n! $$ By Taylor $$ ...


1

The radius of convergence of the following power series $$ e^{z-1}=\sum_{n=0}^\infty \frac{(z-1)^n}{n!} \tag1 $$ is infinite, as may be seen by the ratio test for example. Thus $(1)$ is valid for any finite complex value of $z$, and you get your identity by writing $$ e^{z}=e\:e^{z-1} \tag2 $$ and using $(1)$.


1

If $f'(x) = f(x)$ for all $x \in \mathbb R$ then $f'(z) = f(z)$ for all $z \in \mathbb C$ (Identity Theorem). If follows that the derivative of $g(z):= f(z)e^{-z}$ is identically to zero so that $g$ is a constant and $$ f(z) = C e^z \text{ for some } C \in \mathbb C \, . $$ $C$ is not zero because $f$ is assumed to be non-constant. (If $f$ is real-valued ...


1

Since $z_0$ is inside the unit disc, $\bar{z}_0^{-1}$ is outside the disc, and in particular not inside the contour of integration. Hence, by Cauchy's Theorem, the underlined integral is zero.


1

It seems to me that if you define $$ \mathbf v \cdot \mathbf w = \langle \mathbf w, \mathbf v \rangle $$ you find that both equations are consistent. The main point is that you want $$ \langle c\mathbf w, c\mathbf v \rangle = |c|^2 \langle \mathbf w, \mathbf v \rangle $$ and you get that in both definitions.


7

For the purposes of this exercise we can treat the complex square root, written in polar form like $re^{i\phi}\mapsto \sqrt re^{i\phi/2}$, as a continuous function from the closed upper half plane $\bar{H}$ to itself, so above we specify $r\ge0,\phi\in[0,\pi]$. If $z\in\bar{H}$ then so is $i+z$, and we get a function $f:\bar{H}\to\bar{H}$ that can be ...


0

We can write $$z^2 = i+z$$ Let $z = a+bi$ Now we have $$(a+bi)^2 = i+(a+bi)$$ $$(a^2-b^2)+2abi = a+(b+1)i$$ We have two equations $$a^2-b^2 = a$$ And $$2ab = b+1$$ Use wolfram to do algebra, appears that there are two solutions http://www.wolframalpha.com/input/?i=a%5E2-b%5E2+%3D+a%2C2ab+%3D+b%2B1


9

Hint If we denote $$x = \sqrt{i + \sqrt{i + \sqrt{i + \cdots}}},$$ then formally squaring and rearranging gives that $x$ ought to satisfy the quadratic equation $$x^2 - x - i = 0,$$ leaving two possibilities for $x$. To solve the problem rigorously, you'll need: to treat carefully the usual issue with nonintegral powers of complex numbers---this should ...


1

Assume first that $t \gt 0$. To use Cauchy's theorem, close the integration contour with a circular arc to the left of radius $R$ as follows. ($D$ is opposite $C$ along the imaginary axis.) Now consider $$\oint_{ABCDA} dz \, F(z) e^{z t} $$ where $$\int_{AB} dz \, F(z) e^{z t}= \int_{x-i \sqrt{R^2-x^2}}^{x+i \sqrt{R^2-x^2}} ds \, F(s) e^{s t} $$ ...


0

The assumption of continuity is in the first clause, before "and". The curve $\gamma$ is introduced in the second clause. Therefore, continuity is assumed in $\Omega$. By the way: the assumption of continuity is redundant here, because having a primitive $F$ implies it. Indeed, $F$ is complex-differentiable, hence infinitely smooth, hence so is $f=F'$.


1

Your integral is the imaginary part of $$\int_{-\infty}^\infty f(x)e^{ix}\, dx,$$ where $$f(z) = \frac{z}{2(z^2 + a^2)(z^2 + b^2)}.$$ Show that if $R > \max\{a,b\}$, then for all $z$ in the semicircular arc $\{Re^{i\theta}: 0 \le \theta \le \pi\}$, $$|f(z)| \le \frac{R}{2(R^2 - a^2)(R^2 - b^2)}$$ This implies $\lim_{R\to \infty} \max_{\theta \in ...


0

Denote the only pole inside $|z|\leq 1$ by $a$ and show that it is also the only pole inside $|z-a|\leq 1$. Consider the square $C$ with vertices $a\pm\frac{\sqrt2}{2}$, $a\pm\frac{i\sqrt2}{2}$ inscribed inside the circle $|z-a|\leq 1$, and oriented counterclockwise. Integrate $f$ along $C$: if the pole is simple, the result should be non-zero by residue ...


2

A useful thing to know is that it comes from a power series: indeed, $\sum_{k=0}^{+\infty}{n-4k\choose k}a^{-5k}$ is the $n$-th coefficient of the Taylor series at $0$ of $$\frac{1}{1-x-\left(\frac{x}{a}\right)^5}$$ This is the key. Let $\zeta$ be the positive zero of $p(z) = 1 - z - \left(\frac{z}{a}\right)^5$. Then $\zeta$ is the zero of $p$ with the ...


2

Expanding the series is not so bad really. Rewrite the thing as $$\frac{1}{z^4}\cdot \frac{e^z}{(\sin z)/z}.$$ We want the coefficient of $z^3$ in the expansion of the second quotient. Now $(\sin z)/z = 1 - (z^2/6 + O(z^4)),$ so its reciprocal is $1+(z^2 + O(z^4)).$ So we are looking at $$(1+z+z^2/2 + z^3/6 + O(z^4))(1+z^2/6 + O(z^4)).$$ Finding the ...


1

I don't see any problem defining "set of nonzero complex numbers" as $\mathbb{C}-\{0\}$


2

I wouldn't have written $\Bbb R\ni a+bi\neq 0$, since it seems to imply that $a+bi$ is a real number. Writing $a+bi\neq 0\text{and} a,b\in\Bbb R$ should suffice. Or, since $\Bbb C$ was already defined, $\{x\mid x\in\Bbb C, x\neq 0\}$, or $\{x\in\Bbb C\mid x\neq 0\}$.


0

I think part (i) is a single question. They want you to find the image of the point $z=C+iC$. Otherwise it would have been split into two separate questions.


5

They mean two separate problems there. Do first $x=C$; then, similarly, do $y=C$. EDIT: The best way to understand this problem is with geometry, although you can of course do a bunch of algebraic computation. Note that $f(z)=\dfrac1z=\dfrac{\bar z}{|z|^2}$, so you are reflecting first across the real axis and then reflecting in the unit circle (or vice ...


0

At any point where the CR hold, use the definition of the derivative as a limit to check if the function is differentiable or not. Since you expect it not to be, check what happens along some curves. Usually before going to the definition, one should check fast if the partial derivatives are continuous at that point. Is yes, you know the function is ...


1

Using the limit formula for higher order poles, and the fact that $f(z)=\dfrac{e^z}{z^3\sin z}$ admits an order $4$ pole at $z_0=0$, we get: $$\mathrm{Res}(f,0)=\dfrac{1}{3!}\lim_{z\to 0}\dfrac{d^3}{dz^3}\left[z^4\dfrac{e^z}{z^3\sin z}\right]=\dfrac{1}{6}\cdot 2=\dfrac{1}{3}$$


1

you have $$\sin x \cosh y = 3, \cos x \sinh y = 0 $$ take the second equation. you have $$\sinh y = 0 \to y = 0 \\ \cos x = 0, x = \pm \pi/2 +2k\pi $$ putting $y = 0,$ in the first equation gives $\sin x = 3$ which has no real solution. we are now left with $$x = \pm \pi/2 +2k\pi \to \cosh y = \pm 3.$$ since $\cosh y \ge 1,$ we only need to solve $$\cosh y ...


0

$$\cos z=\pm\sqrt{1-3^2}=\pm2\sqrt2i$$ $$e^{iz}=\cos z+i\sin z=\pm2\sqrt2i+3i$$ $$\implies iz=Log[i(3\pm2\sqrt2)=Log(3\pm2\sqrt2)+Log(i)$$ As $i=\cos\dfrac\pi2+i\sin\dfrac\pi2=e^{i\dfrac\pi2},Log(i)=2n\pi i+i\dfrac\pi2$ where $n$ is any integer


1

\begin{align} (-1)^{2/3} & = (e^{\pi i})^{2/3} \\ & = e^{2/3\pi i} \\ & = \cos\left(\frac{2\pi}3\right) + i\sin\left(\frac{2\pi}3\right) \\ & = -\frac 12+i\frac{\sqrt 3}{2} \end{align} So yes, we did use $(-1)^c = e^{ic\pi}$, which uses the definition of the complex exponent $a^b=e^{b\ln a}$ where the principal branch of the natural ...


0

I have two recommendations for complex analysis: One that I have used extensively that is great is Schaum's outline of complex variables by Murray: http://www.amazon.co.uk/Schaums-Outline-Complex-Variables-2ed/dp/0071615695/ref=sr_1_1?ie=UTF8&qid=1430573254&sr=8-1&keywords=schaum+outline+complex+analysis The cost varies between £10-£15 (that is ...


3

A simple counterexample is $$f(z) = 1 + z^{1/2}.$$ The definition of $z^r$ used here implies that $\operatorname{Re} z^{1/2} \geqslant 0$ for all $z\in \mathbb{C}$, and therefore $\operatorname{Re} f(z) \geqslant 1$, which shows $f$ has no zero. Any exponent $0 < r < 1$ would give a zero-free function in the same manner.


0

Rewrite the integral as $$-\oint_C dz \frac{e^z}{\cos^2{z}} $$ Note that there are double poles at $z=\pm \pi/2$ within $C$. ($z=3 \pi/2$ is not within $C$ and therefore does not contribute to the integral.) The residue at $z=\pi/2$ is therefore given by $$-\left \{ \frac{d}{dz} \left [\left ( z-\frac{\pi}{2} \right )^2 \frac{e^z}{\cos^2{z}}\right ] ...


3

These are elliptic functions. The fragment "except for poles" is just an indication that the author has reservations to write $f(z_0) = f(z_0+a) = f(z_0+b)$ when $z_0$ is a pole of $f$. If you view the functions as having values in $\mathbb{C}$, then the poles don't belong to the domain, and hence the above equation would not make sense. If you view your ...


0

I don't know what your thought process was for the residue calculation. What needs to occur is to compute $Res(f,\frac{\pi}{2}) = \lim_{z\to \pi/2}(z-\frac{\pi}{2})\frac{e^{z}}{\sin(2z)}.$ Since $e^z$ is well behaved, thew limit of interest is really $\lim_{z\to \pi/2}\frac{z-\frac{\pi}{2}}{\sin(2z)}=\lim_{z'\to 0}\frac{z'}{\sin(2z' +\pi)}= -\lim_{z'\to ...


0

$$sin(z)=3$$ Take the inverse sine of both sides, we get: $$z=2\pi n+\pi-sin^{-1}(3)$$ Or: $$z=2\pi n+sin^{-1}(3)$$ (with n is the element of Z -> the set of integers) And if you know maybe from the university, we can write sin^-1(3) in a different way, so we get the following solutions: $$z=2\pi n+\pi-\left(\frac{1}{2}\pi ...


4

Since $w\mapsto\sin w$ is an entire function, we can write it as its Taylor series around $0$: $$ \sin w=\sum_{n=0}^{+\infty}\frac{(-1)^n}{(2n+1)!}w^{2n+1} $$ thus writing $w=\frac1z$ we get $$ \sin\frac1z=\sum_{n=0}^{+\infty}\frac{(-1)^n}{(2n+1)!}\frac1{z^{2n+1}} $$ which is valid on the punctured plane $\Bbb C^{\times}$, hence $$ ...


0

Hint: Just apply to $u'_x$ and $u'_y$ what you proved the line above for $u$.


0

Let you want to solve the equation $sin(z)=\omega$. By the definition of the $sin(z)$ we should have: $$\frac{\mathbb e^{iz}-\mathbb e^{-iz}}{2i}=\omega$$ By multiplying to side of above equation in the $\mathbb e^{iz}$ we get: $$\mathbb e^{2iz}-2i\omega\mathbb e^{iz}-1=0$$ Let $\mathbb e^{iz}=y$, then: $$y^2-2i\omega y-1=0$$ The last equation is a ...


3

Choose the branch cuts as $(-\infty,-1]$ for $(z+1)^{-1/2}$ and $(-\infty,+1]$ for $(z-1)^{-1/2}$. Then, $f(z) =(z^2-1)^{-1/2}$ is continuous across the negative real axis and the "effective" branch cut is $[-1,+1]$. We will integrate $f$ on the clockwise contour $C$, which is the "dog-bone" clock-wise contour that encompasses $z=\pm 1$. To that end, we ...


0

Use the right tool for the right task. In this context, it is much easier to integrate as follows: $$\int_0^1 \dfrac{dx}{\sqrt{x^2-1}} = \underbrace{i\int_0^1\dfrac{dx}{\sqrt{1-x^2}} = i\int_0^{\pi/2} \dfrac{\cos(t)dt}{\cos(t)}}_{x = \sin(t)} = \dfrac{i \pi}2$$


1

Yes it is a real number. But it is also a complex number as well. Every number is a complex number. $$\mathbb N \subset \mathbb Z \subset \mathbb Q \subset \mathbb R \subset \mathbb C$$ $a^2+b^2$ is a complex number where the imaginary part is $0$, i.e., Im$(a^2+b^2)=0$.


2

Think of this way: let $z_1 = r_1 \angle \theta_1$ and $z_2 = r_2 \angle \theta_2$, then $z_1 z_2 = r_1 r_2 \angle (\theta_1+\theta_2)$. Now a picture saves the day!


6

Yes, it is always in $\mathbb R$. In fact it is the square of the length of the vector pointing to the point in the complex plane.


5

What you need to notice is that $$ \frac{f(z)}{z} = \frac{u+iv}{x+iy} = \frac{(u+iv)(x-iy)}{x^2+y^2}, $$ of which the real part is $$ \frac{xu+yv}{x^2+y^2}, $$ which is equal to $e^x \cos{y}$ according to the problem. Now find the harmonic conjugate of $e^x \cos{y}$, which is well-known to be $e^x \sin{y}$ for obvious Euler's formula-related reasons, and ...


2

We have $xu+yv=(x^2+y^2)e^x\cos y$. For notational efficiency, let $G(x,y) = (x^2+y^2)e^x\cos y$. Here is a procedure for finding Step 1: Take the partial derivative with respect to $x$ and multiply by $x$ to find that $$x^2u_x+xu+xyv_x=xG_x$$ where the subscript $x$ means the first partial with respect to $x$. Step 2: Take the partial ...


0

Let $z=x+iy$, $w=x-iy$. Inverting these equations, we find that $$ x = \frac{1}{2}(z+w), \qquad y = \frac{1}{2i}(z-w), $$ so, substituting in, $$ f(z) = u\left( \frac{z+w}{2},\frac{z-w}{2i} \right) + i v\left( \frac{z+w}{2}, \frac{z-w}{2i} \right). \tag{1} $$ Setting $z=z_0$ gives $$ f(z_0) = u\left( \frac{z_0+w}{2},\frac{z_0-w}{2i} \right) + i v\left( ...


0

In general, $f$ need not be holomorphic on the entire plane. Let $K\subset \mathbb{C}$ be a compact set with empty interior and positive Lebesgue measure, such that its complement is connected, e.g. a product of two fat Cantor sets. Define $$f(z) = \int_K \frac{d\overline{\zeta}\wedge d\zeta}{\zeta-z}.$$ Then $f$ is holomorphic on $\mathbb{C}\setminus K$, ...


0

By direct calculation, $\sum_{n\neq0}\frac{\left(-1\right)^{n+1}}{in}e^{in\theta} =\sum_{n=-\infty}^{-1}\frac{\left(-1\right)^{n+1}}{in}e^{in\theta}+\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n+1}}{in}e^{in\theta} =\sum_{n=1}^{\infty}\frac{\left(-1\right)^{-n+1}}{\left(-1\right)in}e^{-in\theta}+\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n+1}}{in}e^{in\theta} ...


0

Proof: Note that the function $f$ has no zeros because of the given inequality. Define a holomorphic function $g:\mathbb{C}\backslash\{0\}\to \mathbb{C}$ as $$ g(z)~:=~\frac{1}{zf(z)^2}.\tag{1}$$ Note that $|g|\leq 1$ is bounded because of the given inequality. The function $g$ has a removable singularity at $z=0$, cf. Riemann's theorem. So the extension ...


1

$$\lim_{z\to \pi/2}(z-\pi/2)\frac{\sin z}{\cos z} = \lim_{z\to \pi/2} \frac{z-\pi/2}{\cos z - \cos (\pi/2)}\sin z$$ $$ = \frac{1}{\cos '(\pi/2)}\cdot 1= -1.$$ It follows that $\tan z$ has a simple pole at $0,$ hence the prinicpal part is $-1/(z-\pi/2).$



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