I have been trying to find the image of $S=\{z\in\mathbb{C}; 0&lt;|z|&lt;r\}$, with $r&gt;0$, under $w=e^{1/z}$.  I started this way: if $z=|z|e^{i\theta}\in S$, by Euler's formula $$     1/z   = |z|^{-1}(\cos\theta-i\sin\theta)   = |z|^{-1}\cos\theta-|z|^{-1}i\sin\theta. $$ Considering $w=|w|e^{i\alpha}$ as image of $z$, $$     |w|e^{i\alpha}   = e^{|z|^{-1}\cos\theta} e^{-|z|^{-1}i\sin\theta}. $$ And then $$            |w| = e^{|z|^{-1}\cos\theta}   \implies \ln|w| = |z|^{-1}\cos\theta   \implies \cos\theta = |z| \ln|w|. $$ And also $$            \alpha =-|z|^{-1}i\sin\theta   \implies \sin\theta =-\alpha |z|. $$  What do I have to do now?
122194
I can't do this exercise of Conway's Book: For $r&gt;0$ let $A=\{w:w=\exp(1/z), 0&lt;|z|&lt;r\}$, determine the set $A$. Any hints?
120562
Is there $n\in \mathbb{N}$ such that $\ln(\ln(n)) \in \mathbb{Q}$? If such $n$ exists, we will get $$\ln(\ln(n))  = \frac{p}{q}, \quad p, q \in \mathbb{Z}.$$  Hence we will get $n = e^{e^{p/q}},$ where the question about the nature of  $e^{e^{p/q}}$ haven't been answered yet.   Is there any other direction ?  Thank you in advance.
21643
Euler’s formula states that $e^{i  x} = \cos(x) + i \sin(x)$.  I can see from the MacLaurin Expansion that this is indeed true; however, I don’t intuitively understand how raising $e$ to the power of $ix$ produces rotation.  Can anyone give me an intuitive understanding?
99386
Show that $\lvert e^{i\theta} - 1\rvert = 2\lvert\sin(\theta/2)\rvert$ by using the geometry of the triangle with vertices 0, 1, and the midpoint of the line joining 0 and $e^{i\theta}$.  I have been able to show this identity through other means, however I am stuck on how to utilize this particular triangle.
43082
Problem :   If $z =re^{i\theta}$, then prove that $|e^{iz} | =e^{-r\sin\theta}$   My working :   $z = re^{i\theta} = r(\cos\theta + i\sin\theta)$   $\Rightarrow iz = ir(\cos\theta +\sin\theta) $ = $-r\sin\theta +ir\cos\theta $  $\Rightarrow e^{iz} =e^{(-r\sin\theta +ir\cos\theta)} = e^{-\sin\theta} e^{ir\cos\theta} $   $\Rightarrow |e^{iz}| = |e^{-r\sin\theta}||e^{ri\cos\theta}|$   Now please guide how to proceed further to get the result... thanks.
10198
Prove if $|z| &lt; 1$ and  $ |w| &lt; 1$, then $|1-zw^*| \neq 0$ and $| {{z-w} \over {1-zw^*}}| &lt; 1$Given that $|1-zw^*|^2 - |z-w|^2 = (1-|z|^2)(1-|w|^2)$I think the first part can be proven by saying $|1-zw^*| = 0$ if and only if $zw^*$ = 1.  And given the conditions that cannot be true. However I don't know if this part is right.
6260
I'm trying to work through the exercises in Otto Forster's book on Riemann Surfaces. While most of them seemed not that hard, this one gives me a headache:  Let $X=\mathbb{C}\setminus\{\pm1\}$  and $Y = \mathbb{C}\setminus\{\frac{\pi}{2}+k\pi~|~k\in\mathbb{Z}\}$. Then $\sin: Y\rightarrow X$ is a covering map. This is clear.  Now consider the following curves:  $$ u: [0,1] \rightarrow X, u(t) = 1-e^{2\pi i t}$$  and  $$ v: [0,1] \rightarrow X, v(t) = -1+e^{2\pi i t} = -u(t)$$  Let $w_1: [0,1]\rightarrow Y$ be the lifting of $u\cdot v$ and $w_2$ be the lifting of $v\cdot u$ with $w_1(0)=w_2(0)=0$.  Show that $w_1(1) = 2\pi$ while $w_2(1) = -2\pi$.  As of now, I don't have many ideas how to solve this. Of course, the inverse sine is a multi-valued function so the two liftings should use different branches. But I don't see how exactly they come into play.  I'm also not sure whether the solution can be done by calculations alone or if there is some more general principle underlying. Any suggestions would be welcome.
27106
The entire problem statement is:  Let $n&gt;1$ and let $U=B(0,1)\subset\mathbb{R}^n$. Show that $u:U\to\mathbb{R}$ given by  $$u(x)=\ln\left(\ln\left(1+\frac{1}{|x|}\right)\right)$$ is in $W^{1,n}(U).$  My attempt at the proof is as follows:  To show that $u\in W^{1,n}(U)$ it suffices to show that $|Du|\in L^{n}(U)$. Consider $$u_{x_i}=\frac{1}{\ln\left(1+\frac{1}{|x|}\right)}\frac{1}{1+\frac{1}{|x|}}|x|^{-2}\frac{x_i}{|x|},$$ which simplifies to, $$u_{x_i}=\frac{1}{\ln\left(1+\frac{1}{|x|}\right)}\frac{1}{1+\frac{1}{|x|}}\frac{x_i}{|x|^3}.$$ Thus, $$|Du|=\frac{1}{\ln\left(1+\frac{1}{|x|}\right)}\frac{1}{1+\frac{1}{|x|}}\frac{1}{|x|^2}$$ which can be manipulated to, $$|Du|=\frac{1}{\ln\left(1+\frac{1}{|x|}\right)}\frac{1}{1+|x|}\frac{1}{|x|}.$$  Moreover, since $U=B(0,1)$ we can bound the second term from above which gives, $$|Du|\leq\frac{1}{\ln\left(1+\frac{1}{|x|}\right)}\frac{1}{|x|}$$  Thus, we can instead show that $$\int_U\left(\frac{1}{\ln\left(1+\frac{1}{|x|}\right)}\frac{1}{|x|}\right)^n&lt;\infty.$$  Converting this into polar coordinates we have with $r=|x|$, $$\int_0^R\int_{\partial B(0,r)}\left(\frac{1}{\ln\left(1+\frac{1}{r}\right)}\frac{1}{r}\right)^n\,dSdr=\int_0^R\left(\frac{1}{\ln\left(1+\frac{1}{r}\right)}\frac{1}{r}\right)^n\int_{\partial B(0,r)}dSdr$$ Note that $\int_{\partial B(0,r)}\,dS=n\alpha(n)r^{n-1}$, which is the surface area of $B(0,r)$. Thus, we have then $$n\alpha(n)\int_0^R\left(\frac{1}{\ln\left(1+\frac{1}{r}\right)}\frac{1}{r}\right)^nr^{n-1}\,dr=n\alpha(n)\int_0^R\left(\frac{1}{\ln\left(1+\frac{1}{r}\right)}\right)^n\frac{1}{r}\,dr$$  And so this is where I get stuck in the problem, since I don't how I can evaluate that last integral to show it's finite.  Thank you for any help or feedback!
91329
How would you go about finding all the solutions to $\cos(z)=0$, where $z\in \mathbb{C}$?  I have $$\cos(z)=0 \implies (e^{iz})^2=-1 \implies \text{Log}(e^{iz})=\text{Log}(e^{i\pi(1/2+2k)})\qquad\text{ where }k\in \mathbb{Z}$$  $$\implies ie^{-y}x=i\pi(1/2+2k)$$  $$\implies z= e^{y}\pi(1/2+2k)+iy \qquad\forall k\in \mathbb{Z}\;\text{ and }\;\forall y\in \mathbb{R}$$  Is this correct?  Thanks!  EDIT:  $e^{2iz}=e^{i\pi(1+2k)} \Rightarrow z=\pi/2+k\pi$ where $k\in \mathbb{Z}$
111516
Is $$\sin^2(z) + \cos^2(z)=1$$ still true for all $z \in \Bbb{C}$? I've tried rewriting it using complex definitions of $\sin$ and $\cos$, and I don't see why it wouldn't, but the text I'm reading asks this question as if it shouldn't hold?
62481
I have posted a proof below, and would appreciate it if someone could review it for accuracy. Thanks!   Problem:   Let n $\in$ $\mathbb{Z}$ with $n$ $\ge$ 3. Prove the following:   (a) Z(D$_{2n}$) = 1 if $n$ is odd.   (b) Z(D$_{2n}$) = {1, r$^k$} if $n$ = $2k$.   Note that $r$ and $s$ generate D$_{2n}$ with the group presentation { $r$,$s$ | $r$$^n$ = $s$$^2$ = 1, $rs$ = $sr$$^{-1}$ }   Proof:  part (a)   Let $n$ $\ge$ 3 where $n$ $\in$ $\mathbb{Z}$ and $n$ is odd.   For any x $\in$ Z(D$_{2n}$), x must commute with both $s$ and $r$, since both are in $D_{2n}$. Note that if $x$ commutes with both $s$ an $r$, then $x$ commutes with any element of $D_{2n}$ since $r$ and $s$ generate D$_{2n}$.    Then for any x$\in$ Z(D$_{2n}$), we have $xr$ = $rx$, where x is of the form $x$ = s$^j$r$^w$ (any such element can be arranged into this form via the relation $rs$ = $sr$$^{-1}$ and the fact that $r$ and $s$ generate D$_{2n}$). Note that since r and s have finite order, the exponents are modulo n and modulo 2 for $r$ and $s$ respectively.   Then we have  s$^j$r$^w$$r$ = $r$s$^j$r$^w$.   Which implies s$^j$r$^{w+1}$ = s$^j$r$^{w +- 1}$  In the case where the power of r on the RHS is $w$ - 1, it is clear we cannot have equality unless |r| = 1, which it is not. Note that if $j$ is even then we have only the $w$ + 1 case. Hence $x$ may be of the form $1$$r$$^w$ = $r$$^w$.   Also we have that x must commute with $s$, hence $s$$x$ = $x$$s$.   Then $s$$r$$^w$ = $r$$^w$$s$.  Which implies $s$$r$$^w$ = $s$$r$$^{-w}$.   Applying s$^{-1}$ to both sides we arrive at the task of finding when r$^w$ = r$^{-w}$ i.e when $w$ = $-w$. But since r has finite order we have:   $$ \bar{w} = (n-1)*w $$   where $\bar{w}$ is the residue class of w modulo n. Hence for some a $\in$ $\mathbb{Z}$$_+$   $$ w + an = wn - w $$  $$ 2w = wn - an $$ $$ 2w = n(w-a) $$   Now since 0 $\lt$ w $\le$ n we have that the LHS is greater than 0. Also since n $\ge$ w then (w-a) $\le$ 2, since otherwise the RHS would be greater than the LHS. So since n is odd we must have that (w-a) is even and hence (w-a) = 2, implying that 2w = 2n and hence n = w.   Hence x = $r$$^w$ = $r$$^n$ = 1. So Z(D$_{2n}$) = 1.   part (b)   From the argument above any x $\in$ Z(D$_{2n}$) must be of the form $s$$^j$$r$$^w$ for j even and j $\in$ $\mathbb{Z}$. Then x is of the form $r$$^w$ for $0$ $\lt$ w $\le$ $n$. Since $n$ is even we have the equation below is satisfied when $2w$ = $n$ or $w$ = $n$, since (w-a) $\le$ 2 but also (w-a) $\ge$ 0 for (w-a) $\in$ $\mathbb{Z}$$_+$.   $$ 2w = n(w - a) $$   Hence $r$$^w$ $\in$ Z(D$_{2n}$) when $2w$ = $n$, as desired.
61177
Question: Let $\theta : \mathbb{Z}_8 → \mathbb{Z}_4$ be defined by $\theta([i]_8) = [i]_4$ for all $i \in \mathbb{Z}$. You may assume that this is a well-defined ring homomorphism.    Find Ker($\theta$) Exhibit the partition of $\mathbb{Z}_8$ into cosets of Ker($\theta$).   My Answer: So I found the Ker($\theta$) as Ker($\theta$) $=\{[0]_8, [4]_8\}$ as $\theta[0]_8=[0]_4$ and $\theta[4]_8=[4]_4=[0]_4$.  But I'm stuck on part 2. I was thinking that I would need to show that the partition of $\mathbb{Z}_8$ are equal to $[0]_4$ or $[4]_4$.   Sorry about how I have typed it, haven't really asked questions on the site before.
19102
$$\overline{z}=z^n$$ where $n\in \mathbb{N}$   So I have started with moving the polar representation as the expression is in the n-th power  $$rcis(-\theta)=r^ncis(n\theta)$$   I can not multiply both sides in $r^{(-n)}$ as $n\in \mathbb{N}$ how should I continue?
2958
Why is $\sin(\alpha+\beta)=\sin\alpha\cos\beta+\sin\beta\cos\alpha$? How can we find the RHS if we don't know what it is? (instead of proving the identity itself) I could find a geometric solution in , but is there any solution that doesnt require drawing something? Edit: I now saw some nice proofs using $e$ and Euler's identity. I would appreciate anything NOT using them too, as a change. Out of curiosity, Is there a proof that says you actually "can't" prove the identity using only other simple 1-variable identities like $\sin^2\alpha+\cos^2\alpha=1$?
115495
I am studying for a qualifier exam in complex analysis and right now I'm solving questions from old exams. I am trying to prove the following:     Prove that if $f$ and $g$ are entire functions such that $f(z)^2 + g(z)^2 = 1$ for all $z \in \mathbb{C}$, then there exists an entire function $h$ such that $f(z) = \cos(h(z))$ and $g(z) = \sin(h(z))$.     My Attempt  The approach that occurred to me is the following. Since $f(z)^2 + g(z)^2 = 1$ then we have $(f(z) + ig(z))(f(z) - ig(z)) =  1$. Then each factor is nonvanishing everywhere in $\mathbb{C}$ and thus by the "holomorphic logarithm theorem" we know that since $\mathbb{C}$ is simply connected, there exists a holomorphic function $H:\mathbb{C} \to \mathbb{C}$ such that  $$e^{H(z)} = f(z) + ig(z)$$  and then we can write $\exp(H(z)) = \exp\left(i\dfrac{H(z)}{i} \right) = \exp(ih(z))$,    where  $h(z) := \dfrac{H(z)}{i}$.  Thus so far we have an entire function $h(z)$ that satisfies  $$e^{ih(z)} = f(z) + ig(z)$$  On the other hand, we also know that $e^{iz} = \cos{z} + i \sin{z}$ for any $z \in \mathbb{C}$, thus we see that   $$e^{ih(z)} = \cos{(h(z))} + i \sin{(h(z))} = f(z) + ig(z)$$  Thus at this point I would like to conclude somehow that we must have $f(z) = \cos(h(z))$ and $g(z) = \sin(h(z))$, but I can't see how and if this is possible.    My questions   Is the approach I have outlined a correct way to proceed, and if so how can I finish my argument? If my argument does not work, how can this be proved?   Thanks for any help.
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I'd like to see a proof why $\varphi \in \operatorname{Hom}{(S^1, S^1)}$ looks like $z^n$ for an integer $n$.  At first I thought I could argue that if I have a homomorphism that maps $e^{ix}$ to some $e^{iy}$ for $x,y \in \mathbb R$ then $z = rx$ for some real number $r$. But on second thought I'm not sure why I can't have $\varphi (e^{ix} ) = e^{g(x)i}$ for a $g$ other than $g(x) = \lambda x$.   I'm also interested in seeing different proofs. I'm sure there are several ways to prove this.  Thanks for your help.
10369
The following is a rigorous proof of De Moivre's theorem by means of mathematical induction. The theorem put simply is that:  Any complex number, $z = a+bi$, on a cartesian plane can be expressed in polar form, where $a=r\cos\theta$ and $b=r\sin\theta$ and $r$ is the absolute distance from the origin to the point $z$. In light of this, the expression for the complex number $z$ in polar form is $z=r(\cos\theta+i\sin\theta)$ In order to find the $n^{th}$ power of $z$, the following rule applies: $$\displaystyle z^n=(r(\cos\theta+i\sin\theta))^n=r^n(\cos n\theta+i\sin n\theta)$$ where $n\in\mathbb N$ (I have confined $n$ to the natural numbers given that I intend on using mathematical induction to prove the theorem.  Base step: We must prove that what is stated is true for $n=1$, hence $$z^1=r^1(\cos 1\cdot\theta+i\sin1\cdot \theta)$$ $$=r(\cos\theta+i\sin\theta)$$ which is true. Next we assume it is true for $n=k$, and therefore proceed to the inductive step. Inductive step: When calculating $z^{n+1}$, it's the same as calculating $z^nz$, hence $$z^{n+1}=(r^n(\cos n\theta + i\sin n\theta))(r(\cos \theta+i\sin \theta))$$ $$=r^nr(\cos n\theta+i\sin n\theta)(\cos\theta +i\sin \theta)$$ $$=r^{n+1}(\cos n\theta\cdot\cos\theta+\cos n\theta \cdot i\sin\theta+i\sin n\theta\cdot\cos\theta+i^2\sin n\theta\cdot\sin\theta)$$ from here, $i^2=-1$, and therefore $+i^2\sin n\theta\cdot\sin\theta$ becomes $-\sin n\theta\cdot\sin\theta$. In addition to this, the following trigonometric identities will be used: $$\sin(n\theta + \theta) = \sin n\theta\cdot\cos\theta+\cos n\theta\cdot\sin\theta$$ $$or$$ $$k\sin(n \theta+\theta)=k\sin n \theta\cdot\cos \theta + \cos n\theta\cdot k\sin\theta$$ $$and$$ $$\cos(n \theta + \theta)=\cos n\theta\cdot\cos\theta-\sin n\theta\cdot\sin\theta$$ by using these, the original equation has now become $$=r^{n+1}(\cos(n \theta + \theta) + i\sin(n \theta + \theta))$$ then, after factoring out $\theta$, the theorem becomes $$=r^{n+1}(\cos((n+1)\theta)+i\sin((n+1)\theta))$$ which completes the proof. Question: is this proof correct all the way through? Or have I missed any conditions? If all is well in the case of this proof, could somebody aid me in proving it for $\{n|n\in\mathbb R,n\geq1\}$