How to determine  $U( Z[i])$  I tried this  $$(a+bi)(c+di) = 1,$$  where $a,b,c,d$ are integers.  and compared real no. With real
43565
How can we show that $\pm 1, \pm i$ are the only units in the ring of Gaussian integers, $\mathbb Z[i]$?  Thank you.
25511
In my algebra class I was given as homework assignment to prove that:     Given an integral domain $A$ and $a,b,c,d,e \in A$. Then if $d = \gcd(b,c)$ and $e = \gcd(ac, ab)$ then $e = ad$.   It is easy to see that $ad \mid ab$ and $ad \mid ac$, so this implies that $\exists q \in A; e = q(ad)$. Now I'm having problems showing that $q = 1_A$. Working through some equalities, I proved that $\forall n\in \mathbb{N}, q^n \mid d$ so this gives me the light suspicion that I might be going the right way, but I'm pretty positive that this does not imply $q = 1_A$.  Does anyone want to give me a hint?  Thanks in advance :)
48783
Find all $z$ for the equation   $$(z+1)^7 = z^7$$    The different solutions can be unsimplified and both rectangular or exponential. I have the lead that I subsitute the $z+1$ term with a root of unity. Then, I got lost when I looked at that $z^7$. A thorough explanation would be great!
33584
If $z_{1},z_{2},z_{3}$ are three complex number such that $|z_{1}| = |z_{2}| = |z_{3}| = 1$ and $\displaystyle \frac{z^2_{1}}{z_{2}z_{3}}+\frac{z^2_{2}}{z_{3}z_{1}}+\frac{z^2_{3}}{z_{1}z_{2}} = -1.$Then possible values of $|z_{1}+z_{2}+z_{3}|$  $\bf{My\; Try::}$ Given $$\displaystyle \frac{z^2_{1}}{z_{2}z_{3}}+\frac{z^2_{2}}{z_{3}z_{1}}+\frac{z^2_{3}}{z_{1}z_{2}} = -1\Rightarrow z^3_{1}+z_{2}^3+z_{3}^3=-z_{1}z_{2}z_{3}$$ Now Using $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$ So $$\left(z_{1}+z_{2}+z_{3}\right)\left[z^2_{1}+z^2_{2}+z^2_{3}-z_{1}z_{2}-z_{2}z_{3}-z_{3}z_{1}\right] = -4z_{1}z_{2}z_{3}$$ So $$(z_{1}+z_{2}+z_{3})\cdot ((z_{1}+z_{2}+z_{3})^2-3(z_{1}z_{2}+z_{2}z_{3}+z_{3}z_{1})) = -4z_{1}z_{2}z_{3}.......(1)$$ Now Let $k=z_{1}+z_{2}+z_{3}.$ Taking Conjugate, we get $\displaystyle k = \frac{1}{z_{1}}+\frac{1}{z_{2}}+\frac{1}{z_{3}} = \frac{z_{1}z_{2}+z_{2}z_{3}+z_{3}z_{1}}{z_{1}z_{2}z_{3}}$ So we get $z_{1}z_{2}+z_{2}z_{3}+z_{3}z_{1}=kz_{1}z_{2}z_{3}$ Put into $(1)\;,$ We get $$(z_{1}+z_{2}+z_{3})\cdot \left[(z_{1}+z_{2}+z_{3})^2-3kz_{1}z_{2}z_{3}\right] = -4z_{1}z_{2}z_{3}$$ Now How can I solve after that, Help Required, Thanks
107147
Is there an identity  for $\Gamma(z) \Gamma(\bar {z})$?  Post-question edits:  Have you come across any, in addition to ? (there, special cases, where the real part of $z$ is integer or half integer, are also listed.)    lists the following under 6.1.45 $$\lim_{|y|\to \infty} (2\pi)^{-1/2} |\Gamma (z)| e^{\pi|y|/2} |y|^{1/2-x} =1$$
6678
Show Möbius transformation which maps $\Im(z)&gt;0$ to itself iff  $$ f(z)= \frac{az+b}{cz+d}\,,\,\,ad-bc&gt;0$$ and $a,b,c,d$ are real.
24665
Prove: if $|z|=1$ than $\frac{z+1}{z-1}$ is an imaginary number    I have tried to look at $$\frac{z+1}{z-1}=\frac{z+1}{z-1}\cdot \overline{\frac{z-1}{\overline{z-1}}}=\frac{z+1}{z-1}\cdot \frac{\overline{z}-1}{\overline{z}-1}$$  But did not get far
28844
Describe the image of the set $\{z=x+iy:x&gt;0,y&gt;0\}$ under the mapping $w=\frac{z-i}{z+i}$  So from this mapping , I can see that $a=1, b=-i, c=1, d=i$ thus $ad-bc=i+i=2i \not =0$ so this is a Mobius transformation. Solving for $z$ I got  $$z=\frac{i+iw}{1-w}$$  for $w=u+iv$, we have   $$z=\frac{-2v+i(1-u^2-v^2)}{(1-u^2)+v^2}$$  so $x=\frac{-2v}{(1-u^2)+v^2}$ and $y=\frac{1-u^2-v^2}{(1-u^2)+v^2}$  Since $x&gt;0$, $v&lt;0$  and since $y&gt;0$, $1-u^2 -v^2 &gt;0$, thus $u^2 +v^2 &lt;1$, this implies that the image is the interior of a unit circle center at the origin, but since $v&lt;0$, we only take the negative part.
105560
Let $U=M_{n}(\mathbb{C})$, and we define a bilinear form $\xi (A,B)=n\cdot tr(AB)-tr(A)tr(B)$. How do I find $dim(U_{\perp })$?  I know that $U_{\perp }=\{A \in M_{n}(\mathbb{C})|\ \forall B\in M_{n}(\mathbb{C}), n\cdot tr(AB)-tr(A)tr(B)=0 \}$. I have no idea how to continue. I'd appreciate any help. Thanks.
115397
Find all units in $\mathbb{Z}[\sqrt n] = \{a + b\sqrt n \mid a, b \in \mathbb Z \}$  and $n \in\mathbb N$, $n\ge 2$.  First let $c +d\sqrt n$ be a unit so $$(a +b\sqrt n)(c +d\sqrt n) = 1,$$ $$ac + bdn +(bc + ad)\sqrt n = 1.$$  What next?
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!
103209
$$ f(z) = \frac{z+1}{i-2z} \{ z ∈ \Bbb C,  f(z) ∈ \Bbb R \}$$  Are there any solutions other than $z = -1$?
74091
Problem: Show that if $(b,c)=1$, then for any integer $a$ we have $(a,bc)=(a,b)(a,c)$. [Hint: Prove that each member of the alleged equation divides the other.]  This is the way I proved it (without the hint): $(a,b)(a,c)=(a(a,b),c(a,b))=(a^2,ab,ac,bc)=(a^2,bc,a(b,c))=(a^2,a,bc)=((a^2,a),bc)=(a,bc)$  But how should I prove it using that hint or at least a similar proof.
96366
I am self studying algebra, and have come across this exercise:     Show that $\mathbb{Z}[i]/(2+3i)$ has 13 elements and is a field.   My proof: Since Z[i] is a euclidian domain with euclidian norm function $f(c+id)= c^2+d^2$, for any element $a+bi \in Z[i]$, we can "divide" it by $2+3i$ to get $a+bi= q (2+3i) + r$ where either $r=0$ or $r=c+di$ for some integers $c,d$ not both $0$ with $c^2 + d^2=f(c+di) &lt; f(2+3i)= 13$. But then $a+bi$ belongs to the coset $ r + (2+3i) $. So every element in the quotient ring belongs to some $r + (2+3i)$ for $r=0$ or $r=c+di$ with $c^2 + d^2&lt;13$. Now we can count all the possible pairs of integers satisfying this condition, ie $(0,0),(0,1),(0,2),(0,3), ..., (3,1)$ to get 13 and so there are 13 elements in the quotient ring.   Is this a correct argument? Also is there an elegant way of showing it is a field rather than showing its an integral domain by showing each element has an inverse by brute force? Thanks in advance!
54133
Find the value of $\large i^{i^{.^{.^.}}}$ ?  How should we start to solve it ?  Also you can see  if it helps.  Thanks
94584
I am trying to prove that in Euclidean domain D with Euclidean function d, u in D is a unit if and only if d(u)=d(1).  Suppose u is a unit, then there exist v in D such that uv=1, this implies u\1 so d(u)&lt;=d(1), but obviously 1 divides u so d(1)&lt;=d(u). Hence, d(u)=d(1).  Conversely, suppose  d(u)=d(1), since u is not zero, there exist q and r in D such that  1=uq+r with r=0 or d(r)&lt; d(u).  If r=0 then u is a unit. Else d(r)&lt; d(u) =d(1), this implies d(r)&lt; d(1). I stop here, because I failed to argue that r must be zero.  Can anyone help me? Thanks.
7074
I tried:  $$|z-2i| = 2|z+3| \Leftrightarrow \\ |x+yi-2i|=2|x+yi+2|\Leftrightarrow \\ \sqrt{x^2+(y-2)^2}=\sqrt{4((x+2)^2+y^2)} \Leftrightarrow \\ \sqrt{x^2+y^2-4y+4} = \sqrt{4x^2+24x+36+4y^2} \Leftrightarrow \\ x^2+y^2-4y+4 = 4x^2+24x+36+4y^2 \Leftrightarrow \\ y^2-4y-4y^2=4x^2+24x+36+x^2 \Leftrightarrow \\ -3y^2-4y=5x^2+24x+26 \Leftrightarrow \\ ???$$  What do I do next?