See explanation. Explanation: To find a product of 2 algebraic sums you have to multiply each term of one sum by each term of the other and then reduce the like terms. \displaystyle{\left({u}+{3}\right)}{\left({u}-{3}\right)}={u}^{{2}}-{3}{u}+{3}{u}-{9}={u}^{{2}}-{9}

The given circle is \;(x-2)^2+y^2=4\; , so take any element \;z=(x,y)\sim x+iy\; in it and apply to it \;w\; : w(z):=\frac{2z+3}{z-4}=\frac{2x+3+2iy}{x-4+iy}\cdot\frac{x-4-iy}{x-4-iy}=\frac{(2x^2-5x-12+2y^2)-11yi}{(x-4)^2+y^2} ...

I'm back-interpreting some of your notation, as the question is imprecise at the moment. But I think that having a reasoned-out answer would be helpful. Firstly, a group G is (roughly) a set of ...

The statement you are trying to prove is, indeed, not true if the field has characteristic 2. If u=(1,0) and v=(0,1) then u and v are linearly independent but u+v=u-v=(1,1).

I'll show you 1), if you get stuck with the others, let me know. U consists of all the vectors for which the first coordinate is equal to the second coordinate. We have to check that for all u,v \in U ...

\begin{align*} 2x + 3y& = u\\ x - 4y &= v \end{align*}\quad \underset{\substack{\text{convert to}\\ \text{matrix language}}}{\leadsto}\quad \begin{bmatrix} 2 & \hphantom{-}3\\ 1 &-4\end{bmatrix}\begin{bmatrix} x\\ y\end{bmatrix}=\begin{bmatrix}u\\ v\end{bmatrix} ...