\noindent{\bf 2.}

If $z = a + bi = (a,b),$ $iz = (0,1)(a,b) = (-b, a).$
${\rm Re}(iz) = -b = -{\rm Im} z$ and ${\rm Im}(iz) = a = {\rm Re} z.$

\noindent{\bf 8.}

    \noindent{\it (a)}

Given $(x,y) + (u,v) = (x,y),$ $(x+u,y+v) = (x,y),$ so $x+u = x$ and
$y+v = y.$ Adding $-x$ and $-y$ to both sides, respectively, $-x+x+u = u
= 0 = -x+x,$ and $-y+y+v = v = 0 = -y+y.$ The real additive identity is
unique, so the complex additive identity is also unique.

    \noindent{\it (b)}

$(x,y)(1,0) = (1x-0y,0x+1y) = (x,y),$ so a multiplicative identity $1$
exists. Let $1'$ also be a multiplicative identity.
$$1 = 1'*1 = 1*1' = 1',$$
by the fact that, if $1$ is a multiplicative identity, $1z = z,$ and
$ab = ba$ under the complex numbers.
and $yu+xv = y \to xv = y-yu = y(1-u).$ If $x\neq0$ and $y\neq0,$
$xv/y = -yv/x$

Let $w = 1$ and $v = 1.$ 

\noindent{\bf 7.}

$${z_1z\over z_2z} = \left({z_1\over z_2}\right) \left({z\over
z}\right).$$

With $z = (x,y),$ from $(4),$
$${z\over z} = zz^{-1} = \left({x^2+y^2\over x^2+y^2},{yx-xy\over
x^2+y^2}\right) = (1,0).$$

\bye