did hw1 in math

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+\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