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diff --git a/kang/hw1.tex b/kang/hw1.tex new file mode 100644 index 0000000..591dead --- /dev/null +++ b/kang/hw1.tex @@ -0,0 +1,36 @@ +\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 |