Math 148: Precalculus

z:= x+yi
w:= a+bi

zw = (xa-yb) + (xb+ya)i

u:= xa-yb
v:= xb+ya

$f: \Bbb C \rightarrow M_2(\Bbb R)$
$f(p+qi) := \begin{pmatrix} p & q \\ -q & p\end{pmatrix}$

$f(z)f(w) = \begin{pmatrix} x & y \\ -y & x\end{pmatrix} \begin{pmatrix} a & b \\ -b & a \end{pmatrix} = \begin{pmatrix} xa-yb & xb+ya \\ -ya-xb & xa-yb \end{pmatrix} = \begin{pmatrix} u & v \\ -v & u\end{pmatrix} = f(zw)$

f(z)+f(w) = f(z+w) trivially.

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Math 148: Precalculus

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