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Verify the associative law for multiplication of complex numbers. That is, show that
$$\left(z_{1} z_{2}\right) z_{3}=z_{1}\left(z_{2} z_{3}\right)$$
for all $z_{1}, z_{2}, z_{3} \in \mathbb{C}$.
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Proof. Let $z_{k}=x_{k}+i y_{k}$ for $k=1,2,3 .$ Then
\begin{aligned} \left(z_{1} z_{2}\right) z_{3}=&\left(\left(x_{1}+y_{1} i\right)\left(x_{2}+y_{2} i\right)\right)\left(x_{3}+y_{3} i\right) \\ =&\left(\left(x_{1} x_{2}-y_{1} y_{2}\right)+i\left(x_{2} y_{1}+x_{1} y_{2}\right)\right)\left(x_{3}+y_{3} i\right) \\ =&\left(x_{1} x_{2} x_{3}-x_{3} y_{1} y_{2}-x_{2} y_{1} y_{3}-x_{1} y_{2} y_{3}\right) \\ &+i\left(x_{2} x_{3} y_{1}+x_{1} x_{3} y_{2}+x_{1} x_{2} y_{3}-y_{1} y_{2} y_{3}\right) \end{aligned}
and
\begin{aligned} z_{1}\left(z_{2} z_{3}\right)=&\left.\left(x_{1}+y_{1} i\right)\left(\left(x_{2}+y_{2} i\right)\right)\left(x_{3}+y_{3} i\right)\right) \\ =&\left(x_{1}+y_{1} i\right)\left(\left(x_{2} x_{3}-y_{2} y_{3}\right)+i\left(x_{2} y_{3}+x_{3} y_{2}\right)\right) \\ =&\left(x_{1} x_{2} x_{3}-x_{3} y_{1} y_{2}-x_{2} y_{1} y_{3}-x_{1} y_{2} y_{3}\right) \\ &+i\left(x_{2} x_{3} y_{1}+x_{1} x_{3} y_{2}+x_{1} x_{2} y_{3}-y_{1} y_{2} y_{3}\right) \end{aligned}
Therefore,
$$\left(z_{1} z_{2}\right) z_{3}=z_{1}\left(z_{2} z_{3}\right)$$
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