czy \(\displaystyle{ arg(z_1 z_2) = arg(z_1) arg(z_2)}\) czy \(\displaystyle{ arg(z_1 z_2) = arg(z_1) + arg(z_2)}\) ?
pozdrawiam!
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Crizz
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Zdecydowanie to drugie
[ Dodano: 4 Maj 2008, 21:32 ]
Niech \(\displaystyle{ z_{1}=|z_{1}|(cos\alpha_{1}+isin\alpha_{1}),z_{2}=|z_{2}|(cos\alpha_{2}+isin\alpha_{2})}\), wtedy \(\displaystyle{ z_{1}z_{2}=|z_{1}z_{2}|((cos\alpha_{1}cos\alpha_{2}-sin\alpha_{1}sin\alpha_{2})+i(sin\alpha_{1}cos\alpha_{2}+cos\alpha_{1}sin\alpha_{2}))=|z_{1}z_{2}|(cos(\alpha_{1}+\alpha_{2})+isin(\alpha_{1}+\alpha_{2}))}\)
[ Dodano: 4 Maj 2008, 21:32 ]
Niech \(\displaystyle{ z_{1}=|z_{1}|(cos\alpha_{1}+isin\alpha_{1}),z_{2}=|z_{2}|(cos\alpha_{2}+isin\alpha_{2})}\), wtedy \(\displaystyle{ z_{1}z_{2}=|z_{1}z_{2}|((cos\alpha_{1}cos\alpha_{2}-sin\alpha_{1}sin\alpha_{2})+i(sin\alpha_{1}cos\alpha_{2}+cos\alpha_{1}sin\alpha_{2}))=|z_{1}z_{2}|(cos(\alpha_{1}+\alpha_{2})+isin(\alpha_{1}+\alpha_{2}))}\)