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anonymous
 one year ago
How it come that these equations below... This is De Moivre's Theorem , any proofs...
anonymous
 one year ago
How it come that these equations below... This is De Moivre's Theorem , any proofs...

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kropot72
 one year ago
Best ResponseYou've already chosen the best response.0The explanations of multiplication and division of complex numbers in polar form do not involve De Moivre's Formula. The explanations begin with the definition of multiplication, as follows: \[\large z _{1}z _{2}=(x _{1}, y _{1})(x _{2}, y _{2})=(x _{1},x _{2}y _{1} y _{2},\ \ x _{1} y _{2}+x _{2}y _{1})\]

kropot72
 one year ago
Best ResponseYou've already chosen the best response.0Let \[\large z _{1}=r _{1}(\cos \theta _{1}+i \sin\theta _{1})\] and \[\large z _{2}=r _{2}(\cos \theta _{2}+i \sin \theta _{2})\] then by the definition of multiplication the product is at first \[\large z _{1}z _{2}=r _{1}r _{2}[(\cos \theta _{1}\cos \theta _{2}\sin\ \theta _{1}\sin \theta _{2})+\] \[\large i(\sin \theta _{1}\cos \theta _{2}+\cos \theta _{1}\sin \theta _{2})]\]

kropot72
 one year ago
Best ResponseYou've already chosen the best response.0The addition rules for sine and cosine now give us: \[\large z _{1}z _{2}=r _{1}r _{2}[\cos(\theta _{1}+\theta _{2})+i \sin(\theta _{1}+\theta _{2})]\]
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