anonymous
  • anonymous
How it come that these equations below... This is De Moivre's Theorem , any proofs...
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
anonymous
  • anonymous
anonymous
  • anonymous
any help ...

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kropot72
  • kropot72
The 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
  • kropot72
Let \[\large z _{1}=r _{1}(\cos \theta _{1}+i \sin\theta _{1})\] 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
  • kropot72
The 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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