DeMoivre's Theorem 2015
z = ( sin 2 0 1 5 π + i cos 2 0 1 5 π ) 2 0 1 5 , ( − z ) 2 0 1 5 = ?
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1 solution
why you common i in first line?
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To obtain a expression of the form cos θ + i sin θ , and then apply DeMoivre's Theorem.
Notice that 1/i = -i
I may be forgetting something, however, shouldn't z be in the form c o s π + i s i n π rather than c o s π − i s i n π ?
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It can also be of the form cos θ − i sin θ , because it comes from cos ( − θ ) + i sin ( − θ ) .
z = ( i ( cos 2 0 1 5 π − i sin 2 0 1 5 π ) ) 2 0 1 5 z = i 2 0 1 5 ( cos π − i sin π ) z = − i ( − 1 ) z = i − z = − i ( − z ) 2 0 1 5 = ( − i ) 2 0 1 5 ( − z ) 2 0 1 5 = i