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V2010
@V2010
July 2022
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3 корень 3+jкорень 3 срочно пожалуйста
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Rechnung
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Z=3+i*√3
|z|=√(3²+(√3)²)=√(9+3)=√12=2√3
cosa=3/(2√3)=3√3/6=√3/2
sina=√3/(2√3)=1/2 => a=π/6
z=|z|(cosa+i*sina)
z=2√3(cosπ/6+i*sinπ/6)
∛z=∛(2√3)(cos (π/6+2πk)/3 + i*sin(π/6+2πk)/3)
k=0;1;2
k=0 ∛z=∛(2√3)(cosπ/18+i*sinπ/18)
k=1 ∛z=∛(2√3)(cos13π/18+i*sin13π/18)
k=2 ∛z=∛(2√3)(cos25π/18+i*sin25π/18)
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Answers & Comments
Verified answer
Z=3+i*√3|z|=√(3²+(√3)²)=√(9+3)=√12=2√3
cosa=3/(2√3)=3√3/6=√3/2
sina=√3/(2√3)=1/2 => a=π/6
z=|z|(cosa+i*sina)
z=2√3(cosπ/6+i*sinπ/6)
∛z=∛(2√3)(cos (π/6+2πk)/3 + i*sin(π/6+2πk)/3)
k=0;1;2
k=0 ∛z=∛(2√3)(cosπ/18+i*sinπ/18)
k=1 ∛z=∛(2√3)(cos13π/18+i*sin13π/18)
k=2 ∛z=∛(2√3)(cos25π/18+i*sin25π/18)