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kotkot666
@kotkot666
August 2022
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Как можно это преобразовать?
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oganesbagoyan
Verified answer
Z =a+bi
z=sin3π/4 +i +icos3π/4 =sin3π/4 +(1 +cos3π/4)i || a=sin3π/4 ; b=
sin3π/4 ||
r =√(sin²3π/4 +(1 +cos3π/4)²) =√(sin²3π/4 +1 +cos²3π/4+2cos3π/4) =
√(2-√2) ;
tqα =(1 +cos3π/4)/sin3π/4 =(1-√2 /2)/√2 /2 =(√2 - 1) ;α =arctq(√2 - 1) .
z =√(2-√2) e^(iα)
-------
sin3π/4 +i +icos3π/4 =sinπ/4 +i- icosπ/4 =√2 /2 +(1 -√2 /2) i
=
(√2-√2) +(1 -√2 /2) i = (√(2-√2) (cosα +isinα) ;
tqα =(1 -√2 /2) / √2 /2 =(√2 -
1).
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Answers & Comments
Verified answer
Z =a+biz=sin3π/4 +i +icos3π/4 =sin3π/4 +(1 +cos3π/4)i || a=sin3π/4 ; b=sin3π/4 ||
r =√(sin²3π/4 +(1 +cos3π/4)²) =√(sin²3π/4 +1 +cos²3π/4+2cos3π/4) =
√(2-√2) ;
tqα =(1 +cos3π/4)/sin3π/4 =(1-√2 /2)/√2 /2 =(√2 - 1) ;α =arctq(√2 - 1) .
z =√(2-√2) e^(iα)
-------
sin3π/4 +i +icos3π/4 =sinπ/4 +i- icosπ/4 =√2 /2 +(1 -√2 /2) i =
(√2-√2) +(1 -√2 /2) i = (√(2-√2) (cosα +isinα) ;
tqα =(1 -√2 /2) / √2 /2 =(√2 - 1).