复数的三角函数如何解这两道的复数Z?
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复数的三角函数如何解这两道的复数Z?
复数的三角函数
如何解这两道的复数Z?
复数的三角函数如何解这两道的复数Z?
(iii) z=x+iy
cosz=cos(x+iy)=cosxcos(iy)-sinx sin(iy)
=cos(x) cosh(y) – sin(x) [i*sinh(y)]
Im cosz>0.5
-sin(x) * sinh(y) > 0.5
sin(x) * sinh(y) < - 0.5
(iv) z=x+iy
sinh(z)=sinh(x+iy)=sinh(x)cosh(iy)+cosh(x)sinh(iy)
=sinh(x)cos(-y) + [cosh(x)]* [-i*sin(-y)]
=sinh(x) cos(y) +i*[cosh(x)] *sin(y)
|sinh z|={ [sinh(x) cos(y) ]^2+[[cosh(x)] *sin(y)]^2}*(1/2)
={ [sinh(x) ]^2 * [cos(y) ]^2+[[cosh(x)] ^2*(1-[cos(y)]^2)}*(1/2)
={ [ cos(y) ]^2 * ([sinh(x)]^2 – [cosh(x)]^2)+[[cosh(x)]^2}*(1/2)
={- [ cos(y) ]^2 +[[cosh(x)]^2}*(1/2) < 2
(iii) z=x+iy
cosz=cos(x+iy)=cosxcos(iy)-sinx sin(iy)
=cos(x) cosh(y) – sin(x) [i*sinh(y)]
Im cosz>0.5
-sin(x) * sinh(y) > 0.5
sin(x) * sinh(y) < - 0.5
(iv) z=x+iy
sinh(z...
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(iii) z=x+iy
cosz=cos(x+iy)=cosxcos(iy)-sinx sin(iy)
=cos(x) cosh(y) – sin(x) [i*sinh(y)]
Im cosz>0.5
-sin(x) * sinh(y) > 0.5
sin(x) * sinh(y) < - 0.5
(iv) z=x+iy
sinh(z)=sinh(x+iy)=sinh(x)cosh(iy)+cosh(x)sinh(iy)
=sinh(x)cos(-y) + [cosh(x)]* [-i*sin(-y)]
=sinh(x) cos(y) +i*[cosh(x)] *sin(y)
|sinh z|={ [sinh(x) cos(y) ]^2+[[cosh(x)] *sin(y)]^2}*(1/2)
={ [sinh(x) ]^2 * [cos(y) ]^2+[[cosh(x)] ^2*(1-[cos(y)]^2)}*(1/2)
={ [ cos(y) ]^2 * ([sinh(x)]^2 – [cosh(x)]^2)+[[cosh(x)]^2}*(1/2)
={- [ cos(y) ]^2 +[[cosh(x)]^2}*(1/2) <2
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