(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這兩題最后的答案就不能再化簡(jiǎn)了么？(iii)sin(x)>0, sinh(y)<-0.5, 00.5,y>0.481211(iv){- [ cos(y) ]^2 +[[cosh(x)]^2}*(1/2) < 2- [ cos(y) ]^2 +[[cosh(x)]^2< 4[[cosh(x)]^2< 5, cosh(x)總是>0cosh(x)<1.443635最后一題- [cos(y)]^2 +[cosh(x)]^2< 4不能直接到[cosh(x)]^2< 5吧...0≤[cos(y)]^2≤1,- [ cos(y) ]^2 +[[cosh(x)]^2+[cos(y)]^2< 4+[cos(y)]^2<5[cosh(x)]^2< 5cosh(x)<平方根5, x<1.44363547我覺得只有當(dāng)y=0+2kπ的時(shí)候x<1.44363547吧，x的范圍應(yīng)該也是與y相關(guān)的范圍是指最大范圍