abs(cos(x+yi))

hbghlyj

$$\abs{\cos(x+yi)}=\frac{\sqrt{\cos (2 x)+\cosh (2 y)}}{\sqrt{2}}=\sqrt{\cos^2x+\sinh^2y}$$证明:
$$\eqalign{\abs{\cos(x+yi)}&=\sqrt{\sin ^2(x) \sinh ^2(y)+\cos ^2(x) \cosh ^2(y)}\\
&=\sqrt{[1-\cos ^2(x)][\cosh^2(y)-1]+\cos ^2(x) \cosh ^2(y)}\\
&=\sqrt{\cos ^2(x)+\cosh^2(y)-1}\\
&=\sqrt{\frac{\cos(2x)+1}2+\frac{\cosh(2y)+1}2-1}\\
&=\frac{\sqrt{\cos (2 x)+\cosh (2 y)}}{\sqrt{2}}
}$$类似等式:
$$\abs{\cosh(x+yi)}=\frac{\sqrt{\cosh (2 x)+\cos (2 y)}}{\sqrt{2}}=\sqrt{\sinh^2x+\cos^2y}$$
$$\abs{\sin(x+yi)}=\frac{\sqrt{\cosh (2 y)-\cos (2 x)}}{\sqrt{2}}$$
$$\abs{\sinh(x+yi)}=\frac{\sqrt{\cosh (2 x)-\cos (2 y)}}{\sqrt{2}}$$

hbghlyj

$|z|\le1$, 求$|\cos z|$的最大值?
$$|\cos(x+yi)|^2=\sin^2x+\sinh^2y\le\sinh^2x+\sinh^2y\le\sinh^21=\frac{1-2 e^2+e^4}{4 e^2}$$