you have that cos cos is a holomorphic function, which coincides with the cosine defined on the real numbers. Therefore, by the uniqueness theorem about holomorphic functions that …

In mathematics, hyperbolic functions are analogues of the ordinary trigonometric functions, but defined using the hyperbola rather than the circle. Just as the points (cos t, sin t) form a circle …

How to show $$\sin (x+iy)=\sin (x) \cosh (y) + i\cos (x) \sinh (y)$$ I begin with $$\sin (x+iy) = \frac {e^ {x+iy}-e^ {-x-iy}} {2i} = \frac {e^xe^ {iy}-e^ {-x}e^ {-iy}} {2i}$$ $$ = \frac {e^xe^ {iy}-e^ {-x}e^ …

\displaystyle \text {cosh}\ x - \text {cosh}\ y = 2 \text {sinh}\ \frac12 (x + y)\ \text {sinh}\ \frac12 (x - y) cosh x−cosh y = 2sinh 21(x+y) sinh 21(x−y)

2011年5月6日 · The trigonometric functions can be defined for complex variables as well as real ones. One way is to use the power series for sin (x) and cos (x), which are convergent for all …

2016年4月15日 · Not sure if I have done this correctly, seems too straight forward, any help is very appreciated. QUESTION: Find the real and imaginary parts of $f (z) = \cos (z)$. …

2022年11月11日 · 证明： cosh (x+y)=coshx\cdot coshy+sinhx\cdot sinhy 欧拉公式法e^ {ix} = (cos x+isin x) （欧拉公式）sinh x = \frac { (e^x-e^ {-x})} {2} cosh x = \frac { (e^x+e^ {-x})} {2} 易得， …

For any z ∈ C, we define the hyperbolic sine function by ez − e−z sinh(z) = cosh(z) = . Proposition 21.1. For any z ∈ C, cosh(z) = sinh(z). Proof. We have. = = sinh(z). and cosh(z) = cos(iz) and …

Assuming that you define the cosine on the complex plane by cosz = 2exp(iz)+exp(−iz) you have that cos is a holomorphic function, which coincides with the cosine defined on the ...

证明当y趋向于无穷大时，sin(x+iy)和cos(x+iy)的绝对值都趋于无穷大。