How to interpret the cosine and sine of complex number?

Question

I know that the $\cos\left(a\right)$ return a point on the unit circle along the $x$ axis, who located at an angle $a$($a$ is real number). The $\sin\left(a\right)$ is analog for $\cos\left(a\right)$ but it return the point along $y$ axis.

Considering this, how to interpret the cosine and sine of complex number?

Given that the cosine and sine of a complex number will return total 4 values, I can suggest a point on the four-dimensional sphere, but it's just a fantasy.

EDIT: I am interested in geometric interpretation, according to the above analogy

Answer 1

Use $e^{ix} =\cos(x)+i\sin(x) $ and $e^{-ix} =\cos(x)-i\sin(x) $ to get $\cos(x) =(e^{ix}+e^{-ix})/2 $ and $\sin(x) =(e^{ix}-e^{-ix})/(2i) $.

These hold for any complex x.

Note that if $x=u+iv$ where $u$ and $v$ are real then $e^x =e^{u+iv} =e^ue^{iv} =e^u(\cos(v)+i\sin(v)) =e^u\cos(v)+ie^u\sin(v) $ and $e^{ix} =e^{i(u+iv)} =e^{iu-v} =e^{-v}e^{iu} =e^{-v}\cos(u)+ie^{-v}\sin(u) $.

Answer 2

Using $\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$ and $\cos(x)=\frac{e^{ix}+e^{-ix}}2$, we get $\sin(ix)=i\sinh(x)$ and $\cos(ix)=\cosh(x)$. Using the sum formulas, we get $$ \sin(x+iy)=\sin(x)\cosh(y)+i\cos(x)\sinh(y) $$ and $$ \cos(x+iy)=\cos(x)\cosh(y)-i\sin(x)\sinh(y) $$