Power series of sin
Just like in the case of the exponential, let us now observe the complex sine function. Again the Taylor series and polynomials centered at zero are considered which now have the following form: \[ \sin\,z = \sum\limits_{k=0}^{\infty} (-1)^k \frac{z^{2k+1}}{(2k+1)!}, \] \[ T_n\sin\,(z) = \sum\limits_{k=0}^n (-1)^k \frac{z^{2k+1}}{(2k+1)!} = z - \frac{1}{3!} z^3 + \frac{1}{5!} z^5 - \dots + (-1)^n \frac{1}{(2n+1)!} z^{2n+1}. \] Let us look at these partial sums one-by-one!
\(f(z) = T_{n}\sin\,(z),\quad z \in \mathbb{C}(0,8)\)
Compare again the Taylor polynomials with the plot of the sine function on the same domain. (The coloring is also shown.)
\(f(z) = \sin\,z,\quad z \in \mathbb{C}(0,8)\)
It is clearly visible also in this case how the partial sums become more and more 'similar' to the original function.