Q: Why is the Taylor expansion of $\sin(z)$ about $z_{0}=-j$ different from what I expect? I am trying to solve this exercise:

Find the Taylor expansion for
$$f(z) = sin(z) , \ about \ z_0 = -j$$

Using the Taylor expansion of $sin(w)$ , I expected the result to be
$$\sum_{n=0}^{\infty}(-1)^n\frac{(z+j)^{2n+1}}{(2n+1)!}$$
but checking the exercise book, the result seems to be
$$f(z) = cosh(1)\sum_{n=0}^{\infty}(-1)^n\frac{(z-j)^{2n+1}}{(2n+1)!} + jsinh(1)\sum_{n=0}^{\infty}(-1)^n\frac{(z-j)^{2n}}{(2n)!}$$
Can someone point out the fault in my logic?

A: When you create a taylor series centered about $a$, you use the formula:
$$ f(x)=\sum_{k=0}^{\infty}\frac{f^{(k)}(a)}{k!}(x-a)^k $$
This means that you have to consider what $\sin^{(k)}(-j)$ is as opposed to just plugging $-j$ into the series for $\sin(x)$
The reason there are two series is because for even values of $k$ the $k$th derivative will be a sine function and for odd values the $k$th derivative will be a cosine function, and those two have different series representations.

A: Your intuition failed you. That's all that happened. You expected, and honestly I'd initially expect, the formula to just have a plus j somewhere in it. The problem is that the Taylor series also depends on the values of the derivatives centered around points. Usually when $j=0$ this works and the sin/cos functions evaluate to 0 or 1 respectively, otherwise you'll get a messier looking formula.
