Because the Taylor Series result in normal polynomials (finite or infinite) we can see what the affect of complex numbers are on them. We will cover a few of them
to see what the affect of applying complex numbers to them are.
Exponential Function
$$
\begin{aligned}
e^x & = \sum_{j=0}^\infty {{x^j} \over j!}\\
e^{xi} & = \sum_{j=0}^\infty {{(xi)^j} \over j!}\\
& = 1 + xi - {{x^2} \over 2!} - {{x^3} \over 3!}i + {{x^4} \over 4!} + {{x^5} \over 5!}i- {{x^6} \over 6!} - {{x^7} \over 7!}i + {{x^8} \over 8!} + {{x^9} \over 9!}i - ...\\
& = \left(1 - {{x^2} \over 2!} + {{x^4} \over 4!} - {{x^6} \over 6!} + {{x^8} \over 8!} - ....\right) + \left(x -{{x^3} \over 3!} + {{x^5} \over 5!}- {{x^7} \over 7!} + {{x^9} \over 9!} - ..\right)i\\
& = \left(\sum_{j=0}^\infty (-1)^j \cdot {{x^{2j}} \over (2j)!}\right) + \left(\sum_{j=0}^\infty (-1)^j \cdot {{x^{2j+1}} \over (2j+1)!}\right)i
\end{aligned}
$$
But these two sums we have already met (
Taylor Series of Trigonometric functions) and can therefore reduce it to:
$$
\begin{aligned}
e^{xi} & = cos(x) + sin(x)i
\end{aligned}
$$
If we want to add a real part to the number we can do it as follow:
$$
\begin{aligned}
e^{a+xi} & = e^a \cdot e^{xi} = e^a \left( cos(x) + sin(x)i \right)
\end{aligned}
$$
Trigonometric Functions
$$
\begin{aligned}
cos(xi) & = \sum_{j=0}^\infty (-1)^j \cdot {{(xi)^{2j}} \over (2j)!}\\
& = \sum_{j=0}^\infty (-1)^j \cdot i^{2j} \cdot {{x^{2j}} \over (2j)!}\\
& = \sum_{j=0}^\infty (-1)^j \cdot (-1)^{j} \cdot {{x^{2j}} \over (2j)!}\\
& = \sum_{j=0}^\infty {{x^{2j}} \over (2j)!}\\
& = cosh(x)\\
sin(xi) & = \sum_{j=0}^\infty (-1)^j \cdot {{(xi)^{2j+1}} \over (2j+1)!}\\
& = \sum_{j=0}^\infty (-1)^j \cdot i^{2j+1} \cdot {{x^{2j+1}} \over (2j+1)!}\\
& = \left( \sum_{j=0}^\infty (-1)^j \cdot i^{2j} \cdot {{x^{2j+1}} \over (2j+1)!} \right)i\\
& = \left( \sum_{j=0}^\infty (-1)^j \cdot (-1)^j \cdot {{x^{2j+1}} \over (2j+1)!} \right)i\\
& = \left( \sum_{j=0}^\infty {{x^{2j+1}} \over (2j+1)!} \right)i\\
& = sinh(x)i\\
\end{aligned}
$$
Hyperbolic Trigonometric Functions
$$
\begin{aligned}
cosh(xi) & = \sum_{j=0}^\infty {{(xi)^{2j}} \over (2j)!}\\
& = \sum_{j=0}^\infty i^{2j} \cdot {{x^{2j}} \over (2j)!}\\
& = \sum_{j=0}^\infty (-1)^{j} \cdot {{x^{2j}} \over (2j)!}\\
& = cos(x)\\
sinh(xi) & = \sum_{j=0}^\infty {{(xi)^{2j+1}} \over (2j+1)!}\\
& = \sum_{j=0}^\infty i^{2j+1} \cdot {{x^{2j+1}} \over (2j+1)!}\\
& = \left( \sum_{j=0}^\infty i^{2j} \cdot {{x^{2j+1}} \over (2j+1)!} \right)i\\
& = \left( \sum_{j=0}^\infty (-1)^j \cdot {{x^{2j+1}} \over (2j+1)!} \right)i\\
& = sin(x)i\\
\end{aligned}
$$
From this we can deduce the following:
$$
\begin{aligned}
e^{xi} & = cos(x)+sin(x)i\\
& = cosh(xi) + sinh(xi)\\
cosh(x) + sinh(x) & = {{e^x + e^{-1}} \over 2} + {{e^x - e^{-1}} \over 2} = e^x\\
\therefore \, e^x & = cos(ix) - sin(ix)i\\
\end{aligned}
$$