Given $A(z)=\sum_{n=0}^{\infty}a_nz^{n}$ and the radius is $R>0$, for every $x\in[0,R)$ $A(x)=e^x-2e^{-x}$:
Prove a. $R=\infty$ b. $A(i\pi)=?$
for a i did:
$$A(x)=\sum_{n=0}^{\infty}a_nx^{n}=e^x-2e^{-x}=\sum_{n=0}^{\infty}x^{n}\frac{1-2(-1)^{n}}{n!}$$ and from here i used the cauchy hadamard theroem and got the needed.
for b. i wasn't sure if i can just plug the $i\pi$ into A.i think i can because we i plug it i will get real numbers because the result of a.
i just need help with the formal explanation
By the Identity Theorem we must have $A(z)=e^{x}-2e^{-z}$ for all $z$ with $|z| <R$. Since this function is an entire function we get a) and the answer to b) is $A(i\pi)=e^{i\pi}-2e^{-i\pi}=(-1)-2(-1)=-1$.