10.3 Find the sum $A=i+i^{2}+i^{3}+\ldots+i^{15}$.

Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
10.3 Let's group the terms as follows:

$$
\begin{aligned}
& A=\left(i+i^{2}+i^{3}+i^{4}\right)+\left(i^{5}+i^{6}+i^{7}+i^{8}\right)+ \\
& +\left(i^{9}+i^{10}+i^{11}+i^{12}\right)+\left(i^{13}+i^{14}+i^{15}\right)
\end{aligned}
$$

from which

$$
A=\left(i+i^{2}+i^{3}+i^{4}\right)\left(1+i^{4}+i^{8}\right)+i^{13}+i^{14}+i^{15}
$$

Since

$$
\begin{aligned}
& i+i^{2}+i^{3}+i^{4}=i-1-i+1=0 \\
& A=i^{13}+i^{14}+i^{15}=i-1-i=-1
\end{aligned}
$$

Answer: -1.