Solution 3.1:1f

Let's begin by calculating some powers of i:

\displaystyle \begin{align}i^2&=i\cdot i=-1,\\ i^3&=i^2\cdot i = (-1)\cdot i = -i,\\ i^4&=i^2\cdot i^2 = (-1)\cdot (-1) = 1.\end{align}

Now, we observe that because \displaystyle i^4=1, we can try to factorize \displaystyle i^{11} and \displaystyle i^{20} in terms of \displaystyle i^4,

\displaystyle \begin{align}i^2&=i\cdot i=-1,\\ i^{11}&=i^{4+4+3} = i^4\cdot i^4\cdot i^3 = 1\cdot 1 \cdot (-i)=-i\\ i^{20}&=i^{4+4+4+4+4} = i^4\cdot i^4\cdot i^4\cdot i^4\cdot i^4 = 1\cdot 1 \cdot 1\cdot 1 \cdot 1=1\end{align}

The answer becomes

\displaystyle i^{20}+i^{11}=1-i