Q.Evaluate $i^{50}+i^{23}$.

Powers of $i$ repeat every four steps, so reduce each exponent modulo $4$.

$50=4(12)+2\Rightarrow i^{50}=i^{2}=-1$; $23=4(5)+3\Rightarrow i^{23}=i^{3}=-i$. Sum $=-1-i$.

Use the period-4 cycle $i^{1}=i,\ i^{2}=-1,\ i^{3}=-i,\ i^{4}=1$, so $i^{n}=i^{\,n\bmod 4}$.

Term 1: $50\div4$ leaves remainder $2$ (since $50=4\times12+2$), hence

Term 2: $23\div4$ leaves remainder $3$ (since $23=4\times5+3$), hence

Add:

Check (independent recomputation): $i^{50}=(i^{2})^{25}=(-1)^{25}=-1$ and $i^{23}=(i^{4})^{5}\cdot i^{3}=1^{5}\cdot(-i)=-i$; sum $=-1-i$, matching.

Alternatively, $i^{50}=(i^{4})^{12}\cdot i^{2}=1\cdot(-1)=-1$ and $i^{23}=(i^{4})^{5}\cdot i^{3}=1\cdot(-i)=-i$, giving the same $-1-i$.

Taking the remainder of $50$ as $0$ or $1$ (arithmetic slip), or forgetting that $i^{3}=-i$ (writing $+i$). Always confirm $n=4q+r$ with $0\le r\le3$.