11TH•Mathematics•Complex Numbers•Question 9

Solution

Show that for $n\in\mathbb{N}$, $n>4$, we have $i^n=i^r$, where $r$ is the remainder when $n$ is divided by $4$.

\begin{aligned} & \boxed{\text{Let } n\in\mathbb{N},\ n>4.\ \text{Write } n=4q+r,\ 0\le r<4.} \\ \\ & \boxed{\text{Then:}} \\ \\ & i^n = i^{4q+r} = i^{4q}\,i^r = (i^4)^q\,i^r \\ \\ & \boxed{\text{Since } i^4=1:} \\ \\ & (i^4)^q = 1^q = 1 \\ \\ & \Rightarrow i^n = i^r \end{aligned}

Also:

\begin{aligned} & \boxed{r=0\Rightarrow i^n=1} \\ & \boxed{r=1\Rightarrow i^n=i} \\ & \boxed{r=2\Rightarrow i^n=-1} \\ & \boxed{r=3\Rightarrow i^n=-i} \end{aligned}

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