Question Number 8341 by Rasheed Soomro last updated on 09/Oct/16
$$\mathrm{What}\:\mathrm{are}\:\mathrm{necessary}\:\mathrm{and}\:\mathrm{sufficient}\:\mathrm{conditions} \\ $$$$\mathrm{that}\:\left(\mathrm{a}+\mathrm{ib}\right)^{\mathrm{n}} \:\mathrm{is}\:\mathrm{cyclic}\:\mathrm{for}\:\mathrm{an}\:\:\mathrm{n}\:\mathrm{not}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{0}? \\ $$
Answered by prakash jain last updated on 09/Oct/16
![∣a+ib∣=1⇒(√(a^2 +b^2 ))=1 The above is necessary and?sufficient condition. then if arctan(b,a)=((2π)/k) then (a+bi)^(j+mk) =(a+bi), j,m,k∈Z, j∈[0,k−1]](https://www.tinkutara.com/question/Q8344.png)
$$\mid{a}+{ib}\mid=\mathrm{1}\Rightarrow\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\mathrm{The}\:\mathrm{above}\:\mathrm{is}\:\mathrm{necessary}\:\mathrm{and}?\mathrm{sufficient} \\ $$$$\mathrm{condition}. \\ $$$$\mathrm{then}\:\mathrm{if} \\ $$$${arctan}\left({b},{a}\right)=\frac{\mathrm{2}\pi}{{k}}\:\mathrm{then} \\ $$$$\left({a}+{bi}\right)^{{j}+{mk}} =\left({a}+{bi}\right),\:{j},{m},{k}\in\mathbb{Z},\:{j}\in\left[\mathrm{0},{k}−\mathrm{1}\right] \\ $$