Show that: `{i^(19)+(1/i)^(25)}^2=-4` (ii) `{i^(17)-(1/i)^(34)}^2=2i` (iii) `{i^(18)+(1/i)^(24)}^3=0` (iv) `i^n+i^(n+1)+i^(n+2)+i^(n+3)=0` for all `n in Ndot`

Explanation

(i) `{i^{19}+(1 / i)^{25}}^{2}`


`={i^{19}+frac{1}{i^{25}}}^{2}={i^{3}+frac{1}{i}}^{2}={-i+frac{i^{3}}{i^{4}}}^{2}`


`=[-i+i^{3}]^{2}=(-i-i)^{2}=4 i^{2}=-4`


(ii) `{i^{17}-(1 / i)^{34}}^{2}`


`={i^{17}-frac{1}{i^{34}}}^{2}={i-frac{1}{i^{2}}}^{2}={i-frac{1}{(-1)}}^{2}`


`=(i+1)^{2}=i^{2}+2 i+1=-1+2 i+1=2 i`


(iii) `{i^{18}+(1 / i)^{24}}^{3}`


`={i^{18}+frac{1}{i^{24}}}^{3}=(i^{2}+frac{1}{1})^{3}=(-1+1)^{3}=0`


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