Question

Evaluate the following:
(i) i⁴⁵⁷
(ii) i⁵²⁸
(iii) $\frac{1}{i^{58}}$
(iv) $i^{37}+\frac{1}{i^{67}}$
(v) ${\left(i^{41}+\frac{1}{i^{257}}\right)}^9$
(vi) $(i^{77}+i^{70}+i^{87}+i^{414}{)}^3$
(vii) $i^{30}+i^{40}+i^{60}$
(viii) $i^{49}+i^{68}+i^{89}+i^{110}$

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right)i^{457}=i^{4\times 114+1} \\ ={\left(i^4\right)}^{114}\times i \\ =i\left(\because i^4=1\right) \\ \left(\mathrm{ii}\right)i^{528}=i^{4\times 132} \\ ={\left(i^4\right)}^{132} \\ =1\left(\because i^4=1\right) \\ \left(\mathrm{iii}\right)\frac{1}{i^{58}}=\frac{1}{i^{4\times 14+2}} \\ =\frac{1}{{\left(i^4\right)}^{14}\times i^2} \\ =\frac{1}{i^2}\left(\because i^4=1\right) \\ =-1\left(\because i^2=-1\right) \\ \left(\mathrm{iv}\right)i^{37}+\frac{1}{i^{67}}=i^{4\times 9+1}+\frac{1}{i^{4\times 16+3}} \\ ={\left(i^4\right)}^9\times i+\frac{1}{{\left(i^4\right)}^{16}\times i^3} \\ =i-\frac{1}{i}\left(\because i^3=-i\right) \\ =i-\frac{1}{i}\times \frac{i}{i} \\ =i-\frac{i}{i^2} \\ =i-\left(-i\right)\left(\because i^2=-1\right) \\ =2i \\ \left(\mathrm{v}\right){\left(i^{41}+\frac{1}{i^{257}}\right)}^9={\left(i^{4\times 10+1}+\frac{1}{i^{4\times 64+1}}\right)}^9 \\ ={\left[{\left(i^4\right)}^{10}\times i+\frac{1}{{\left(i^4\right)}^{64}\times i}\right]}^9 \\ ={\left(i+\frac{1}{i}\right)}^9\left(\because i^4=1 \\ \right) \end{gathered}$$

$$\begin{gathered} ={\left(i+\frac{i}{i^2}\right)}^9 \\ ={\left(i-i\right)}^9\left(\because i^2=-1\right) \\ =0 \end{gathered}$$

$$\begin{gathered} \left(\mathrm{vi}\right){\left(i^{77}+i^{70}+i^{87}+i^{414}\right)}^3={\left(i^{4\times 19+1}+i^{4\times 17+2}+i^{4\times 21+3}+i^{4\times 103+2}\right)}^3 \\ =\left[\left\{{\left(i^4\right)}^{19}\times i\right\}+\left\{{\left(i^4\right)}^{17}\times i^2\right\}+\left\{{\left(i^4\right)}^{21}\times i^3\right\}+\left\{{\left(i^4\right)}^{103}\times i^2\right\}\right] \\ ={\left(i-1-i-1\right)}^3\left(\because i^4=1,i^3=-i\mathrm{and}{i}^2=-1\right) \\ ={\left(-2\right)}^3 \\ =-8 \\ \left(\mathrm{vii}\right)i^{30}+i^{40}+i^{60}=i^{4\times 7+2}+i^{4\times 10}+i^{4\times 15} \\ =\left[{\left(i^4\right)}^7\times i^2\right]+\left[{\left(i^4\right)}^{10}\right]+\left[{\left(i^4\right)}^{15}\right] \\ =-1+1+1\left(\because i^4=1,i^2=-1\right) \\ =1 \\ \left(\mathrm{viii}\right)i^{49}+i^{68}+i^{89}+i^{110}=i^{4\times 12+1}+i^{4\times 17}+i^{4\times 22+1}+i^{4\times 27+2} \\ =\left[{\left(i^4\right)}^{12}\times i\right]+\left[{\left(i^4\right)}^{17}\right]+\left[{\left(i^4\right)}^{22}\times i\right]+\left[{\left(i^4\right)}^{27}\times i^2\right] \\ =i+1+i-1\left(\because i^4=1,i^2=-1\right) \\ =2i \end{gathered}$$