Evaluate the following:

Solution:

(i) $i^{457}=i^{4 \times 114+1}$

$=\left(i^{4}\right)^{114} \times i$

$=i \quad\left(\because i^{4}=1\right)$

(ii) $i^{528}=i^{4 \times 132}$

$=\left(i^{4}\right)^{132}$

$=1 \quad\left(\because i^{4}=1\right)$

(iii) $\frac{1}{i^{58}}=\frac{1}{i^{4 \times 14+2}}$

$=\frac{1}{\left(i^{4}\right)^{14} \times i^{2}}$

$=\frac{1}{i^{2}} \quad\left(\because i^{4}=1\right)$

$=-1 \quad\left(\because i^{2}=-1\right)$

(iv) $i^{37}+\frac{1}{i^{67}}=i^{4 \times 9+1}+\frac{1}{i^{4 \times 16+3}}$

$=\left(i^{4}\right)^{9} \times i+\frac{1}{\left(i^{4}\right)^{16} \times i^{3}}$

$=i-\frac{1}{i} \quad\left(\because i^{3}=-i\right)$

$=i-\frac{1}{i} \times \frac{i}{i}$

$=i-\frac{i}{i^{2}}$

$=i-(-i) \quad\left(\because i^{2}=-1\right)$

$=2 i$

(v) $\left(i^{41}+\frac{1}{i^{257}}\right)^{9}=\left(i^{4 \times 10+1}+\frac{1}{i^{4 \times 64+1}}\right)^{9}$

$=\left[\left(i^{4}\right)^{10} \times i+\frac{1}{\left(i^{4}\right)^{64} \times i}\right]^{9}$

$=\left(i+\frac{1}{i}\right)^{9} \quad\left(\because i^{4}=1\right.$

$=\left(i+\frac{i}{i^{2}}\right)^{9}$

$=(i-i)^{9} \quad\left(\because i^{2}=-1\right)$

= 0

(vi) $\left(i^{77}+i^{70}+i^{87}+i^{414}\right)^{3}=\left(i^{4 \times 19+1}+i^{4 \times 17+2}+i^{4 \times 21+3}+i^{4 \times 103+2}\right)^{3}$

$=\left[\left\{\left(i^{4}\right)^{19} \times i\right\}+\left\{\left(i^{4}\right)^{17} \times i^{2}\right\}+\left\{\left(i^{4}\right)^{21} \times i^{3}\right\}+\left\{\left(i^{4}\right)^{103} \times i^{2}\right\}\right]$

$=(i-1-i-1)^{3} \quad\left(\because i^{4}=1, i^{3}=-i\right.$ and $\left.i^{2}=-1\right)$

$=(-2)^{3}$

$=-8$

(vii) $i^{30}+i^{40}+i^{60}=i^{4 \times 7+2}+i^{4 \times 10}+i^{4 \times 15}$

$=\left[\left(i^{4}\right)^{7} \times i^{2}\right]+\left[\left(i^{4}\right)^{10}\right]+\left[\left(i^{4}\right)^{15}\right]$

$=-1+1+1 \quad\left(\because i^{4}=1, i^{2}=-1\right)$

= 1

(viii) $i^{49}+i^{68}+i^{89}+i^{110}=i^{4 \times 12+1}+i^{4 \times 17}+i^{4 \times 22+1}+i^{4 \times 27+2}$

$=\left[\left(i^{4}\right)^{12} \times i\right]+\left[\left(i^{4}\right)^{17}\right]+\left[\left(i^{4}\right)^{22} \times i\right]+\left[\left(i^{4}\right)^{27} \times i^{2}\right]$

$=i+1+i-1 \quad\left(\because i^{4}=1, i^{2}=-1\right)$

$=2 i$