Question

Find the values of the following expressions:
(i) i⁴⁹ + i⁶⁸ + i⁸⁹ + i¹¹⁰
(ii) i³⁰ + i⁸⁰ + i¹²⁰
(iii) i + i² + i³ + i⁴
(iv) i⁵ + i¹⁰ + i¹⁵
(v) $\frac{i^{592}+i^{590}+i^{588}+i^{586}+i^{584}}{i^{582}+i^{580}+i^{578}+i^{576}+i^{574}}$
(vi) 1+ i² + i⁴ + i⁶ + i⁸ + ... + i²⁰
(vii) (1 + i)⁶ + (1 − i)³

Answer 1

$$\begin{gathered} \left(\mathrm{i}\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+i^2\left[\because i^4=1\right] \\ =2i+1-1\left[\because i^2=-1\right] \\ =2i \\ \left(\mathrm{ii}\right)i^{30}+i^{80}+i^{120} \\ =i^{4\times 7+2}+i^{4\times 20}+i^{4\times 30} \\ =\left\{{\left(i^4\right)}^7\times i^2\right\}+\left\{{\left(i^4\right)}^{20}\right\}+\left\{{\left(i^4\right)}^{30}\right\} \\ =i^2+1+1\left[\because i^4=1\right] \\ =-1+2\left[\because i^2=-1\right] \\ =1 \\ \left(\mathrm{iii}\right)i+i^2+i^3+i^4 \\ =i-1-i+1\left[\because i^2=-1,i^3=-i\mathrm{and}{i}^4=1\right] \\ =0 \\ \left(\mathrm{iv}\right)i^5+i^{10}+i^{15} \\ =i^{4\times 1+1}+i^{4\times 2+2}+i^{4\times 3+3} \\ =\left\{{\left(i^4\right)}^1\times i\right\}+\left\{{\left(i^4\right)}^2\times i^2\right\}+\left\{{\left(i^4\right)}^3\times i^3\right\} \\ =i+i^2+i^3\left[\because i^4=1\right] \\ =i-1-i\left[\because i^2=-1,i^3=-i\right] \\ =-1 \\ \left(\mathrm{v}\right)\frac{i^{592}+i^{590}+i^{588}+i^{586}+i^{584}}{i^{582}+i^{580}+i^{578}+i^{576}+i^{574}} \\ =\frac{i^{4\times 148}+i^{4\times 147+2}+i^{4\times 147}+i^{4\times 146+2}+i^{4\times 146}}{i^{4\times 145+2}+i^{4\times 145}+i^{4\times 144+2}+i^{4\times 144}+i^{4\times 143+2}} \\ =\frac{{\left(i^4\right)}^{148}+\left\{{\left(i^4\right)}^{147}\times i^2\right\}+\left\{{\left(i^4\right)}^{146}\right\}+\left\{{\left(i^4\right)}^{146}\times i^2\right\}+\left\{{\left(i^4\right)}^{146}\right\}}{\left\{{\left(i^4\right)}^{145}\times i^2\right\}+\left\{{\left(i^4\right)}^{145}\right\}+\left\{{\left(i^4\right)}^{144}\times i^2\right\}+\left\{{\left(i^4\right)}^{144}\right\}+\left\{{\left(i^4\right)}^{143}\times i^2\right\}} \\ =\frac{1+i^2+1+i^2+1}{i^2+1+i^2+1+i^2}\left[\because i^4=1\right] \\ =\frac{1-1+1-1+1}{-1+1-1+1-1}\left[\because i^2=-1\right] \\ =-1 \end{gathered}$$


$$\begin{gathered} \left(\mathrm{vi}\right)1+i^2+i^4+i^6+i^8+...+i^{20} \\ \because i^2=-1, \\ {i}^4=1, \\ {i}^6=-1, \\ {i}^8=1, \\ . \\ . \\ . \\ . \\ . \\ {i}^{20}=1 \\ \therefore 1+i^2+i^4+i^6+i^8+...+i^{20} \\ =\left[1+\left(-1\right)\right]+\left[1+\left(-1\right)\right]+\left[1+\left(-1\right)\right]+...+\left[1+\left(-1\right)\right]+1 \\ =5\times \left[1+\left(-1\right)\right]+1\left[\mathrm{As},\mathrm{there}\mathrm{are}11\mathrm{terms}\right] \\ =5\times 0+1 \\ =1 \end{gathered}$$


(vii) (1 + i)⁶ + (1 − i)³
= [(1 + i)²]³ + (1 − i)³
= [1² + i² + 2i]³ + (1³ − i³ + 3i² − 3i)
= [1 − 1 + 2i]³ + (1 + i − 3 − 3i) [∵ i² = −1, i³ = −i]
= (2i)³ + (−2 − 2i)
= 8i³ − 2 − 2i
= −8i − 2 − 2i [∵ i³ = −i]
= −10i − 2