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Class 11 Complex Numbers Miscellaneous Exercise Solution

28. Evaluate: ${[i^{18} + {(\frac{1}{i})}^{23}]}^{3}$

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Payal Gupta | Contributor-Level 10

4 months ago

28. ${[i^{18} + {(\frac{1}{i})}^{23}]}^{3}$

= $[i 4?$
$4 + 2 + 1 i 4?$
${6 + 1]}^{3}$

= ${[? 1 + \frac{1}{i}]}^{3}$ [as i^{4 *k + 2} = –1 and i^{4 *k + 1} = i]

= ${[? 1 + \frac{i}{i^{2}}]}^{3}$

= [–1 – i]³ [as i² = –1]

= (-1)³ (1 + i)³

= –1 [1³ + i³ + 3 * 1 *i (1 + i)] [since, (a + b)³ = a³ + b³ + 3ab (a + b)]

= –1 [1 – i³ + 3i (1 + i)]

= –1 [1 – i³ + 3i + 3i²]

= –1 [1 – i + 3i – 3] &nb

...more

28. ${[i^{18} + {(\frac{1}{i})}^{23}]}^{3}$

= $[i 4?$
$4 + 2 + 1 i 4?$
${6 + 1]}^{3}$

= ${[? 1 + \frac{1}{i}]}^{3}$ [as i^{4 *k + 2} = –1 and i^{4 *k + 1} = i]

= ${[? 1 + \frac{i}{i^{2}}]}^{3}$

= [–1 – i]³ [as i² = –1]

= (-1)³ (1 + i)³

= –1 [1³ + i³ + 3 * 1 *i (1 + i)] [since, (a + b)³ = a³ + b³ + 3ab (a + b)]

= –1 [1 – i³ + 3i (1 + i)]

= –1 [1 – i³ + 3i + 3i²]

= –1 [1 – i + 3i – 3] [Since, i² = –1 and i³ = i².i = –i]

= –1 [–2 + 2i]

= 2 – 2i

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Class 11 Complex Numbers Miscellaneous Exercise Solution

28. Evaluate: ${[i^{18} + {(\frac{1}{i})}^{23}]}^{3}$