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Evaluate : $\;[i^{18} +(\large\frac{1}{i})^{25}]^{3}$

$(a)\;3i\qquad(b)\;2i\qquad(c)\;2-3i\qquad(d)\;2-2i$

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Answer : $\;2-2i$
Explanation :
$\;[i^{18} +(\large\frac{1}{i})^{25}]^{3} = [i^{4\times4+2} +\large\frac{1}{i^{4 \times 6 +1 }}]^{3}$
$= [(i^{4})^{4} \;.i^{2} + \large\frac{1}{(i^{4})^{6}}]^{3}$
$= [i^{2} + \large\frac{1}{i}]^{3} \quad [i^{4} =1]$
$= [-1 + \large\frac{1}{i} \times \large\frac{1}{i}] \quad [i^{2} =-1]$
$= [-1 + \large\frac{i}{i^{2}}]^{3}$
$= [-1-i]^{3}$
$= (-1)^{3} [1+i]^{3}$
$= -[1^{3} + (i)^{3} + 3 \;. 1\;.i\; (1+i)]$
$ = - [1+ i^{3} +3i +3i^{2}]$
$ = -[1 -i + 3i -3]$
$= - [-2+2i]$
$= 2-2i$
answered Apr 11, 2014 by yamini.v
 

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