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Q : 1       Evaluate     \small \left [ i^{18}+\left ( \frac{1}{i} \right )^2^5\right ]^3.

Answers (1)

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The given problem is 
\small \left [ i^{18}+\left ( \frac{1}{i} \right )^2^5\right ]^3
Now, we will reduce it into
\small \left [ i^{18}+\left ( \frac{1}{i} \right )^2^5\right ]^3= \left [ (i^4)^4.i^2+\frac{1}{(i^4)^6.i} \right ]^3
                                =\left [ 1^4.(-1)+\frac{1}{1^6.i} \right ]^3                                                              (\because i^4 = 1, i^2 = -1 )
                                = \left [ -1+\frac{1}{i} \right ]^3
                                = \left [ -1+\frac{1}{i} \times \frac{i}{i}\right ]^3
                                = \left [ -1+\frac{i}{i^2} \right ]^3
                                 = \left [ -1+\frac{i}{-1} \right ]^3 = \left [ -1-i \right ]^3
Now,
-(1+i)^3=-(1^3+i^3+3.1^2.i+3.1.i^2)                                                            (using \ (a+b)^3=a^3+b^3+3.a^2.b+3.a.b^2)
                     = -(1 - i +3i+3(-1))                                                                      (\because i^3=-i , i^2 = -1)
                     = -(1 - i +3i-3)= -(-2+2i)
                     =2-2i
Therefore, answer is 2-2i

Posted by

Gautam harsolia

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