Evaluate. [i^18+(1/i)^25]^3

Name: Evaluate. [i^18+(1/i)^25]^3
Uploaded: Mon, 2019-06-17 16:46
Description: Evaluate. [i^18+(1/i)^25]^3

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

Answers (1)

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 $2-2i$