Question

Evaluate ${\lceil i^{18}+{\left(\frac{1}{i}\right)}^{25}\rceil }^3$

Answer 1

Given data : $z={[i^{18}+{\left(\frac{1}{i}\right)}^{25}]}^3$

Using exponent property

$z={\left[\right(i{)}^{4\times 4+2}+\frac{1}{(i{)}^{4\times 6+1}}]}^3$

$={\left[\right(i{)}^2+\frac{1}{i}]}^3[\because i^4=1]$

Arrange in numerator form

$z={\left[\right(i{)}^2+\frac{1}{i}\times \frac{i}{i}]}^3$

$={\left[\right(i{)}^2+{\frac{i}{i}}^2]}^3\because i^2=-1$

$={[-1-i]}^3$

$z={[i^{18}+{\left(\frac{1}{i}\right)}^{25}]}^3={[-1-i]}^3$

$=-(1+i{)}^3$

Apply $(A+B{)}^3$ formula

$z=-\left[\right(1{)}^3+(i{)}^3+3(1{)}^2+3\left(1\right)\left(i{)}^2\right]$

$=-[1+i^3+3i+3i+3i^2]$

$=-[1-i+3i-3]$

$=-[-2+2i]$

$=2-2i$