Question

Simplest form of ${\left[i^{18}+{\left(\frac{1}{i}\right)}^{25}\right]}^3$ is

• A. $2+3i$
• B. $2-2i$
• C. $2+2i$
• D. $2-3i$

Answer 1

Answer: (B)

${\left[i^{18}+{\left(\frac{1}{i}\right)}^{25}\right]}^3={\left[{\left(i^2\right)}^9+\frac{1}{i^{25}}\right]}^3$
$={\left[{\left(i^2\right)}^9+\frac{1}{i\cdot i^{21}}\right]}^3$
$={\left[(-1{)}^9+\frac{1}{i{\left(i^4\right)}^6}\right]}^3\left(\because i^2=-1\right)$
$={\left[-1+\frac{1}{i}\right]}^3\left(\because i^4=1\right)$
$={\left[-1+\frac{1}{i}\times \frac{i}{i}\right]}^3={\left[-1+\frac{i}{i^2}\right]}^3$
$=[-1-i{]}^3$
$\left(\because i^2=-1\right)$
$=(-1{)}^3[1+i{]}^3$
$=-\left[(1{)}^3+i^3+3\times 1\times i(1+i)\right]$
$\left[\because (a+b{)}^3=a^3+b^3+3ab(a+b)\right]$
$=-\left[1-i+3i+3i^2\right]\left(\because i^3=-i\right)$
$=-[1-i+3i-3]\left(\because i^2=-1\right)$
$=-[-2+2i]=2-2i$
Students are suggested not to open the cube of expression at initial stage because the expression becomes complex by doing so.