# Evaluate the following powers of i. { a) i^8 \\b) i^{11} \\c) i^{42} \\ d) i^{105} }

## Question:

Evaluate the following powers of i.

{eq}a) i^8 \\b) i^{11} \\c) i^{42} \\ d) i^{105} {/eq}

## Imaginary Number:

The square root of any negative number is called an imaginary number.

{eq}\iota=\sqrt{-1} {/eq} is an imaginary unit iota, that is denoted by {eq}\iota {/eq}

Here, in our problem we have to evaluate the integral powers of {eq}\iota {/eq}

Integrals Powers of {eq}\iota {/eq}:

{eq}\iota=\sqrt{-1}\\ \left ( \iota \right )^2=-1\\ \left ( \iota \right )^3=-\iota\\ \left ( \iota \right )^4=1 {/eq}

When {eq}\iota {/eq} is raised to an integral power, the answer is always {eq}\iota, -1, -\iota, 1 {/eq}

As we know:

{eq}\iota=\sqrt{-1}\\ \left ( \iota \right )^2=-1\\ \left ( \iota \right )^3=-\iota\\ \left ( \iota \right )^4=1 {/eq}

a)

{eq}\begin{align} \iota ^{8} &=\left (\iota ^{4} \right )^{2} \ & \left [ \because (a^{b})^{c}=a^{bc} \right ]\\ &=(1)^2 \ & \left [ \because \iota^4=1 \right ]\\ &=1 \end{align} {/eq}

b)

{eq}\begin{align} \iota ^{11} &=\left (\iota ^{8+3} \right )\\ &=\iota^8 .\iota ^3\ & \left [ \because (a)^{b+c}=a^b.a^c \right ]\\ &=(\iota^4 )^2.\iota ^3\ & \left [ \because \left (a^{b} \right )^c=a^{bc} \right ]\\ &=(1)^2(-\iota)\ & \left [ \because \iota^4=1, \iota^3=-\iota \right ]\\ &=-\iota \end{align} {/eq}

c)

{eq}\begin{align} \iota ^{42} &=\left (\iota ^{40+2} \right )\\ &=\iota^{40} .\iota ^2\ & \left [ \because (a)^{b+c}=a^b.a^c \right ]\\ &=(\iota^4 )^{10}.\iota ^2\ & \left [ \because \left (a^{b} \right )^c=a^{bc} \right ]\\ &=(1)^{10}(-1)\ & \left [ \because \iota^4=1, \iota^2=-1 \right ]\\ &=-1 \end{align} {/eq}

d)

{eq}\begin{align} \iota ^{105} &=\left (\iota ^{104+1} \right )\\ &=\iota^{104} .\iota \ & \left [ \because (a)^{b+c}=a^b.a^c \right ]\\ &=(\iota^4 )^{26}.\iota \ & \left [ \because \left (a^{b} \right )^c=a^{bc} \right ]\\ &=(1)^{26}(\iota)\ & \left [ \because \iota^4=1 \right ]\\ &=\iota \end{align} {/eq} 