Question

i^99=........ find the value ?

Answer 1

Answer:

(-i)

Step-by-step explanation:

(i^96)(i^3)

(i^4×24)(-i)

(1)(-i)

-i

Answer 2

Answer:

So value of ${i}^{99}$is $- i$

Step-by-step explanation:

Given term is ${i}^{99}$

This is math of Algebra. Here we can use rules of power of indices rule and rules of complex number.

Here power is 99 and base is i.

Here we need to break 99 into prime numbers.

99 is a odd number.

$99 = 3 \times 3 \times 11$

So,${i}^{3 \times 3 \times 11} = {i}^{ {3}^{ {3}^{11} } }$

Now ${i}^{3} = {i}^{2 + 1} = {i}^{2} \times {i}^{1} = ( - 1) \times i = - i$

So,$( {i}^{ {3})^{( {3})^{11} } } = {( - i)}^{( {3})^{11} }$

Now ${( - i)}^{3} = - {i}^{3} = - ( - i) = i$

So,${( - i)}^{ {(3)}^{11} } = {i}^{11}$

Now ${i}^{11} = {i}^{(2 + 2 + 2 + 2 + 2 + 1)}$

$= {i}^{2} \times {i}^{2} \times {i}^{2} \times {i}^{2} \times {i}^{2} \times i$

$= ( - 1) \times ( - 1) \times ( - 1) \times ( - 1) \times ( - 1) \times i = - i$

So value of ${i}^{99}$is $- i$

Here applied formulas are

${i}^{2} = - 1$

${x}^{ {(y)}^{(a)} } = {x}^{y \times a}$

${x}^{y} \times {x}^{a} = {x}^{y + a}$