Question

Evaluate:- i^41+1/i^71​

Answer 1

Answer:

$\large \bold\red{ \frac{ - 2}{i} }$

Step-by-step explanation:

Given,

${i}^{41} + \frac{1}{ {i}^{71} }$

Here,

$i$ is called iota.

Further simplifying,

We get,

$= {i}^{(40 + 1)} + \frac{1}{ {i}^{(68 + 3)} }$

But,

We know that,

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

Therefore,

We get,

$= ({i}^{40} \times {i}^{1} ) + \frac{1}{( {i}^{68} \times {i}^{3}) } \\ \\ = {( {i}^{4}) }^{10} \times i + \frac{1}{ { ({i}^{4}) }^{17} \times {i}^{3} }$

But,

We know that,

• ${i}^{4} = 1$
• ${i}^{3} = - i$

Therefore,

Substituting the values,

We get,

$= {1}^{10} \times i + \frac{1}{1 \times ( - i)} \\ \\ = i - \frac{1}{i} \\ \\ = \frac{ {i}^{2} - 1 }{i}$

But,

We know that,

• ${i}^{2} = - 1$

Therefore,

We get,

$= \frac{ - 1 - 1}{i} \\ \\ = \large \bold{ \frac{ - 2}{i} }$