Question

The value of ${\left( { - i} \right)^{ - i}}$ equals?
A.${e^{4n - 1\dfrac{\pi }{2}}},n \in I$
B.${e^{i4n - 1\dfrac{\pi }{2}}},n \in I$
C.${e^{4n + 1\pi /2}},n \in I$
D.${e^{ - i4n - 1\dfrac{\pi }{2}}},n \in I$

Answer 1

Here we need to find the value of ${\left( { - i} \right)^{ - i}}$. For that, we will first calculate the value of $- i$ by using Euler’s identity. Then we will put the value of $- i$ in the given expression and solving further we will get the value of the given expression.

Complete step-by-step answer:
We know the value of iota is the square root of $- 1$ i.e. $i = \sqrt { - 1}$.
We will first calculate the value of negative iota i.e. $- i$.
We know from Euler’s identity;
${e^{ix}} = \cos x + i\sin x$
Now, we will substitute the value $x = - \dfrac{\pi }{2}$ in the Euler’s identity.
${e^{ - i\dfrac{\pi }{2}}} = \cos \dfrac{\pi }{2} - i\sin \dfrac{\pi }{2}$…………….$\left( 1 \right)$
We know the
$\sin \dfrac{\pi }{2} = 1$ and $\cos \dfrac{\pi }{2} = 0$
Putting values in equation 1, we get
$\Rightarrow$ ${e^{i\dfrac{-\pi }{2}}} = 0 - i \times 1$
After further simplification, we get
$\Rightarrow - i = {e^{i\dfrac{-\pi }{2}}}$………………$\left( 2 \right)$
We have got the value of $- i$, we will put this value in the given expression.
Now, we will find the value of ${\left( { - i} \right)^{ - i}}$
Taking $- i$ power on both sides in equation 2, we get
$\Rightarrow$ ${\left( { - i} \right)^{ - i}} = {e^{ - i \times \dfrac{\pi }{2} \times - i}}$
Simplifying the given expression further, we get
$\Rightarrow$ ${\left( { - i} \right)^{ - i}} = {e^{{i^2} \times \dfrac{\pi }{2}}}$………………….$\left( 3 \right)$
We know, $i = \sqrt { - 1}$,
Squaring both sides, we get
$\Rightarrow$ ${i^2} = {\left( {\sqrt { - 1} } \right)^2} = - 1$
We will put the value of ${i^2}$ in equation $\left( 3 \right)$.
${\left( { - i} \right)^{ - i}} = {e^{-1 \times \dfrac{\pi }{2}}} = {e^{\dfrac{-\pi }{2}}}$
As the power of e is the odd multiple of $\dfrac{\pi }{2}$ , we can write the above equation as
$\Rightarrow$ ${\left( { - i} \right)^{ - i}} = {e^{\left( {4n - 1} \right)\dfrac{\pi }{2}}},n \in I$
Hence the value of ${\left( { - i} \right)^{ - i}}$is ${e^{\left( {4n - 1} \right)\dfrac{\pi }{2}}}$, $n \in I$.
Therefore, the correct option is A.

Note: We have used the Euler’s formula to obtain the value of ${\left( { - i} \right)^{ - i}}$. Euler’s identity is defined as a formula that gives the relationship between the complex exponential function and the trigonometric functions.
For any real value of $x$, Euler’s formula is written as
${e^{ix}} = \cos x + i\sin x$