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

Hint: 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
\[{e^{ix}} = \cos x + i\sin x\]