Question

The imaginary part of \[{i^i}\] is:
A. 0
B. 1
C. 2
D. -1

Answer 1

Hint: In the above question, we are given a complex number \[{i^i}\] . We have to find the imaginary part of that complex number. In order to approach our solution, we have to use a complex number formula which is given as \[{e^{i\theta }} = \cos \theta + i\sin \theta \] . After finding the value of \[{i^i}\] , we can write it in the form of \[a + ib\] and then can compare the real and imaginary parts to find what is the imaginary part of \[{i^i}\] .

Complete step by step solution:
Given complex number is \[{i^i}\] .
We have to find its imaginary part.
Since, we know the complex number formula, which is written as
\[ \Rightarrow {e^{i\theta }} = \cos \theta + i\sin \theta \]
Putting \[\theta = \dfrac{\pi }{2}\] in above formula, we can write it as
\[ \Rightarrow {e^{i\dfrac{\pi }{2}}} = \cos \dfrac{\pi }{2} + i\sin \dfrac{\pi }{2}\]
Since, \[\cos \dfrac{\pi }{2} = 0\] and \[\sin \dfrac{\pi }{2} = 1\] ,
Therefore, we get
\[ \Rightarrow {e^{i\dfrac{\pi }{2}}} = i\]
Now, let
\[ \Rightarrow a + ib = {i^i}\]
Taking log both sides, we get
\[ \Rightarrow \log \left( {a + ib} \right) = \log {i^i}\]
\[ \Rightarrow \log \left( {a + ib} \right) = i\log i\]
Putting \[{e^{i\dfrac{\pi }{2}}} = i\] in above equation, we get
\[ \Rightarrow \log \left( {a + ib} \right) = i\log {e^{i\dfrac{\pi }{2}}}\]
We can write the above equation as,
\[ \Rightarrow \log \left( {a + ib} \right) = i \cdot i\dfrac{\pi }{2}\log e\]
Since, \[\log e = 1\] , therefore we get,
\[ \Rightarrow \log \left( {a + ib} \right) = {i^2}\dfrac{\pi }{2}\]
Again, putting \[{i^2} = - 1\] we get
\[ \Rightarrow \log \left( {a + ib} \right) = - \dfrac{\pi }{2}\]
Shifting logarithmic sign to the RHS, we get
\[ \Rightarrow \left( {a + ib} \right) = {e^{ - \dfrac{\pi }{2}}}\]
We can write the above equation as
\[ \Rightarrow a + ib = {e^{ - \dfrac{\pi }{2}}} + i0\]
Now, comparing the real and imaginary parts of LHS and RHS, we get the imaginary part of the complex number as \[0\] .
Therefore, the imaginary part of \[{i^i}\] is \[0\] .
Hence, the correct option is 1) \[0\] .

Note:
Complex numbers are superset of both real numbers and complex/imaginary numbers. That means all the real numbers also fall in the category of complex numbers. The difference is just that their imaginary parts are equal to zero. A real number \[a\] can be written as \[a + i0\] . Therefore, the above given complex number \[{i^i}\] is actually a real number because its imaginary part is zero.