Question

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

Answer 1

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.