As we all know about the topic of complex numbers, we are familiar with the term iota (i), where i = √(-1). A question arises that is there any value possible for i^{i}.

So, its simple answer is yes, there is a value for i^{i}. The below solution is mentioned for it.

We have to find the value of i^{i}. So, Let y = i^{i}

Taking ln on both sides,

ln(y)= i ln(i) ----- ( i ) [ ln (a^{b}) = b*ln(a) ]

Now, for solving ln(i), we have to understand the following concept :

In the polar representation of complex numbers, we write **z = re ^{iθ}**, where –

z = a + ib, r = |a^{2}+ b^{2}| θ = tan^{-1}(b/a), So, taking log on both sides of the equation z = re^{iθ}ln(z) = ln(r) + iθ [ln(e^{a}) = a, and ln(a*b) = ln(a) + ln(b)] Putting the value of z, r and θ in the above equation ln(a+ib) = ln(|a^{2 + b2|) + i*tan-1(b/a)}

So, writing ln(i) = ln(0 + 1i), and applying the above formula

ln(0+1i) = ln(|0^{2}+ 1^{2}|) + i*tan^{-1}(1/0) ln(i) = ln1 + i*∏/2 [ tan^{-1}(1/0) = tan^{-1}(∞) = ∏/2 ] ln(i) = i*∏/2 [ ln1 = 0 ]

Now putting the value of ln(i) in equation ( i )

ln(y) = i * ( i*∏/2 ) ln(y) = i^{2 }* ∏/2 ln(y) = -1 * ∏/2 [i^{2}= -1] ln(y) = -∏/2 y = e^{-∏/2 }[ln(a) = b ⇒ a = e^{b}]

As we have assumed y = i^{i}.

So,

i^{i}= e^{ -∏/2}

If we calculate the value for** e ^{ -∏/2 }**with the help of a calculator, we get its approximate value as 0.20788.

i^{i}= 0.20788