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Akshay

is i^i a real number?

what is its value?

guys, i think i've got the answer.........but i'd like to get it verified.................

answer i've got:e^( -pi/2)(euler form)

Rahul Roy

Re:Complex Numbers........e^(pi/2)^i=e^iSq.pi/2=e^-pi/2so purely real

Neeraj Agarwal

yup...both correct

Rahul Raghavendra

I would just like to add a bit ........

To find $i^i$ , lets look at it in this way

Let $e^z=i$ , where z= a+ib=any complex number

Now taking complex powers (i) on both sides .We get

$(e^z)^i=i^i=e^{iz}$ .

Now

$e^z=e^{a+ib}=e^{a}e^{ib}=e^{a}(cos(b)+isin(b))$ which is in turn = i

Therefore By comparing

We get

a=0 and sinb=1

Thus $b=\frac{\pi}{2}+2n\pi$ n belongs to I

Therefore

$e^{iz}=e^{i(i(\frac{\pi}{2}+2n\pi))}$ = $i^i$

Therefore $i^i=e^{\frac{-\pi}{2}-2n\pi}$

The smallest being = $e^{-\frac{\pi}{2}}$