i-i-Whis-is-it-plz-explain-it-i-1-

Question Number 53688 by kwonjun1202 last updated on 25/Jan/19

i^{i} ? ?

W h i s i s i t,, p l z e x p l a i n i t

i = \sqrt{- 1}

Commented by Abdo msup. last updated on 25/Jan/19

w e h a v e i = e^{\frac{i π}{2}} \Rightarrow i^{i} = e^{- \frac{π}{2}} .

Answered by MJS last updated on 25/Jan/19

i = 1 \times e^{i \frac{π}{2}}

i^{i} = 1^{i} \times e^{i^{2} \frac{π}{2}} = 1 \times e^{- \frac{π}{2}} = e^{- \frac{π}{2}} = \frac{1}{\sqrt{e^{π}}}

Answered by tanmay.chaudhury50@gmail.com last updated on 25/Jan/19

w e k n o w a^{x} = e^{{x l o g}_{e} a}

n o w i^{i} = e^{{i L o g}_{e} i} = e^{i L o g (0 + 1 \times i)}

n o w f o r m u l a

{L o g}_{e} (α + i β) = {l o g}_{e} (α + i β) + i 2 π n

= [{l o g}_{e} \sqrt{α^{2} + β^{2}} + {i t a n}^{- 1} (\frac{β}{α})] + i 2 π n

{L o g}_{e} (α + i β)

= \frac{1}{2} {l o g}_{e} (α^{2} + β^{2}) + i {2 π n + {t a n}^{- 1} (\frac{β}{α})}

s o a s p e r p r o b l e m α = 0 β = 1

s o v a l u e o f {L o g}_{e} (0 + i)

= \frac{1}{2} {l o g}_{e} (0^{2} + 1^{2}) + i {2 π n + {t a n}^{- 1} (\frac{1}{0})}

= \frac{1}{2} \times 0 + i {2 π n + \frac{π}{2}}

n o w

i^{i} = e^{{i L o g}_{e} i}

= e^{i {(2 π n + \frac{π}{2}) i}}

= e^{(2 π n + \frac{π}{2}) \times - 1}

= e^{- (2 π n + \frac{π}{2})}

= e^{- \frac{π}{2}} p r i n c i p l e v a l u e