IIT JEE , AIEEE Forum - Mathematics , Algebra

ritwalok

-1=i & i^i=e^(-pi/2) . i is imaginary no. i^i which is imaginary raise to imaginary ,then how it is a very large real no.

kghedriu

say f(x)= e^(sin-1x) ....it is e to the power of sin inverse x...

now... f(1/2)= e^pi/6 and f(-1/2)= e^-pi/6

cubing this....we get e^pi/2 or e^-pi/2 in RHS.....

by expansions...

f(x)= 1+x+ x^2/2! +2x^3/3!+ 5x^4/4!+.......

={1 +

t 2k+1}+ {x+

t 2k+2} where k runs frm 1 to infinity and mod x less dan 1..

and t2k+1= (1)(2^2+1)(4^2+1).....[(2k-1)^2+1]x^2k/(2k)!

and

t 2k+2 = (1^2+1)(3^2+1)........x^2k+1/(2k+1)!

now...use these formulas to calculate 85 terms in each with each term rounded to 54 decimal places..this processs gives fifty two decimal places of e^pi/6

simly calc for e^-pi/6

den multiply to get e ^-pi/3

calc e^-pi/3

multiply to get e^-pi/2 which is LHS and i^i is our RHS

this is the only way to do this...

it was given in 1922 by UHLER....

finally the value of i^i is as follows...

0.2078795763

elessar_iitkgp

i = cos

/2 + i sin

/2 = e^i/2
iⁱ = (e^i/2)ⁱ = e^-/2 (As i² = -1)
= 1/e^/2
Now e lies between 2 and 3.

/2 is approximately = 1.57
Hence e^/2 isn't a very large value. Infact it will be less than 6 but definitely greater than 1.
Hence its reciprocal 1/e^/2should be less than unity
And hence iⁱ is a real number less than unity.

Please check the question. It shouldn't be a large real number.

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