What is the value of iota^iota?
Answers
Answered by
1
e^-π/2........................
Anonymous:
Is there any proof to that?
yes
Bro..can u provide it?
cos@ + i.sin@ = e^(i.@)
if we take @ = pi/2:
i = e^(i.pi*/2)
Raise both sides to exponent i:
i^i = e^ (i.pi/2 . i)
i^i = e^(-pi/2)
if we take @ = pi/2:
i = e^(i.pi*/2)
Raise both sides to exponent i:
i^i = e^ (i.pi/2 . i)
i^i = e^(-pi/2)
Ok so this works at pie/2 !!...Thnk u ..nice explanation
welcome
Answered by
1
cosθ + isinθ= e^(i.θ)
Consider θ = π/2
then,
i = e^(iπ*/2)
Raising both the sides to exponent i
i^i = e^ (i.π/2.i)
i^i = e^(-π/2)
Consider θ = π/2
then,
i = e^(iπ*/2)
Raising both the sides to exponent i
i^i = e^ (i.π/2.i)
i^i = e^(-π/2)
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