e^ix = cosx + isinx - Math Central

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Question from trale, a student:

Can we use e^ix = cosx + isinx for finding i^i like that: x= pi/2 =>
e^(ipi/2)=0+i then [e^(ipi/2)]^i=i^i.then we find i^i= 0,207879576....
is it true? can we give value for x for free?thank you.

Hi Trale,

What you have done is correct. Euler's Formula

e^ix = cosx + isinx

is true for any real number x. You have chosen to evaluate this expression when x = π/2 which gives

e^-π/2 = iⁱ

and this gives a way to find a decimal approximation for iⁱ. What does this expression give when x = π? What about x = 2π?

You should also read RD's response to a similar question?

Harley

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