can someone tell me how can we calculate (expi)^i.
there is one thought to say that it is equal to exp(-1) but this is not true.
i think that one has to use the euler's equation : exp(ix)=cosx+isinx. but in this case, x=1? and how can we calculate after (cos1+isin1)^i?
anyway that was my way of thinking, of course someone specialised on maths knows better than me? Any ideas?
To calculate the expression (expi)^i, we indeed need to use Euler's equation: exp(ix) = cos(x) + isin(x).
First, we can rewrite (expi)^i as (exp(i * ln(e)))^i. Here, we use the fact that exp(x) and ln(x) are inverse functions, so ln(exp(x)) = x.
Next, we substitute x with i, which gives us: (exp(i * ln(e)))^i = (exp(i * i * ln(e))).
Since i * i = -1, we have: (exp(-ln(e))) = exp(-1).
Therefore, the expression (expi)^i simplifies to exp(-1).