what is the value of iota^iota
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Answer:
Step-by-step explanation:
e^ix =cos x +isin x;
if x=Ï€/2
e^iπ/2=cosπ/2+isinπ/2=i
i^i= (e^iπ/2)^i
i^i=e^-Ï€/2
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