Hello.
Can someone please check if I got the answer right? I don't have the right answers for my exercise so I need someone to check it for me, please.
Please find the principle value of (1+i)^-i
My attempt:
√2^(- i) = (e^ln√2)^(- i) = e^-i•ln√2
= cos(- ln√2) + i•sin(- ln√2)
= cos(ln√2) - i•sin(ln√2)
(1 + i) = √2[(1/√2) + i(1/√2)]
= √2[cos(π/4) + isin(π/4)]
= √2•e^i(π/4)
(1 + i)^(- i) = [√2•e^i(π/4)]^(- i)
=[e^(π/4)] [√2^(- i)]
=[e^(π/4)][cos(ln√2) - i•sin(ln√2)]
Therefore (1 + i)^(- i) = [e^(π/4)][cos(ln√2) - i•sin(ln√2)]
Can someone please check if I got the answer right? I don't have the right answers for my exercise so I need someone to check it for me, please.
Please find the principle value of (1+i)^-i
My attempt:
√2^(- i) = (e^ln√2)^(- i) = e^-i•ln√2
= cos(- ln√2) + i•sin(- ln√2)
= cos(ln√2) - i•sin(ln√2)
(1 + i) = √2[(1/√2) + i(1/√2)]
= √2[cos(π/4) + isin(π/4)]
= √2•e^i(π/4)
(1 + i)^(- i) = [√2•e^i(π/4)]^(- i)
=[e^(π/4)] [√2^(- i)]
=[e^(π/4)][cos(ln√2) - i•sin(ln√2)]
Therefore (1 + i)^(- i) = [e^(π/4)][cos(ln√2) - i•sin(ln√2)]
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