What is e^-iPi/2 and how can it be represented using cos and sin notation?

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The discussion centers on the expression e^-iPi/2 and its representation in terms of cosine and sine. It is established that e^iPi/2 equals i, leading to the conclusion that e^-iPi/2 equals -i. The formula e^(ix) = cos(x) + i*sin(x) is applied, specifically with x set to -Pi/2, resulting in e^-iPi/2 being expressed as cos(-Pi/2) + i*sin(-Pi/2). The conversation emphasizes understanding this transformation as a rotation in the complex plane. The explanation clarifies the relationship between these complex exponential forms and their geometric interpretations.
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If e^iPi/2=i, what does e^-iPi/2 equal? And could you please write it out in cos + sin notation? For some reason I am not geting this. This is not homework, by the way, its come up in an article on spinors I'm reading Thanks.
 
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A simple google check would do the trick.

Anywho, e^(ix)=cos(x)+i*sin(x) for complex variable x.
 
Klungo said:
A simple google check would do the trick.

Anywho, e^(ix)=cos(x)+i*sin(x) for complex variable x.

Continuing with this. If x=-\pi /2, then

e^{-i\pi /2}=\cos(-\pi/2)+i\sin(-\pi/2)=-i.
 
I got the first one Klungo. Thanks for the follow up micromass. Now I got it, its just like spinning the dial from the top of the complex plane to the bottom.
 
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