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.
 
Good morning I have been refreshing my memory about Leibniz differentiation of integrals and found some useful videos from digital-university.org on YouTube. Although the audio quality is poor and the speaker proceeds a bit slowly, the explanations and processes are clear. However, it seems that one video in the Leibniz rule series is missing. While the videos are still present on YouTube, the referring website no longer exists but is preserved on the internet archive...

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