Question

how do you express 2i in trigonometric form?

Answer 1

\(2i\) is actually \(0+2i\)...does that help?

Answer 2

@yummydum would that mean that its cos90+isin90?

Answer 3

if you multiply by 2, yes

Answer 4

since \(\cos(\frac{\pi}{2})=0\) and \(\sin(\frac{\pi}{2})=1\) you have
\[2(\cos(\frac{\pi}{2})+i\sin(\frac{\pi}{2}))=2\times (0+i)=2i\]

Answer 5

you can turn that into trigonometric form bt finding \(r\) by using \(r=\sqrt{a^2+b^2}\) and then find \(\theta\)using \(\theta=tan^{-1}\left(b\over a\right)\) so do:\[r=\sqrt{0+2^2}\]\[r=\sqrt{4}\]\[r=2\]\[\theta=\tan^{-1}\left({2\over0}\right)\]\[\theta=\tan^{-1}(0)\]\[\theta=90~degrees\]now write it in \(r(\cos\theta+i~\sin\theta)\) form which should give you:\[2(\cos90+i\sin90)\]

Answer 6

hope that helps! @serinaaa :)

Answer 7

@satellite73 @yummydum thank you so much!

Answer 8

@yummydum don't forget to be careful when using \(\theta=\tan^{-1}(\frac{b}{a})\) because that is not always the case

Answer 9

i figured out how to use it though, thanks btw :)