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    http://www.ocr.org.uk/Images/64993-q...hematics-3.pdf

    Question 4. I get a bit stuck on simplifying:

     \frac{2-3i}{13}(e^{(3i+2)}^\frac{\pi}{2}-e^0) how does the  (e^{(3i+2)}^\frac{\pi}{2}-e^0) simplify down to -ie^\pi-1 it's the -i part i don't get.
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    (Original post by Music99)
    http://www.ocr.org.uk/Images/64993-q...hematics-3.pdf

    Question 4. I get a bit stuck on simplifying:

     \frac{2-3i}{13}(e^{(3i+2)}^\frac{\pi}{2}-e^0) how does the  (e^{(3i+2)}^\frac{\pi}{2}-e^0) simplify down to -ie^\pi-1 it's the -i part i don't get.
    e^{\frac{3i\pi}{2}}=\cos \frac{3\pi}{2}+i \sin \frac{3\pi}{2}
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    (Original post by Mr M)
    e^{\frac{3i\pi}{2}}=\cos \frac{3\pi}{2}+i \sin {3\pi}{2}
    Ahh of course thanks!
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    Because when you multiply out the brackets you get e^(3ipi/2 + pi) then you use rules of indices to get e^3ipi/2 x e^pi and e^3ipi/2 is the same as -i hope that makes sense I'm awful at explaining! I did aqa and this was part of fp2 for us
 
 
 
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