Year 12s: what will be the first way you research your future options? >>

Year 12s: what will be the first way you research your future options? >>

Year 12s: what will be the first way you research your future options? >>

Fp3 integration question

http://www.ocr.org.uk/Images/64993-question-paper-unit-4727-further-pure-mathematics-3.pdf

Question 4. I get a bit stuck on simplifying:

Unparseable latex formula:

\frac{2-3i}{13}(e^{(3i+2)}^\frac{\pi}{2}-e^0)

how does the

Unparseable latex formula:

(e^{(3i+2)}^\frac{\pi}{2}-e^0)

simplify down to

-ie^\pi-1

it's the -i part i don't get.

Study Forum Helper

Original post by Music99

http://www.ocr.org.uk/Images/64993-question-paper-unit-4727-further-pure-mathematics-3.pdf

Question 4. I get a bit stuck on simplifying:

Unparseable latex formula:

\frac{2-3i}{13}(e^{(3i+2)}^\frac{\pi}{2}-e^0)

how does the

Unparseable latex formula:

(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}

(edited 11 years ago)

OP

Original post by Mr M

e^{\frac{3i\pi}{2}}=\cos \frac{3\pi}{2}+i \sin {3\pi}{2}

Ahh of course thanks!

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 :smile:

:smile:

hope that makes sense I'm awful at explaining! I did aqa and this was part of fp2 for us :smile:

:smile:

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