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Complex numbers

help here
simplify;cosA-isinA/cos4A-isin4A
(edited 8 years ago)
Reply 1
Avatar for Zacken
Zacken
22
Original post by shady2.0
help here
simplify;cosA-isinA/cos4A-isinA


cos⁡A−isin⁡A=e−iA\cos A - i \sin A = e^{-iA} and cos⁡4A−isin⁡4A=e−4iA\cos 4A - i\sin 4A = e^{-4iA}

So your thing is equivalent to e−ie−4iA\displaystyle \frac{e^{-i}}{e^{-4iA}}, simply that using the rule of indices, then move it back into trig. form.
Original post by Zacken
...


Is it −isin⁡4A-i\sin 4A in the denominator though?

Probably should be, and then I'd agree with your working, but it's not showing as such in the initial post.
Reply 3
Avatar for shady2.0
shady2.0
OP
3
Original post by ghostwalker
Is it −isin⁡4A-i\sin 4A in the denominator though?

Probably should be, and then I'd agree with your working, but it's not showing as such in the initial post.
yes of course it is -isin4A in denominator ..but can u tell me how how simplify the trig identities in the exponential form like the way u did it?
Original post by shady2.0
yes of course it is -isin4A in denominator


Then I suggest you correct your original post.


..but can u tell me how how simplify the trig identities in the exponential form like the way u did it?


You want to quote Zacken, not me.

That said:

cos⁡x−isin⁡x=cos⁡(−x)+isin⁡(−x)=e−ix\cos x - i \sin x=\cos (-x) +i\sin(-x)=e^{-ix}

etc.
Reply 5
Avatar for Zacken
Zacken
22
Original post by ghostwalker
Is it −isin⁡4A-i\sin 4A in the denominator though?

Probably should be, and then I'd agree with your working, but it's not showing as such in the initial post.


It would have looked a little out of place if it wasn't in the denominator so I just assumed it was. But yeah, ambiguous on the OP's part.

@OP, for future reference, you may want to use brackets to make your equations clear. What you wrote could be mis-interpreted as cos⁡A−isin⁡Acos⁡4A−isin⁡4A\frac{\cos A - i\sin A}{\cos 4A} - i \sin 4A, it would have been much clear had you written (cos A - i sin A)/(cos 4A - i sin 4A).
Original post by Zacken
It would have looked a little out of place if it wasn't in the denominator so I just assumed it was. But yeah, ambiguous on the OP's part.


My issue was with the value, rather than it being in the denominator or not.

Original post was showing −isin⁡A-i\sin A in the denominator.

And has now been corrected.
(edited 8 years ago)
Reply 7
Avatar for Zacken
Zacken
22
Original post by ghostwalker
My issue was with the value, rather than it being in the denominator or not.

Original post was showing −isin⁡A-i\sin A in the denominator.

And has now been corrected.


Ah, I didn't even notice that. :facepalm:

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