- #1

binbagsss

- 1,266

- 11

Member warned about posting without the homework template

Im to solve ##(k+l)^{2}e^{-ila}-(k-l)^{2}e^{ila}=0##, for ## k##,

The solution is ##k=l(e^{ial}-1)/(e^{ial}+1)=il tan(al/2)##

FIRST QUESTION

So it's a quadratic in k, should be simple enough, my working so far using the quad. formula is ##k= (4l^{2}(e^{-ila}+e^{ila})\pm \sqrt{4l^{2}(e^{-ila}+e^{ila})^{2}-4l^{2}(e^{-ila}-e^{ila})^{2}})/2(e^{-ila}-e^{ila})##

=##(4l^{2}(e^{-ila}+e^{ila})\pm \sqrt{16l^{2}})/2(e^{-ila}-e^{ila})##

... which doesn't give the correct answer.

SECOND QUESTION.

Even after attaining the correct solution, I'm struggling to show that ##l(e^{ial}-1)/(e^{ial}+1)=il tan(al/2)##

LHS: ##cos (al) + isin (al) -1 / cos (al) + isin (al) +1=(cos^{2}(al/2)-sin^{2}(al/2)+isin(al/2)cos(al/2)-1)/(cos^{2}(al/2)-sin^{2}(al/2)+isin(al/2)cos(al/2)+1)##

using## cos 2x= cos^{2}x-sin^{2}x##

and ##sin 2x=2sinxcosx,##

but this doesn't seem to be going anywhere...

Any help on either question really appreciated,

Thanks in advance.

The solution is ##k=l(e^{ial}-1)/(e^{ial}+1)=il tan(al/2)##

FIRST QUESTION

So it's a quadratic in k, should be simple enough, my working so far using the quad. formula is ##k= (4l^{2}(e^{-ila}+e^{ila})\pm \sqrt{4l^{2}(e^{-ila}+e^{ila})^{2}-4l^{2}(e^{-ila}-e^{ila})^{2}})/2(e^{-ila}-e^{ila})##

=##(4l^{2}(e^{-ila}+e^{ila})\pm \sqrt{16l^{2}})/2(e^{-ila}-e^{ila})##

... which doesn't give the correct answer.

SECOND QUESTION.

Even after attaining the correct solution, I'm struggling to show that ##l(e^{ial}-1)/(e^{ial}+1)=il tan(al/2)##

LHS: ##cos (al) + isin (al) -1 / cos (al) + isin (al) +1=(cos^{2}(al/2)-sin^{2}(al/2)+isin(al/2)cos(al/2)-1)/(cos^{2}(al/2)-sin^{2}(al/2)+isin(al/2)cos(al/2)+1)##

using## cos 2x= cos^{2}x-sin^{2}x##

and ##sin 2x=2sinxcosx,##

but this doesn't seem to be going anywhere...

Any help on either question really appreciated,

Thanks in advance.

Last edited: