urgent help on math question - Singapore Forums by SGClub.com  Home Photos Member List Register Mark Forums Read   Home » urgent help on math question
 Why aren't you a member of SGClub.com yet?? » Join 130,000+ other members in chatting. » Make lots of new friends here. » Keep up-to-date with current events. » Participate in Club outings. » Download lots of Free Stuff! Registration just takes 2mins and is absolutely free so join our community today! I Want to Choose my Own Personal Nickname Now! 4th May 2013, 12:10 AM #1 (permalink) Experienced SGClubber  Posts: 1,634 Join Date: Dec 2009 Likes: 21 Liked 39 Times in 26 Posts Gender: urgent help on math question Tweet i am having my paper on monday and there are questions which i cannot figure out how to do... please assist... 1) Using the identity e^(iz) = cos z + i sin z, where z = x+iy, express f(z) = sin z in the form of f(z) = u(x,y) + iv(x,y), where u(x,y) and v(x,y) are real functions of x and y. my attempt on this question can only reach till: f(z) = [e^(ix)] / [2i*e^y] - e^y /2i *e^(-ix) Sponsors:    4th May 2013, 04:45 PM #2 (permalink) Experienced SGClubber    Posts: 2,404 Join Date: Jul 2009 Likes: 0 Liked 211 Times in 174 Posts Gender: Re: urgent help on math question Tweet f(z)=sin(z) =sin(x+iy) = sin(x) cos(iy) +cos(x) sin(iy) ------------(1) Recognizing that in general, cos(y) =[ e^(iy)+ e^(-iy) ]/2, then cos(iy) = { e^[i(iy)]+ e^-[i(iy)] } /2 (simply replace y by iy on the RHS) = [e^(-y) +e^(y)]/2 = cosh(y) Similarly, recognizing that in general, sin(y) =[ e^(iy)- e^(-iy) ]/ (2i) , then sin(iy) ={ e^[i(iy)] - e^-[i(iy)] } /(2i) = [e^(-y) - e^(y)]/ (2i) = (i)* [e^(y) - e^(-y)]/2 = i sinh(y) Substituting these into (1), we have f(z)= sin(x) cosh(y) + i cos(x) sinh(y) , where u(x,y) = sin(x) cosh(y) and v(x,y) =cos(x) sinh(y) (shown) Hope this helps. Good luck for your upcoming test. Peace. __________________ White Group Mathematics Free A Level H2 Maths Resource Site Last edited by whitecorp; 4th May 2013 at 04:47 PM.    4th May 2013, 04:57 PM #3 (permalink) Cool SGClubber  Posts: 129 Join Date: Nov 2011 Likes: 7 Liked 11 Times in 10 Posts Gender: Re: urgent help on math question Tweet okay; sin z = [e^iz-e^(-iz)]/[2i]=i[e^(-iz)-e^iz]/2=...=i[e^(-ix+y)-e^(ix-y)]/2 Here's the tricky part; sin z =i[e^(-ix+y)-e^(ix+y)+e^(ix+y)-e^(ix-y)]/2 =i[-e^y(e^ix-e^(-ix))+e^(ix)(e^y-e^-y)]/2 =i[-(2isinx)e^y+(cosx+isinx)(e^y-e^-y)]/2 =i(cosx(e^y-e^-y)-isinx(e^y+e^-y)]/2 =icosxsinhy+sinxcoshy    4th May 2013, 05:00 PM #4 (permalink) Cool SGClubber  Posts: 129 Join Date: Nov 2011 Likes: 7 Liked 11 Times in 10 Posts Gender: Re: urgent help on math question Tweet okay; sin z = [e^iz-e^(-iz)]/[2i]=i[e^(-iz)-e^iz]/2=...=i[e^(-ix+y)-e^(ix-y)]/2 Here's the tricky part; sin z =i[e^(-ix+y)-e^(ix+y)+e^(ix+y)-e^(ix-y)]/2 =i[-e^y(e^ix-e^(-ix))+e^(ix)(e^y-e^-y)]/2 =i[-(2isinx)e^y+(cosx+isinx)(e^y-e^-y)]/2 =i(cosx(e^y-e^-y)-isinx(e^y+e^-y)]/2 =icosxsinhy+sinxcoshy, sinh(y)=(e^y-e^y)/2 and cosh(y)=(e^y+e^-y)/2   Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post pokkax Math Help 11 3rd September 2010 08:03 PM Angelicia SGClub Cafe 11 19th July 2010 05:16 PM qwertylol42 Math Help 4 29th October 2009 08:34 PM thankyou Math Help 12 12th July 2009 02:34 PM kelecthor Math Help 1 14th May 2009 09:40 PM Featured Photos by marisoljames322 · · · Member Galleries 20359 photos13619 comments by marisoljames322 · · · Member Galleries 20359 photos13619 comments by pdsubbu · · · Member Galleries 20359 photos13619 comments by Vikas Dhar · · · Photography 35 photos36 comments by aaudreygan · · · Guy Photos 1581 photos946 comments by aaudreygan · · · Girl Photos 4351 photos28606 comments by a8paris · · · Guy Photos 1581 photos946 comments by a8paris · · · Guy Photos 1581 photos946 comments by aaricia · · · Girl Photos 4351 photos28606 comments by aaricia · · · Girl Photos 4351 photos28606 comments Copyright© 2004-2013 SGClub.com. All rights reserved.