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Old 4th May 2013, 12:10 AM   #1 (permalink)
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urgent help on math question

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)

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Old 4th May 2013, 04:45 PM   #2 (permalink)
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Re: urgent help on math question

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.


Last edited by whitecorp; 4th May 2013 at 04:47 PM.
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Old 4th May 2013, 04:57 PM   #3 (permalink)
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Re: urgent help on math question

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

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Old 4th May 2013, 05:00 PM   #4 (permalink)
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Re: urgent help on math question

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

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