Hint to show $tanh(z)=frac{sinh(2x)+isin(2y)}{cosh(2x)+cos(2y)}$?












1












$begingroup$


I really can't figure out how to do this at all. I've been trying to show this for nearly 4 hours now. I've tried working from $tanh(z)=frac{sinh(z)}{cosh(z)}$ and expanding the top and bottom, but that just becomes a mess that, after trying so hard to put into the desired form, didn't work. I also tried working from the identity



$$tanh(z)=tanh(x+iy)=frac{tanh(x)+tanh(iy)}{1+tanh(x)tanh(iy)}$$



but I wasn't able to put that into the desired form as well. I've tried working from the right side to the left, using every formula for $sin(2y)$, $sinh(2x)$, $cosh(2x)$, and $cos(2y)$ I could derive/find. I even tried multiplying the right-side by $coth(z)$ and working it out to show that it's equal to $1$, but that didn't work.



Could you please give me some hints?






algebra-precalculus complex-numbers hyperbolic-functions







share|cite|improve this question









$endgroup$












  • $begingroup$
    In $tanh(z) = frac{sinh(z)}{cosh(z)}$, write out the ratio in terms of exponentials, and then multiply and divide by the conjugate of the denominator (i.e $cosh(overline{z})$),
    $endgroup$
    – vnd
    Nov 5 '15 at 3:42












  • $begingroup$
    @vnd Yeah I tried that and it worked. I just saw that somebody had already worked out the solution in an answer (below)!
    $endgroup$
    – Arturo don Juan
    Nov 5 '15 at 4:07
















1












$begingroup$


I really can't figure out how to do this at all. I've been trying to show this for nearly 4 hours now. I've tried working from $tanh(z)=frac{sinh(z)}{cosh(z)}$ and expanding the top and bottom, but that just becomes a mess that, after trying so hard to put into the desired form, didn't work. I also tried working from the identity



$$tanh(z)=tanh(x+iy)=frac{tanh(x)+tanh(iy)}{1+tanh(x)tanh(iy)}$$



but I wasn't able to put that into the desired form as well. I've tried working from the right side to the left, using every formula for $sin(2y)$, $sinh(2x)$, $cosh(2x)$, and $cos(2y)$ I could derive/find. I even tried multiplying the right-side by $coth(z)$ and working it out to show that it's equal to $1$, but that didn't work.



Could you please give me some hints?






algebra-precalculus complex-numbers hyperbolic-functions







share|cite|improve this question









$endgroup$












  • $begingroup$
    In $tanh(z) = frac{sinh(z)}{cosh(z)}$, write out the ratio in terms of exponentials, and then multiply and divide by the conjugate of the denominator (i.e $cosh(overline{z})$),
    $endgroup$
    – vnd
    Nov 5 '15 at 3:42












  • $begingroup$
    @vnd Yeah I tried that and it worked. I just saw that somebody had already worked out the solution in an answer (below)!
    $endgroup$
    – Arturo don Juan
    Nov 5 '15 at 4:07














1












1








1





$begingroup$


I really can't figure out how to do this at all. I've been trying to show this for nearly 4 hours now. I've tried working from $tanh(z)=frac{sinh(z)}{cosh(z)}$ and expanding the top and bottom, but that just becomes a mess that, after trying so hard to put into the desired form, didn't work. I also tried working from the identity



$$tanh(z)=tanh(x+iy)=frac{tanh(x)+tanh(iy)}{1+tanh(x)tanh(iy)}$$



but I wasn't able to put that into the desired form as well. I've tried working from the right side to the left, using every formula for $sin(2y)$, $sinh(2x)$, $cosh(2x)$, and $cos(2y)$ I could derive/find. I even tried multiplying the right-side by $coth(z)$ and working it out to show that it's equal to $1$, but that didn't work.



Could you please give me some hints?






algebra-precalculus complex-numbers hyperbolic-functions







share|cite|improve this question









$endgroup$




I really can't figure out how to do this at all. I've been trying to show this for nearly 4 hours now. I've tried working from $tanh(z)=frac{sinh(z)}{cosh(z)}$ and expanding the top and bottom, but that just becomes a mess that, after trying so hard to put into the desired form, didn't work. I also tried working from the identity



$$tanh(z)=tanh(x+iy)=frac{tanh(x)+tanh(iy)}{1+tanh(x)tanh(iy)}$$



but I wasn't able to put that into the desired form as well. I've tried working from the right side to the left, using every formula for $sin(2y)$, $sinh(2x)$, $cosh(2x)$, and $cos(2y)$ I could derive/find. I even tried multiplying the right-side by $coth(z)$ and working it out to show that it's equal to $1$, but that didn't work.



Could you please give me some hints?






algebra-precalculus complex-numbers hyperbolic-functions




algebra-precalculus complex-numbers hyperbolic-functions






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 5 '15 at 3:25









Arturo don JuanArturo don Juan

1,8111032




1,8111032












  • $begingroup$
    In $tanh(z) = frac{sinh(z)}{cosh(z)}$, write out the ratio in terms of exponentials, and then multiply and divide by the conjugate of the denominator (i.e $cosh(overline{z})$),
    $endgroup$
    – vnd
    Nov 5 '15 at 3:42












  • $begingroup$
    @vnd Yeah I tried that and it worked. I just saw that somebody had already worked out the solution in an answer (below)!
    $endgroup$
    – Arturo don Juan
    Nov 5 '15 at 4:07


















  • $begingroup$
    In $tanh(z) = frac{sinh(z)}{cosh(z)}$, write out the ratio in terms of exponentials, and then multiply and divide by the conjugate of the denominator (i.e $cosh(overline{z})$),
    $endgroup$
    – vnd
    Nov 5 '15 at 3:42












  • $begingroup$
    @vnd Yeah I tried that and it worked. I just saw that somebody had already worked out the solution in an answer (below)!
    $endgroup$
    – Arturo don Juan
    Nov 5 '15 at 4:07
















$begingroup$
In $tanh(z) = frac{sinh(z)}{cosh(z)}$, write out the ratio in terms of exponentials, and then multiply and divide by the conjugate of the denominator (i.e $cosh(overline{z})$),
$endgroup$
– vnd
Nov 5 '15 at 3:42






$begingroup$
In $tanh(z) = frac{sinh(z)}{cosh(z)}$, write out the ratio in terms of exponentials, and then multiply and divide by the conjugate of the denominator (i.e $cosh(overline{z})$),
$endgroup$
– vnd
Nov 5 '15 at 3:42














$begingroup$
@vnd Yeah I tried that and it worked. I just saw that somebody had already worked out the solution in an answer (below)!
$endgroup$
– Arturo don Juan
Nov 5 '15 at 4:07




$begingroup$
@vnd Yeah I tried that and it worked. I just saw that somebody had already worked out the solution in an answer (below)!
$endgroup$
– Arturo don Juan
Nov 5 '15 at 4:07










1 Answer
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$begingroup$

Let me try. We have $$tanh (z) = frac{sinh z}{cosh z} = frac{e^z - e^{-z}}{e^z + e^{-z}} = frac{e^{2z} -1 }{e^{2z}+1} = frac{e^{2x}(cos(y) + isin(y))^2 -1}{e^{2x}(cos(y) + isin(y))^2 + 1} = frac{e^{2x}cos 2y -1 + ie^xsin 2y}{e^{2x}cos 2y + 1 + ie^{2x}sin 2y} = frac{(e^{2x}cos 2y -1 + ie^{2x}sin 2y)(e^{2x}cos 2y +1 - ie^{2x}sin 2y)}{(e^{2x}cos 2y +1)^2 + e^{4x}sin^22y} = frac{e^{4x}cos^2 2y - (1-ie^{2x}sin 2y)^2}{e^{4x}+1 + 2e^{2x}cos 2y} = frac{e^{4x}-1 + 2ie^{2x}sin 2y}{e^{4x}+1 + 2e^{2x}cos 2y} = frac{e^{2x}-e^{-2x} + 2isin 2y}{e^{2x}+e^{-2x} + 2cos 2y} = frac{sinh 2x + isin 2y}{cosh 2x + cos 2y}$$






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    $begingroup$

    Let me try. We have $$tanh (z) = frac{sinh z}{cosh z} = frac{e^z - e^{-z}}{e^z + e^{-z}} = frac{e^{2z} -1 }{e^{2z}+1} = frac{e^{2x}(cos(y) + isin(y))^2 -1}{e^{2x}(cos(y) + isin(y))^2 + 1} = frac{e^{2x}cos 2y -1 + ie^xsin 2y}{e^{2x}cos 2y + 1 + ie^{2x}sin 2y} = frac{(e^{2x}cos 2y -1 + ie^{2x}sin 2y)(e^{2x}cos 2y +1 - ie^{2x}sin 2y)}{(e^{2x}cos 2y +1)^2 + e^{4x}sin^22y} = frac{e^{4x}cos^2 2y - (1-ie^{2x}sin 2y)^2}{e^{4x}+1 + 2e^{2x}cos 2y} = frac{e^{4x}-1 + 2ie^{2x}sin 2y}{e^{4x}+1 + 2e^{2x}cos 2y} = frac{e^{2x}-e^{-2x} + 2isin 2y}{e^{2x}+e^{-2x} + 2cos 2y} = frac{sinh 2x + isin 2y}{cosh 2x + cos 2y}$$






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      $begingroup$

      Let me try. We have $$tanh (z) = frac{sinh z}{cosh z} = frac{e^z - e^{-z}}{e^z + e^{-z}} = frac{e^{2z} -1 }{e^{2z}+1} = frac{e^{2x}(cos(y) + isin(y))^2 -1}{e^{2x}(cos(y) + isin(y))^2 + 1} = frac{e^{2x}cos 2y -1 + ie^xsin 2y}{e^{2x}cos 2y + 1 + ie^{2x}sin 2y} = frac{(e^{2x}cos 2y -1 + ie^{2x}sin 2y)(e^{2x}cos 2y +1 - ie^{2x}sin 2y)}{(e^{2x}cos 2y +1)^2 + e^{4x}sin^22y} = frac{e^{4x}cos^2 2y - (1-ie^{2x}sin 2y)^2}{e^{4x}+1 + 2e^{2x}cos 2y} = frac{e^{4x}-1 + 2ie^{2x}sin 2y}{e^{4x}+1 + 2e^{2x}cos 2y} = frac{e^{2x}-e^{-2x} + 2isin 2y}{e^{2x}+e^{-2x} + 2cos 2y} = frac{sinh 2x + isin 2y}{cosh 2x + cos 2y}$$






      share|cite|improve this answer











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        3








        3





        $begingroup$

        Let me try. We have $$tanh (z) = frac{sinh z}{cosh z} = frac{e^z - e^{-z}}{e^z + e^{-z}} = frac{e^{2z} -1 }{e^{2z}+1} = frac{e^{2x}(cos(y) + isin(y))^2 -1}{e^{2x}(cos(y) + isin(y))^2 + 1} = frac{e^{2x}cos 2y -1 + ie^xsin 2y}{e^{2x}cos 2y + 1 + ie^{2x}sin 2y} = frac{(e^{2x}cos 2y -1 + ie^{2x}sin 2y)(e^{2x}cos 2y +1 - ie^{2x}sin 2y)}{(e^{2x}cos 2y +1)^2 + e^{4x}sin^22y} = frac{e^{4x}cos^2 2y - (1-ie^{2x}sin 2y)^2}{e^{4x}+1 + 2e^{2x}cos 2y} = frac{e^{4x}-1 + 2ie^{2x}sin 2y}{e^{4x}+1 + 2e^{2x}cos 2y} = frac{e^{2x}-e^{-2x} + 2isin 2y}{e^{2x}+e^{-2x} + 2cos 2y} = frac{sinh 2x + isin 2y}{cosh 2x + cos 2y}$$






        share|cite|improve this answer











        $endgroup$



        Let me try. We have $$tanh (z) = frac{sinh z}{cosh z} = frac{e^z - e^{-z}}{e^z + e^{-z}} = frac{e^{2z} -1 }{e^{2z}+1} = frac{e^{2x}(cos(y) + isin(y))^2 -1}{e^{2x}(cos(y) + isin(y))^2 + 1} = frac{e^{2x}cos 2y -1 + ie^xsin 2y}{e^{2x}cos 2y + 1 + ie^{2x}sin 2y} = frac{(e^{2x}cos 2y -1 + ie^{2x}sin 2y)(e^{2x}cos 2y +1 - ie^{2x}sin 2y)}{(e^{2x}cos 2y +1)^2 + e^{4x}sin^22y} = frac{e^{4x}cos^2 2y - (1-ie^{2x}sin 2y)^2}{e^{4x}+1 + 2e^{2x}cos 2y} = frac{e^{4x}-1 + 2ie^{2x}sin 2y}{e^{4x}+1 + 2e^{2x}cos 2y} = frac{e^{2x}-e^{-2x} + 2isin 2y}{e^{2x}+e^{-2x} + 2cos 2y} = frac{sinh 2x + isin 2y}{cosh 2x + cos 2y}$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 17 '18 at 16:47









        gammatester

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        16.7k21632










        answered Nov 5 '15 at 4:01









        GAVDGAVD

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        6,66611129






























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