Is this the correct way to think about the comlplex function $f(z)=cos(z)$?












0












$begingroup$


When considering complex functions I know that say $f(z)=e^z=e^{x+iy}=e^x(cos(y)+isin(y))$. So now consider $f(z)=cos(z).$



My question is :



Following the same logic used for $e^z$ does this the mean that $cos(z)=cos(x+iy)=cos(x)cos(iy)-sin(x)sin(iy)$ (by trig identities) ?






complex-analysis trigonometry complex-numbers







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    When considering complex functions I know that say $f(z)=e^z=e^{x+iy}=e^x(cos(y)+isin(y))$. So now consider $f(z)=cos(z).$



    My question is :



    Following the same logic used for $e^z$ does this the mean that $cos(z)=cos(x+iy)=cos(x)cos(iy)-sin(x)sin(iy)$ (by trig identities) ?






    complex-analysis trigonometry complex-numbers







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      When considering complex functions I know that say $f(z)=e^z=e^{x+iy}=e^x(cos(y)+isin(y))$. So now consider $f(z)=cos(z).$



      My question is :



      Following the same logic used for $e^z$ does this the mean that $cos(z)=cos(x+iy)=cos(x)cos(iy)-sin(x)sin(iy)$ (by trig identities) ?






      complex-analysis trigonometry complex-numbers







      share|cite|improve this question











      $endgroup$




      When considering complex functions I know that say $f(z)=e^z=e^{x+iy}=e^x(cos(y)+isin(y))$. So now consider $f(z)=cos(z).$



      My question is :



      Following the same logic used for $e^z$ does this the mean that $cos(z)=cos(x+iy)=cos(x)cos(iy)-sin(x)sin(iy)$ (by trig identities) ?






      complex-analysis trigonometry complex-numbers




      complex-analysis trigonometry complex-numbers






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 12 at 23:08









      José Carlos Santos

      175k24134243




      175k24134243










      asked Jan 12 at 23:04









      can'tcauchycan'tcauchy

      1,025417




      1,025417






















          1 Answer
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          $begingroup$

          Yes, that is correct. I mean, that's not the definition, but it is correct nevertheless. Now, you should add to it that$$cos(iy)=1+frac{y^2}{2!}+frac{y^4}{4!}+cdots=cosh(y)$$and that$$sin(iy)=iy+ifrac{y^3}{3!}+ifrac{y^5}{5!}+cdots=isinh(y).$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Ah this addition to what I was thinking will prove to be very helpful in calculating residues and the like I think . Thank you :) I will accept the answer when it allows me it says I have to wait six minutes
            $endgroup$
            – can'tcauchy
            Jan 12 at 23:13












          • $begingroup$
            one more small query , What is the usual definition if mine is not the standard ?
            $endgroup$
            – can'tcauchy
            Jan 12 at 23:14






          • 1




            $begingroup$
            I think that the most usual one is$$cos z=1-frac{z^2}{2!}+frac{z^4}{4!}-cdotstext{ and }sin z=z-frac{z^3}{3!}+frac{z^5}{5!}-cdots$$Another possibility is$$cos z=frac{e^{iz}+e^{-iz}}2text{ and }sin z=frac{e^{iz}-e^{-iz}}{2i}.$$
            $endgroup$
            – José Carlos Santos
            Jan 12 at 23:17












          • $begingroup$
            ah yes these look familiar , now that I see them again. thanks.
            $endgroup$
            – can'tcauchy
            Jan 12 at 23:17












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          $begingroup$

          Yes, that is correct. I mean, that's not the definition, but it is correct nevertheless. Now, you should add to it that$$cos(iy)=1+frac{y^2}{2!}+frac{y^4}{4!}+cdots=cosh(y)$$and that$$sin(iy)=iy+ifrac{y^3}{3!}+ifrac{y^5}{5!}+cdots=isinh(y).$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Ah this addition to what I was thinking will prove to be very helpful in calculating residues and the like I think . Thank you :) I will accept the answer when it allows me it says I have to wait six minutes
            $endgroup$
            – can'tcauchy
            Jan 12 at 23:13












          • $begingroup$
            one more small query , What is the usual definition if mine is not the standard ?
            $endgroup$
            – can'tcauchy
            Jan 12 at 23:14






          • 1




            $begingroup$
            I think that the most usual one is$$cos z=1-frac{z^2}{2!}+frac{z^4}{4!}-cdotstext{ and }sin z=z-frac{z^3}{3!}+frac{z^5}{5!}-cdots$$Another possibility is$$cos z=frac{e^{iz}+e^{-iz}}2text{ and }sin z=frac{e^{iz}-e^{-iz}}{2i}.$$
            $endgroup$
            – José Carlos Santos
            Jan 12 at 23:17












          • $begingroup$
            ah yes these look familiar , now that I see them again. thanks.
            $endgroup$
            – can'tcauchy
            Jan 12 at 23:17
















          1












          $begingroup$

          Yes, that is correct. I mean, that's not the definition, but it is correct nevertheless. Now, you should add to it that$$cos(iy)=1+frac{y^2}{2!}+frac{y^4}{4!}+cdots=cosh(y)$$and that$$sin(iy)=iy+ifrac{y^3}{3!}+ifrac{y^5}{5!}+cdots=isinh(y).$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Ah this addition to what I was thinking will prove to be very helpful in calculating residues and the like I think . Thank you :) I will accept the answer when it allows me it says I have to wait six minutes
            $endgroup$
            – can'tcauchy
            Jan 12 at 23:13












          • $begingroup$
            one more small query , What is the usual definition if mine is not the standard ?
            $endgroup$
            – can'tcauchy
            Jan 12 at 23:14






          • 1




            $begingroup$
            I think that the most usual one is$$cos z=1-frac{z^2}{2!}+frac{z^4}{4!}-cdotstext{ and }sin z=z-frac{z^3}{3!}+frac{z^5}{5!}-cdots$$Another possibility is$$cos z=frac{e^{iz}+e^{-iz}}2text{ and }sin z=frac{e^{iz}-e^{-iz}}{2i}.$$
            $endgroup$
            – José Carlos Santos
            Jan 12 at 23:17












          • $begingroup$
            ah yes these look familiar , now that I see them again. thanks.
            $endgroup$
            – can'tcauchy
            Jan 12 at 23:17














          1












          1








          1





          $begingroup$

          Yes, that is correct. I mean, that's not the definition, but it is correct nevertheless. Now, you should add to it that$$cos(iy)=1+frac{y^2}{2!}+frac{y^4}{4!}+cdots=cosh(y)$$and that$$sin(iy)=iy+ifrac{y^3}{3!}+ifrac{y^5}{5!}+cdots=isinh(y).$$






          share|cite|improve this answer









          $endgroup$



          Yes, that is correct. I mean, that's not the definition, but it is correct nevertheless. Now, you should add to it that$$cos(iy)=1+frac{y^2}{2!}+frac{y^4}{4!}+cdots=cosh(y)$$and that$$sin(iy)=iy+ifrac{y^3}{3!}+ifrac{y^5}{5!}+cdots=isinh(y).$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 12 at 23:08









          José Carlos SantosJosé Carlos Santos

          175k24134243




          175k24134243












          • $begingroup$
            Ah this addition to what I was thinking will prove to be very helpful in calculating residues and the like I think . Thank you :) I will accept the answer when it allows me it says I have to wait six minutes
            $endgroup$
            – can'tcauchy
            Jan 12 at 23:13












          • $begingroup$
            one more small query , What is the usual definition if mine is not the standard ?
            $endgroup$
            – can'tcauchy
            Jan 12 at 23:14






          • 1




            $begingroup$
            I think that the most usual one is$$cos z=1-frac{z^2}{2!}+frac{z^4}{4!}-cdotstext{ and }sin z=z-frac{z^3}{3!}+frac{z^5}{5!}-cdots$$Another possibility is$$cos z=frac{e^{iz}+e^{-iz}}2text{ and }sin z=frac{e^{iz}-e^{-iz}}{2i}.$$
            $endgroup$
            – José Carlos Santos
            Jan 12 at 23:17












          • $begingroup$
            ah yes these look familiar , now that I see them again. thanks.
            $endgroup$
            – can'tcauchy
            Jan 12 at 23:17


















          • $begingroup$
            Ah this addition to what I was thinking will prove to be very helpful in calculating residues and the like I think . Thank you :) I will accept the answer when it allows me it says I have to wait six minutes
            $endgroup$
            – can'tcauchy
            Jan 12 at 23:13












          • $begingroup$
            one more small query , What is the usual definition if mine is not the standard ?
            $endgroup$
            – can'tcauchy
            Jan 12 at 23:14






          • 1




            $begingroup$
            I think that the most usual one is$$cos z=1-frac{z^2}{2!}+frac{z^4}{4!}-cdotstext{ and }sin z=z-frac{z^3}{3!}+frac{z^5}{5!}-cdots$$Another possibility is$$cos z=frac{e^{iz}+e^{-iz}}2text{ and }sin z=frac{e^{iz}-e^{-iz}}{2i}.$$
            $endgroup$
            – José Carlos Santos
            Jan 12 at 23:17












          • $begingroup$
            ah yes these look familiar , now that I see them again. thanks.
            $endgroup$
            – can'tcauchy
            Jan 12 at 23:17
















          $begingroup$
          Ah this addition to what I was thinking will prove to be very helpful in calculating residues and the like I think . Thank you :) I will accept the answer when it allows me it says I have to wait six minutes
          $endgroup$
          – can'tcauchy
          Jan 12 at 23:13






          $begingroup$
          Ah this addition to what I was thinking will prove to be very helpful in calculating residues and the like I think . Thank you :) I will accept the answer when it allows me it says I have to wait six minutes
          $endgroup$
          – can'tcauchy
          Jan 12 at 23:13














          $begingroup$
          one more small query , What is the usual definition if mine is not the standard ?
          $endgroup$
          – can'tcauchy
          Jan 12 at 23:14




          $begingroup$
          one more small query , What is the usual definition if mine is not the standard ?
          $endgroup$
          – can'tcauchy
          Jan 12 at 23:14




          1




          1




          $begingroup$
          I think that the most usual one is$$cos z=1-frac{z^2}{2!}+frac{z^4}{4!}-cdotstext{ and }sin z=z-frac{z^3}{3!}+frac{z^5}{5!}-cdots$$Another possibility is$$cos z=frac{e^{iz}+e^{-iz}}2text{ and }sin z=frac{e^{iz}-e^{-iz}}{2i}.$$
          $endgroup$
          – José Carlos Santos
          Jan 12 at 23:17






          $begingroup$
          I think that the most usual one is$$cos z=1-frac{z^2}{2!}+frac{z^4}{4!}-cdotstext{ and }sin z=z-frac{z^3}{3!}+frac{z^5}{5!}-cdots$$Another possibility is$$cos z=frac{e^{iz}+e^{-iz}}2text{ and }sin z=frac{e^{iz}-e^{-iz}}{2i}.$$
          $endgroup$
          – José Carlos Santos
          Jan 12 at 23:17














          $begingroup$
          ah yes these look familiar , now that I see them again. thanks.
          $endgroup$
          – can'tcauchy
          Jan 12 at 23:17




          $begingroup$
          ah yes these look familiar , now that I see them again. thanks.
          $endgroup$
          – can'tcauchy
          Jan 12 at 23:17


















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