Find every $z$s that fit $cos(z) = -2$












3












$begingroup$


I couldn't find any, I tried to write $cos(z)$ as $cos(x)cos(iy)-sin(x)sin(iy)$ which then gave me



$cos(x)cosh(y) - isin(x)sinh(y) = -2$



$sin(x)=0$ so that imaginary part become $0$



now we have to find $cosh(y) = -2$ which is not true for no $y$.



is it right or i made a mistake in my substitutions?






complex-numbers







share|cite|improve this question











$endgroup$

















    3












    $begingroup$


    I couldn't find any, I tried to write $cos(z)$ as $cos(x)cos(iy)-sin(x)sin(iy)$ which then gave me



    $cos(x)cosh(y) - isin(x)sinh(y) = -2$



    $sin(x)=0$ so that imaginary part become $0$



    now we have to find $cosh(y) = -2$ which is not true for no $y$.



    is it right or i made a mistake in my substitutions?






    complex-numbers







    share|cite|improve this question











    $endgroup$















      3












      3








      3


      3



      $begingroup$


      I couldn't find any, I tried to write $cos(z)$ as $cos(x)cos(iy)-sin(x)sin(iy)$ which then gave me



      $cos(x)cosh(y) - isin(x)sinh(y) = -2$



      $sin(x)=0$ so that imaginary part become $0$



      now we have to find $cosh(y) = -2$ which is not true for no $y$.



      is it right or i made a mistake in my substitutions?






      complex-numbers







      share|cite|improve this question











      $endgroup$




      I couldn't find any, I tried to write $cos(z)$ as $cos(x)cos(iy)-sin(x)sin(iy)$ which then gave me



      $cos(x)cosh(y) - isin(x)sinh(y) = -2$



      $sin(x)=0$ so that imaginary part become $0$



      now we have to find $cosh(y) = -2$ which is not true for no $y$.



      is it right or i made a mistake in my substitutions?






      complex-numbers




      complex-numbers






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 2 at 21:45









      El borito

      664216




      664216










      asked Jan 2 at 21:31









      no0obno0ob

      788




      788






















          3 Answers
          3






          active

          oldest

          votes


















          1












          $begingroup$

          Hint: start with
          $$
          cos z=frac{e^{iz}+e^{-iz}}{2}
          $$
          and obtain a quadratic equation for $e^{iz}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            i got $exp(iz) = -2+sqrt(3)$ and $exp(iz) = -2-sqrt(3)$ which gets me that $e^x = -2+sqrt(3)$ which is negative on the right side and impossible :-?
            $endgroup$
            – no0ob
            Jan 2 at 21:41












          • $begingroup$
            If $e^{iz}=-2pmsqrt{3}$ then $$e^{iz+ipi}=e^{ipi}(-2pmsqrt{3})=(-1)(-2pmsqrt{3})=2pmsqrt{3}$$ whith positive RHS.
            $endgroup$
            – A.Γ.
            Jan 2 at 21:43












          • $begingroup$
            What should $z$ be btw?
            $endgroup$
            – no0ob
            Jan 2 at 21:48










          • $begingroup$
            @no0ob Do you follow the argumentation? If $$e^{i(z+pi)}=2pmsqrt{3}$$ then how to find $i(z+pi)$?
            $endgroup$
            – A.Γ.
            Jan 2 at 21:52






          • 1




            $begingroup$
            @no0ob Right, sorry, it should be every second zero of $sin x=0$ to make $cos x=-1$, i.e. $x=pi+2pi k$.
            $endgroup$
            – A.Γ.
            Jan 2 at 22:10





















          2












          $begingroup$

          I should have used $x=(2k+1)pi$ for the $x$ and then had the



          $cos( (2k+1)pi ) cosh(y) = -2$



          which made it $cosh(y) = 2$






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            Right, and here you get two solutions for $y$ with $pm$ (recall $2pmsqrt{3}$ in another approach).
            $endgroup$
            – A.Γ.
            Jan 2 at 22:12






          • 1




            $begingroup$
            Yes; since $cos xcosh y<0$, you need to have $cos x<0$ and therefore $x=pi+2kpi$.
            $endgroup$
            – egreg
            Jan 2 at 22:33



















          0












          $begingroup$

          Be:
          $$
          cos z=frac{e^{iz}+e^{-iz}}{2}=-2
          $$
          Then:
          begin{eqnarray}
          frac{e^{iz}+e^{-iz}}{2} &=& -2 \
          e^{iz}+e^{-iz} &=& -4 \
          e^{2iz} + 1 &=& -4e^{iz} \
          left(e^{iz}right)^2 + 4left(e^{iz}right) + 1 &=& 0 \
          e^{iz} &=& frac{-4 pm sqrt{16-4}}{2} \
          e^{iz} &=& -2 pm sqrt{3} \
          end{eqnarray}
          Then since $e^{iz}=cos(z)+isin(z)$, $e^{iz}=e^{i(z+2kpi)}$ with $kinmathbb{Z}$:
          begin{eqnarray}
          e^{i(z+2k_1pi)} &=& -2 pm sqrt{3} \
          i(z +2k_1pi) &=& lnleft(-2 pm sqrt{3}right) \
          z &=& 2k_1pi - ilnleft(-2 pm sqrt{3}right) \
          end{eqnarray}
          There are two set solution to differents solutions:
          begin{eqnarray}
          z_{+} &=& 2k_1pi - ilnleft(-2 + sqrt{3}right) \
          z_{-} &=& 2k_1pi - ilnleft(-2 - sqrt{3}right) \
          end{eqnarray}
          For $z_{-}$:
          begin{eqnarray}
          z_{-} &=& 2k_1pi - ilnleft(left(2 + sqrt{3}right)e^{-i(1+2k_2)pi}right) \
          z_{-} &=& 2k_1pi - ileft[lnleft(2 + sqrt{3}right)-i(1+2k_2)piright] \
          z_{-} &=& 2k_1pi - (1+2k_2)pi - ilnleft(2 + sqrt{3}right) \
          z_{-} &=& (1+2k)pi - ilnleft(2 + sqrt{3}right) \
          end{eqnarray}
          Then, with $kinmathbb{Z}$ the solutions are:
          begin{eqnarray}
          z_{+} &=& 2kpi - ilnleft(-2 + sqrt{3}right) \
          z_{-} &=& (1+2k)pi - ilnleft(2 + sqrt{3}right) \
          end{eqnarray}






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            you mean $2kpi + 1$
            $endgroup$
            – no0ob
            Jan 2 at 21:59










          • $begingroup$
            Why $2kpi +1$?
            $endgroup$
            – El borito
            Jan 2 at 22:02










          • $begingroup$
            Cause i had answered my approach using $(2k+1)pi$ there should be one solution i believe :-?
            $endgroup$
            – no0ob
            Jan 2 at 22:04












          • $begingroup$
            oh, okey, I'm going to fix it
            $endgroup$
            – El borito
            Jan 2 at 22:10








          • 2




            $begingroup$
            $|-2pmsqrt{3}|=sqrt{7}$ is nonsense. Are you thinking of $-2pmcolor{red}{i}sqrt{3}$ here?
            $endgroup$
            – A.Γ.
            Jan 2 at 22:54












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          3 Answers
          3






          active

          oldest

          votes








          3 Answers
          3






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          Hint: start with
          $$
          cos z=frac{e^{iz}+e^{-iz}}{2}
          $$
          and obtain a quadratic equation for $e^{iz}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            i got $exp(iz) = -2+sqrt(3)$ and $exp(iz) = -2-sqrt(3)$ which gets me that $e^x = -2+sqrt(3)$ which is negative on the right side and impossible :-?
            $endgroup$
            – no0ob
            Jan 2 at 21:41












          • $begingroup$
            If $e^{iz}=-2pmsqrt{3}$ then $$e^{iz+ipi}=e^{ipi}(-2pmsqrt{3})=(-1)(-2pmsqrt{3})=2pmsqrt{3}$$ whith positive RHS.
            $endgroup$
            – A.Γ.
            Jan 2 at 21:43












          • $begingroup$
            What should $z$ be btw?
            $endgroup$
            – no0ob
            Jan 2 at 21:48










          • $begingroup$
            @no0ob Do you follow the argumentation? If $$e^{i(z+pi)}=2pmsqrt{3}$$ then how to find $i(z+pi)$?
            $endgroup$
            – A.Γ.
            Jan 2 at 21:52






          • 1




            $begingroup$
            @no0ob Right, sorry, it should be every second zero of $sin x=0$ to make $cos x=-1$, i.e. $x=pi+2pi k$.
            $endgroup$
            – A.Γ.
            Jan 2 at 22:10


















          1












          $begingroup$

          Hint: start with
          $$
          cos z=frac{e^{iz}+e^{-iz}}{2}
          $$
          and obtain a quadratic equation for $e^{iz}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            i got $exp(iz) = -2+sqrt(3)$ and $exp(iz) = -2-sqrt(3)$ which gets me that $e^x = -2+sqrt(3)$ which is negative on the right side and impossible :-?
            $endgroup$
            – no0ob
            Jan 2 at 21:41












          • $begingroup$
            If $e^{iz}=-2pmsqrt{3}$ then $$e^{iz+ipi}=e^{ipi}(-2pmsqrt{3})=(-1)(-2pmsqrt{3})=2pmsqrt{3}$$ whith positive RHS.
            $endgroup$
            – A.Γ.
            Jan 2 at 21:43












          • $begingroup$
            What should $z$ be btw?
            $endgroup$
            – no0ob
            Jan 2 at 21:48










          • $begingroup$
            @no0ob Do you follow the argumentation? If $$e^{i(z+pi)}=2pmsqrt{3}$$ then how to find $i(z+pi)$?
            $endgroup$
            – A.Γ.
            Jan 2 at 21:52






          • 1




            $begingroup$
            @no0ob Right, sorry, it should be every second zero of $sin x=0$ to make $cos x=-1$, i.e. $x=pi+2pi k$.
            $endgroup$
            – A.Γ.
            Jan 2 at 22:10
















          1












          1








          1





          $begingroup$

          Hint: start with
          $$
          cos z=frac{e^{iz}+e^{-iz}}{2}
          $$
          and obtain a quadratic equation for $e^{iz}$.






          share|cite|improve this answer









          $endgroup$



          Hint: start with
          $$
          cos z=frac{e^{iz}+e^{-iz}}{2}
          $$
          and obtain a quadratic equation for $e^{iz}$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 2 at 21:36









          A.Γ.A.Γ.

          22.9k32656




          22.9k32656












          • $begingroup$
            i got $exp(iz) = -2+sqrt(3)$ and $exp(iz) = -2-sqrt(3)$ which gets me that $e^x = -2+sqrt(3)$ which is negative on the right side and impossible :-?
            $endgroup$
            – no0ob
            Jan 2 at 21:41












          • $begingroup$
            If $e^{iz}=-2pmsqrt{3}$ then $$e^{iz+ipi}=e^{ipi}(-2pmsqrt{3})=(-1)(-2pmsqrt{3})=2pmsqrt{3}$$ whith positive RHS.
            $endgroup$
            – A.Γ.
            Jan 2 at 21:43












          • $begingroup$
            What should $z$ be btw?
            $endgroup$
            – no0ob
            Jan 2 at 21:48










          • $begingroup$
            @no0ob Do you follow the argumentation? If $$e^{i(z+pi)}=2pmsqrt{3}$$ then how to find $i(z+pi)$?
            $endgroup$
            – A.Γ.
            Jan 2 at 21:52






          • 1




            $begingroup$
            @no0ob Right, sorry, it should be every second zero of $sin x=0$ to make $cos x=-1$, i.e. $x=pi+2pi k$.
            $endgroup$
            – A.Γ.
            Jan 2 at 22:10




















          • $begingroup$
            i got $exp(iz) = -2+sqrt(3)$ and $exp(iz) = -2-sqrt(3)$ which gets me that $e^x = -2+sqrt(3)$ which is negative on the right side and impossible :-?
            $endgroup$
            – no0ob
            Jan 2 at 21:41












          • $begingroup$
            If $e^{iz}=-2pmsqrt{3}$ then $$e^{iz+ipi}=e^{ipi}(-2pmsqrt{3})=(-1)(-2pmsqrt{3})=2pmsqrt{3}$$ whith positive RHS.
            $endgroup$
            – A.Γ.
            Jan 2 at 21:43












          • $begingroup$
            What should $z$ be btw?
            $endgroup$
            – no0ob
            Jan 2 at 21:48










          • $begingroup$
            @no0ob Do you follow the argumentation? If $$e^{i(z+pi)}=2pmsqrt{3}$$ then how to find $i(z+pi)$?
            $endgroup$
            – A.Γ.
            Jan 2 at 21:52






          • 1




            $begingroup$
            @no0ob Right, sorry, it should be every second zero of $sin x=0$ to make $cos x=-1$, i.e. $x=pi+2pi k$.
            $endgroup$
            – A.Γ.
            Jan 2 at 22:10


















          $begingroup$
          i got $exp(iz) = -2+sqrt(3)$ and $exp(iz) = -2-sqrt(3)$ which gets me that $e^x = -2+sqrt(3)$ which is negative on the right side and impossible :-?
          $endgroup$
          – no0ob
          Jan 2 at 21:41






          $begingroup$
          i got $exp(iz) = -2+sqrt(3)$ and $exp(iz) = -2-sqrt(3)$ which gets me that $e^x = -2+sqrt(3)$ which is negative on the right side and impossible :-?
          $endgroup$
          – no0ob
          Jan 2 at 21:41














          $begingroup$
          If $e^{iz}=-2pmsqrt{3}$ then $$e^{iz+ipi}=e^{ipi}(-2pmsqrt{3})=(-1)(-2pmsqrt{3})=2pmsqrt{3}$$ whith positive RHS.
          $endgroup$
          – A.Γ.
          Jan 2 at 21:43






          $begingroup$
          If $e^{iz}=-2pmsqrt{3}$ then $$e^{iz+ipi}=e^{ipi}(-2pmsqrt{3})=(-1)(-2pmsqrt{3})=2pmsqrt{3}$$ whith positive RHS.
          $endgroup$
          – A.Γ.
          Jan 2 at 21:43














          $begingroup$
          What should $z$ be btw?
          $endgroup$
          – no0ob
          Jan 2 at 21:48




          $begingroup$
          What should $z$ be btw?
          $endgroup$
          – no0ob
          Jan 2 at 21:48












          $begingroup$
          @no0ob Do you follow the argumentation? If $$e^{i(z+pi)}=2pmsqrt{3}$$ then how to find $i(z+pi)$?
          $endgroup$
          – A.Γ.
          Jan 2 at 21:52




          $begingroup$
          @no0ob Do you follow the argumentation? If $$e^{i(z+pi)}=2pmsqrt{3}$$ then how to find $i(z+pi)$?
          $endgroup$
          – A.Γ.
          Jan 2 at 21:52




          1




          1




          $begingroup$
          @no0ob Right, sorry, it should be every second zero of $sin x=0$ to make $cos x=-1$, i.e. $x=pi+2pi k$.
          $endgroup$
          – A.Γ.
          Jan 2 at 22:10






          $begingroup$
          @no0ob Right, sorry, it should be every second zero of $sin x=0$ to make $cos x=-1$, i.e. $x=pi+2pi k$.
          $endgroup$
          – A.Γ.
          Jan 2 at 22:10













          2












          $begingroup$

          I should have used $x=(2k+1)pi$ for the $x$ and then had the



          $cos( (2k+1)pi ) cosh(y) = -2$



          which made it $cosh(y) = 2$






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            Right, and here you get two solutions for $y$ with $pm$ (recall $2pmsqrt{3}$ in another approach).
            $endgroup$
            – A.Γ.
            Jan 2 at 22:12






          • 1




            $begingroup$
            Yes; since $cos xcosh y<0$, you need to have $cos x<0$ and therefore $x=pi+2kpi$.
            $endgroup$
            – egreg
            Jan 2 at 22:33
















          2












          $begingroup$

          I should have used $x=(2k+1)pi$ for the $x$ and then had the



          $cos( (2k+1)pi ) cosh(y) = -2$



          which made it $cosh(y) = 2$






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            Right, and here you get two solutions for $y$ with $pm$ (recall $2pmsqrt{3}$ in another approach).
            $endgroup$
            – A.Γ.
            Jan 2 at 22:12






          • 1




            $begingroup$
            Yes; since $cos xcosh y<0$, you need to have $cos x<0$ and therefore $x=pi+2kpi$.
            $endgroup$
            – egreg
            Jan 2 at 22:33














          2












          2








          2





          $begingroup$

          I should have used $x=(2k+1)pi$ for the $x$ and then had the



          $cos( (2k+1)pi ) cosh(y) = -2$



          which made it $cosh(y) = 2$






          share|cite|improve this answer











          $endgroup$



          I should have used $x=(2k+1)pi$ for the $x$ and then had the



          $cos( (2k+1)pi ) cosh(y) = -2$



          which made it $cosh(y) = 2$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 2 at 22:55









          El borito

          664216




          664216










          answered Jan 2 at 21:58









          no0obno0ob

          788




          788








          • 1




            $begingroup$
            Right, and here you get two solutions for $y$ with $pm$ (recall $2pmsqrt{3}$ in another approach).
            $endgroup$
            – A.Γ.
            Jan 2 at 22:12






          • 1




            $begingroup$
            Yes; since $cos xcosh y<0$, you need to have $cos x<0$ and therefore $x=pi+2kpi$.
            $endgroup$
            – egreg
            Jan 2 at 22:33














          • 1




            $begingroup$
            Right, and here you get two solutions for $y$ with $pm$ (recall $2pmsqrt{3}$ in another approach).
            $endgroup$
            – A.Γ.
            Jan 2 at 22:12






          • 1




            $begingroup$
            Yes; since $cos xcosh y<0$, you need to have $cos x<0$ and therefore $x=pi+2kpi$.
            $endgroup$
            – egreg
            Jan 2 at 22:33








          1




          1




          $begingroup$
          Right, and here you get two solutions for $y$ with $pm$ (recall $2pmsqrt{3}$ in another approach).
          $endgroup$
          – A.Γ.
          Jan 2 at 22:12




          $begingroup$
          Right, and here you get two solutions for $y$ with $pm$ (recall $2pmsqrt{3}$ in another approach).
          $endgroup$
          – A.Γ.
          Jan 2 at 22:12




          1




          1




          $begingroup$
          Yes; since $cos xcosh y<0$, you need to have $cos x<0$ and therefore $x=pi+2kpi$.
          $endgroup$
          – egreg
          Jan 2 at 22:33




          $begingroup$
          Yes; since $cos xcosh y<0$, you need to have $cos x<0$ and therefore $x=pi+2kpi$.
          $endgroup$
          – egreg
          Jan 2 at 22:33











          0












          $begingroup$

          Be:
          $$
          cos z=frac{e^{iz}+e^{-iz}}{2}=-2
          $$
          Then:
          begin{eqnarray}
          frac{e^{iz}+e^{-iz}}{2} &=& -2 \
          e^{iz}+e^{-iz} &=& -4 \
          e^{2iz} + 1 &=& -4e^{iz} \
          left(e^{iz}right)^2 + 4left(e^{iz}right) + 1 &=& 0 \
          e^{iz} &=& frac{-4 pm sqrt{16-4}}{2} \
          e^{iz} &=& -2 pm sqrt{3} \
          end{eqnarray}
          Then since $e^{iz}=cos(z)+isin(z)$, $e^{iz}=e^{i(z+2kpi)}$ with $kinmathbb{Z}$:
          begin{eqnarray}
          e^{i(z+2k_1pi)} &=& -2 pm sqrt{3} \
          i(z +2k_1pi) &=& lnleft(-2 pm sqrt{3}right) \
          z &=& 2k_1pi - ilnleft(-2 pm sqrt{3}right) \
          end{eqnarray}
          There are two set solution to differents solutions:
          begin{eqnarray}
          z_{+} &=& 2k_1pi - ilnleft(-2 + sqrt{3}right) \
          z_{-} &=& 2k_1pi - ilnleft(-2 - sqrt{3}right) \
          end{eqnarray}
          For $z_{-}$:
          begin{eqnarray}
          z_{-} &=& 2k_1pi - ilnleft(left(2 + sqrt{3}right)e^{-i(1+2k_2)pi}right) \
          z_{-} &=& 2k_1pi - ileft[lnleft(2 + sqrt{3}right)-i(1+2k_2)piright] \
          z_{-} &=& 2k_1pi - (1+2k_2)pi - ilnleft(2 + sqrt{3}right) \
          z_{-} &=& (1+2k)pi - ilnleft(2 + sqrt{3}right) \
          end{eqnarray}
          Then, with $kinmathbb{Z}$ the solutions are:
          begin{eqnarray}
          z_{+} &=& 2kpi - ilnleft(-2 + sqrt{3}right) \
          z_{-} &=& (1+2k)pi - ilnleft(2 + sqrt{3}right) \
          end{eqnarray}






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            you mean $2kpi + 1$
            $endgroup$
            – no0ob
            Jan 2 at 21:59










          • $begingroup$
            Why $2kpi +1$?
            $endgroup$
            – El borito
            Jan 2 at 22:02










          • $begingroup$
            Cause i had answered my approach using $(2k+1)pi$ there should be one solution i believe :-?
            $endgroup$
            – no0ob
            Jan 2 at 22:04












          • $begingroup$
            oh, okey, I'm going to fix it
            $endgroup$
            – El borito
            Jan 2 at 22:10








          • 2




            $begingroup$
            $|-2pmsqrt{3}|=sqrt{7}$ is nonsense. Are you thinking of $-2pmcolor{red}{i}sqrt{3}$ here?
            $endgroup$
            – A.Γ.
            Jan 2 at 22:54
















          0












          $begingroup$

          Be:
          $$
          cos z=frac{e^{iz}+e^{-iz}}{2}=-2
          $$
          Then:
          begin{eqnarray}
          frac{e^{iz}+e^{-iz}}{2} &=& -2 \
          e^{iz}+e^{-iz} &=& -4 \
          e^{2iz} + 1 &=& -4e^{iz} \
          left(e^{iz}right)^2 + 4left(e^{iz}right) + 1 &=& 0 \
          e^{iz} &=& frac{-4 pm sqrt{16-4}}{2} \
          e^{iz} &=& -2 pm sqrt{3} \
          end{eqnarray}
          Then since $e^{iz}=cos(z)+isin(z)$, $e^{iz}=e^{i(z+2kpi)}$ with $kinmathbb{Z}$:
          begin{eqnarray}
          e^{i(z+2k_1pi)} &=& -2 pm sqrt{3} \
          i(z +2k_1pi) &=& lnleft(-2 pm sqrt{3}right) \
          z &=& 2k_1pi - ilnleft(-2 pm sqrt{3}right) \
          end{eqnarray}
          There are two set solution to differents solutions:
          begin{eqnarray}
          z_{+} &=& 2k_1pi - ilnleft(-2 + sqrt{3}right) \
          z_{-} &=& 2k_1pi - ilnleft(-2 - sqrt{3}right) \
          end{eqnarray}
          For $z_{-}$:
          begin{eqnarray}
          z_{-} &=& 2k_1pi - ilnleft(left(2 + sqrt{3}right)e^{-i(1+2k_2)pi}right) \
          z_{-} &=& 2k_1pi - ileft[lnleft(2 + sqrt{3}right)-i(1+2k_2)piright] \
          z_{-} &=& 2k_1pi - (1+2k_2)pi - ilnleft(2 + sqrt{3}right) \
          z_{-} &=& (1+2k)pi - ilnleft(2 + sqrt{3}right) \
          end{eqnarray}
          Then, with $kinmathbb{Z}$ the solutions are:
          begin{eqnarray}
          z_{+} &=& 2kpi - ilnleft(-2 + sqrt{3}right) \
          z_{-} &=& (1+2k)pi - ilnleft(2 + sqrt{3}right) \
          end{eqnarray}






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            you mean $2kpi + 1$
            $endgroup$
            – no0ob
            Jan 2 at 21:59










          • $begingroup$
            Why $2kpi +1$?
            $endgroup$
            – El borito
            Jan 2 at 22:02










          • $begingroup$
            Cause i had answered my approach using $(2k+1)pi$ there should be one solution i believe :-?
            $endgroup$
            – no0ob
            Jan 2 at 22:04












          • $begingroup$
            oh, okey, I'm going to fix it
            $endgroup$
            – El borito
            Jan 2 at 22:10








          • 2




            $begingroup$
            $|-2pmsqrt{3}|=sqrt{7}$ is nonsense. Are you thinking of $-2pmcolor{red}{i}sqrt{3}$ here?
            $endgroup$
            – A.Γ.
            Jan 2 at 22:54














          0












          0








          0





          $begingroup$

          Be:
          $$
          cos z=frac{e^{iz}+e^{-iz}}{2}=-2
          $$
          Then:
          begin{eqnarray}
          frac{e^{iz}+e^{-iz}}{2} &=& -2 \
          e^{iz}+e^{-iz} &=& -4 \
          e^{2iz} + 1 &=& -4e^{iz} \
          left(e^{iz}right)^2 + 4left(e^{iz}right) + 1 &=& 0 \
          e^{iz} &=& frac{-4 pm sqrt{16-4}}{2} \
          e^{iz} &=& -2 pm sqrt{3} \
          end{eqnarray}
          Then since $e^{iz}=cos(z)+isin(z)$, $e^{iz}=e^{i(z+2kpi)}$ with $kinmathbb{Z}$:
          begin{eqnarray}
          e^{i(z+2k_1pi)} &=& -2 pm sqrt{3} \
          i(z +2k_1pi) &=& lnleft(-2 pm sqrt{3}right) \
          z &=& 2k_1pi - ilnleft(-2 pm sqrt{3}right) \
          end{eqnarray}
          There are two set solution to differents solutions:
          begin{eqnarray}
          z_{+} &=& 2k_1pi - ilnleft(-2 + sqrt{3}right) \
          z_{-} &=& 2k_1pi - ilnleft(-2 - sqrt{3}right) \
          end{eqnarray}
          For $z_{-}$:
          begin{eqnarray}
          z_{-} &=& 2k_1pi - ilnleft(left(2 + sqrt{3}right)e^{-i(1+2k_2)pi}right) \
          z_{-} &=& 2k_1pi - ileft[lnleft(2 + sqrt{3}right)-i(1+2k_2)piright] \
          z_{-} &=& 2k_1pi - (1+2k_2)pi - ilnleft(2 + sqrt{3}right) \
          z_{-} &=& (1+2k)pi - ilnleft(2 + sqrt{3}right) \
          end{eqnarray}
          Then, with $kinmathbb{Z}$ the solutions are:
          begin{eqnarray}
          z_{+} &=& 2kpi - ilnleft(-2 + sqrt{3}right) \
          z_{-} &=& (1+2k)pi - ilnleft(2 + sqrt{3}right) \
          end{eqnarray}






          share|cite|improve this answer











          $endgroup$



          Be:
          $$
          cos z=frac{e^{iz}+e^{-iz}}{2}=-2
          $$
          Then:
          begin{eqnarray}
          frac{e^{iz}+e^{-iz}}{2} &=& -2 \
          e^{iz}+e^{-iz} &=& -4 \
          e^{2iz} + 1 &=& -4e^{iz} \
          left(e^{iz}right)^2 + 4left(e^{iz}right) + 1 &=& 0 \
          e^{iz} &=& frac{-4 pm sqrt{16-4}}{2} \
          e^{iz} &=& -2 pm sqrt{3} \
          end{eqnarray}
          Then since $e^{iz}=cos(z)+isin(z)$, $e^{iz}=e^{i(z+2kpi)}$ with $kinmathbb{Z}$:
          begin{eqnarray}
          e^{i(z+2k_1pi)} &=& -2 pm sqrt{3} \
          i(z +2k_1pi) &=& lnleft(-2 pm sqrt{3}right) \
          z &=& 2k_1pi - ilnleft(-2 pm sqrt{3}right) \
          end{eqnarray}
          There are two set solution to differents solutions:
          begin{eqnarray}
          z_{+} &=& 2k_1pi - ilnleft(-2 + sqrt{3}right) \
          z_{-} &=& 2k_1pi - ilnleft(-2 - sqrt{3}right) \
          end{eqnarray}
          For $z_{-}$:
          begin{eqnarray}
          z_{-} &=& 2k_1pi - ilnleft(left(2 + sqrt{3}right)e^{-i(1+2k_2)pi}right) \
          z_{-} &=& 2k_1pi - ileft[lnleft(2 + sqrt{3}right)-i(1+2k_2)piright] \
          z_{-} &=& 2k_1pi - (1+2k_2)pi - ilnleft(2 + sqrt{3}right) \
          z_{-} &=& (1+2k)pi - ilnleft(2 + sqrt{3}right) \
          end{eqnarray}
          Then, with $kinmathbb{Z}$ the solutions are:
          begin{eqnarray}
          z_{+} &=& 2kpi - ilnleft(-2 + sqrt{3}right) \
          z_{-} &=& (1+2k)pi - ilnleft(2 + sqrt{3}right) \
          end{eqnarray}







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 3 at 5:46

























          answered Jan 2 at 21:56









          El boritoEl borito

          664216




          664216












          • $begingroup$
            you mean $2kpi + 1$
            $endgroup$
            – no0ob
            Jan 2 at 21:59










          • $begingroup$
            Why $2kpi +1$?
            $endgroup$
            – El borito
            Jan 2 at 22:02










          • $begingroup$
            Cause i had answered my approach using $(2k+1)pi$ there should be one solution i believe :-?
            $endgroup$
            – no0ob
            Jan 2 at 22:04












          • $begingroup$
            oh, okey, I'm going to fix it
            $endgroup$
            – El borito
            Jan 2 at 22:10








          • 2




            $begingroup$
            $|-2pmsqrt{3}|=sqrt{7}$ is nonsense. Are you thinking of $-2pmcolor{red}{i}sqrt{3}$ here?
            $endgroup$
            – A.Γ.
            Jan 2 at 22:54


















          • $begingroup$
            you mean $2kpi + 1$
            $endgroup$
            – no0ob
            Jan 2 at 21:59










          • $begingroup$
            Why $2kpi +1$?
            $endgroup$
            – El borito
            Jan 2 at 22:02










          • $begingroup$
            Cause i had answered my approach using $(2k+1)pi$ there should be one solution i believe :-?
            $endgroup$
            – no0ob
            Jan 2 at 22:04












          • $begingroup$
            oh, okey, I'm going to fix it
            $endgroup$
            – El borito
            Jan 2 at 22:10








          • 2




            $begingroup$
            $|-2pmsqrt{3}|=sqrt{7}$ is nonsense. Are you thinking of $-2pmcolor{red}{i}sqrt{3}$ here?
            $endgroup$
            – A.Γ.
            Jan 2 at 22:54
















          $begingroup$
          you mean $2kpi + 1$
          $endgroup$
          – no0ob
          Jan 2 at 21:59




          $begingroup$
          you mean $2kpi + 1$
          $endgroup$
          – no0ob
          Jan 2 at 21:59












          $begingroup$
          Why $2kpi +1$?
          $endgroup$
          – El borito
          Jan 2 at 22:02




          $begingroup$
          Why $2kpi +1$?
          $endgroup$
          – El borito
          Jan 2 at 22:02












          $begingroup$
          Cause i had answered my approach using $(2k+1)pi$ there should be one solution i believe :-?
          $endgroup$
          – no0ob
          Jan 2 at 22:04






          $begingroup$
          Cause i had answered my approach using $(2k+1)pi$ there should be one solution i believe :-?
          $endgroup$
          – no0ob
          Jan 2 at 22:04














          $begingroup$
          oh, okey, I'm going to fix it
          $endgroup$
          – El borito
          Jan 2 at 22:10






          $begingroup$
          oh, okey, I'm going to fix it
          $endgroup$
          – El borito
          Jan 2 at 22:10






          2




          2




          $begingroup$
          $|-2pmsqrt{3}|=sqrt{7}$ is nonsense. Are you thinking of $-2pmcolor{red}{i}sqrt{3}$ here?
          $endgroup$
          – A.Γ.
          Jan 2 at 22:54




          $begingroup$
          $|-2pmsqrt{3}|=sqrt{7}$ is nonsense. Are you thinking of $-2pmcolor{red}{i}sqrt{3}$ here?
          $endgroup$
          – A.Γ.
          Jan 2 at 22:54


















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