A faster solution to this equation: $e^{-ix}(1+e^{-ix}) = -1$












2














I want to find $x$ such that



$$e^{-ix}(1+e^{-ix}) = -1$$



I was able to see that it was the same as $cos x = -0.5$ with a bit of work, but I think there must be a faster way since usually questions from GATE exam have elegant solutions.



Let me give some context, I needed to solve this equation in order to find the frequencies rejected by a digital system. Question 10.2 (GATE IN 2003 digital systems rejection of frequencies cause gain is zero)



question 10.2 from GATE IN 2003



The following is my attempt



my attempt to find x



So I wanted an alternative way to solve it, it would be nice if it was faster by hand and also any fast way to find the general form for x once we it's a set of angles?






algebra-precalculus trigonometry polynomials complex-numbers alternative-proof







share|cite|improve this question





























    2














    I want to find $x$ such that



    $$e^{-ix}(1+e^{-ix}) = -1$$



    I was able to see that it was the same as $cos x = -0.5$ with a bit of work, but I think there must be a faster way since usually questions from GATE exam have elegant solutions.



    Let me give some context, I needed to solve this equation in order to find the frequencies rejected by a digital system. Question 10.2 (GATE IN 2003 digital systems rejection of frequencies cause gain is zero)



    question 10.2 from GATE IN 2003



    The following is my attempt



    my attempt to find x



    So I wanted an alternative way to solve it, it would be nice if it was faster by hand and also any fast way to find the general form for x once we it's a set of angles?






    algebra-precalculus trigonometry polynomials complex-numbers alternative-proof







    share|cite|improve this question



























      2












      2








      2







      I want to find $x$ such that



      $$e^{-ix}(1+e^{-ix}) = -1$$



      I was able to see that it was the same as $cos x = -0.5$ with a bit of work, but I think there must be a faster way since usually questions from GATE exam have elegant solutions.



      Let me give some context, I needed to solve this equation in order to find the frequencies rejected by a digital system. Question 10.2 (GATE IN 2003 digital systems rejection of frequencies cause gain is zero)



      question 10.2 from GATE IN 2003



      The following is my attempt



      my attempt to find x



      So I wanted an alternative way to solve it, it would be nice if it was faster by hand and also any fast way to find the general form for x once we it's a set of angles?






      algebra-precalculus trigonometry polynomials complex-numbers alternative-proof







      share|cite|improve this question















      I want to find $x$ such that



      $$e^{-ix}(1+e^{-ix}) = -1$$



      I was able to see that it was the same as $cos x = -0.5$ with a bit of work, but I think there must be a faster way since usually questions from GATE exam have elegant solutions.



      Let me give some context, I needed to solve this equation in order to find the frequencies rejected by a digital system. Question 10.2 (GATE IN 2003 digital systems rejection of frequencies cause gain is zero)



      question 10.2 from GATE IN 2003



      The following is my attempt



      my attempt to find x



      So I wanted an alternative way to solve it, it would be nice if it was faster by hand and also any fast way to find the general form for x once we it's a set of angles?






      algebra-precalculus trigonometry polynomials complex-numbers alternative-proof




      algebra-precalculus trigonometry polynomials complex-numbers alternative-proof






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 27 at 0:19









      Servaes

      22.3k33793




      22.3k33793










      asked Nov 26 at 8:12









      Aditya

      268314




      268314






















          3 Answers
          3






          active

          oldest

          votes


















          6














          This just a quadratic equation in $z=e^{-ix}$, where $cos(x)=operatorname{Re}(z)$. Use the quadratic formula to solve
          $$z^2+z+1=0,$$
          to find that $z=frac{-1pmsqrt{-3}}{2}=-frac{1}{2}pmfrac{1}{2}sqrt{3}i$, and hence that $cos(x)=operatorname{Re}(z)=-frac{1}{2}$.






          share|cite|improve this answer























          • Thanks a lot, I think I can use this idea to solve a whole class of problems.
            – Aditya
            Nov 26 at 8:38








          • 1




            There is a typo: ‘$x=text{Re}(z)$’.
            – Szeto
            Nov 26 at 8:47












          • @Szeto Fixed, thanks for spotting!
            – Servaes
            Nov 26 at 8:48



















          5














          $$begin{align}
          e^{-ix}left(1+e^{-ix}right)=-1 &quadtoquad1 + e^{-ix} = -e^{ix} \[6pt]
          &quadtoquad e^{ix}+e^{-ix}=-1 \[6pt]
          &quadtoquad frac12left(e^{ix}+e^{-ix}right) = -frac12 \[6pt]
          &quadtoquad cos x = -frac12
          end{align}$$






          share|cite|improve this answer

















          • 1




            Really neat! I knew cos x could be written in this form, but I never saw that the equation could be manipulated like this!
            – Aditya
            Nov 26 at 8:38



















          3














          Just take modulus and square the equation. You will get $cos, x=-0.5$ immediately.



          $$|e^{-ix}|^{2} |1+e^{-ix}|^{2}= |-1|^{2}$$



          $$1 times left ( (1+cos x)^2 +(-sin x)^2 right ) = 1$$



          $$ 2 + 2 cos x = 1 $$



          So $cos, x=frac{1-2}{2}$!!






          share|cite|improve this answer



















          • 2




            Taking the modulus might give false solutions. For example, $e^{ix}=1$ has only one point on the unit circle as a solution, but $|e^{ix}|=1$ is satisfied by every real $x$.
            – A.Γ.
            Nov 26 at 8:27






          • 1




            What I am saying is that taking modulus tells you that $cos , x$ must be $-0.5$ if it all there is a solution $x$. This proves rigorously that there is no solution with $cos , x neq -0.5$ and then yo u have to go back to the equation and verify that this indeed gives a solution.
            – Kavi Rama Murthy
            Nov 26 at 8:33













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          3 Answers
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          3 Answers
          3






          active

          oldest

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          active

          oldest

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          active

          oldest

          votes









          6














          This just a quadratic equation in $z=e^{-ix}$, where $cos(x)=operatorname{Re}(z)$. Use the quadratic formula to solve
          $$z^2+z+1=0,$$
          to find that $z=frac{-1pmsqrt{-3}}{2}=-frac{1}{2}pmfrac{1}{2}sqrt{3}i$, and hence that $cos(x)=operatorname{Re}(z)=-frac{1}{2}$.






          share|cite|improve this answer























          • Thanks a lot, I think I can use this idea to solve a whole class of problems.
            – Aditya
            Nov 26 at 8:38








          • 1




            There is a typo: ‘$x=text{Re}(z)$’.
            – Szeto
            Nov 26 at 8:47












          • @Szeto Fixed, thanks for spotting!
            – Servaes
            Nov 26 at 8:48
















          6














          This just a quadratic equation in $z=e^{-ix}$, where $cos(x)=operatorname{Re}(z)$. Use the quadratic formula to solve
          $$z^2+z+1=0,$$
          to find that $z=frac{-1pmsqrt{-3}}{2}=-frac{1}{2}pmfrac{1}{2}sqrt{3}i$, and hence that $cos(x)=operatorname{Re}(z)=-frac{1}{2}$.






          share|cite|improve this answer























          • Thanks a lot, I think I can use this idea to solve a whole class of problems.
            – Aditya
            Nov 26 at 8:38








          • 1




            There is a typo: ‘$x=text{Re}(z)$’.
            – Szeto
            Nov 26 at 8:47












          • @Szeto Fixed, thanks for spotting!
            – Servaes
            Nov 26 at 8:48














          6












          6








          6






          This just a quadratic equation in $z=e^{-ix}$, where $cos(x)=operatorname{Re}(z)$. Use the quadratic formula to solve
          $$z^2+z+1=0,$$
          to find that $z=frac{-1pmsqrt{-3}}{2}=-frac{1}{2}pmfrac{1}{2}sqrt{3}i$, and hence that $cos(x)=operatorname{Re}(z)=-frac{1}{2}$.






          share|cite|improve this answer














          This just a quadratic equation in $z=e^{-ix}$, where $cos(x)=operatorname{Re}(z)$. Use the quadratic formula to solve
          $$z^2+z+1=0,$$
          to find that $z=frac{-1pmsqrt{-3}}{2}=-frac{1}{2}pmfrac{1}{2}sqrt{3}i$, and hence that $cos(x)=operatorname{Re}(z)=-frac{1}{2}$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 26 at 8:48

























          answered Nov 26 at 8:20









          Servaes

          22.3k33793




          22.3k33793












          • Thanks a lot, I think I can use this idea to solve a whole class of problems.
            – Aditya
            Nov 26 at 8:38








          • 1




            There is a typo: ‘$x=text{Re}(z)$’.
            – Szeto
            Nov 26 at 8:47












          • @Szeto Fixed, thanks for spotting!
            – Servaes
            Nov 26 at 8:48


















          • Thanks a lot, I think I can use this idea to solve a whole class of problems.
            – Aditya
            Nov 26 at 8:38








          • 1




            There is a typo: ‘$x=text{Re}(z)$’.
            – Szeto
            Nov 26 at 8:47












          • @Szeto Fixed, thanks for spotting!
            – Servaes
            Nov 26 at 8:48
















          Thanks a lot, I think I can use this idea to solve a whole class of problems.
          – Aditya
          Nov 26 at 8:38






          Thanks a lot, I think I can use this idea to solve a whole class of problems.
          – Aditya
          Nov 26 at 8:38






          1




          1




          There is a typo: ‘$x=text{Re}(z)$’.
          – Szeto
          Nov 26 at 8:47






          There is a typo: ‘$x=text{Re}(z)$’.
          – Szeto
          Nov 26 at 8:47














          @Szeto Fixed, thanks for spotting!
          – Servaes
          Nov 26 at 8:48




          @Szeto Fixed, thanks for spotting!
          – Servaes
          Nov 26 at 8:48











          5














          $$begin{align}
          e^{-ix}left(1+e^{-ix}right)=-1 &quadtoquad1 + e^{-ix} = -e^{ix} \[6pt]
          &quadtoquad e^{ix}+e^{-ix}=-1 \[6pt]
          &quadtoquad frac12left(e^{ix}+e^{-ix}right) = -frac12 \[6pt]
          &quadtoquad cos x = -frac12
          end{align}$$






          share|cite|improve this answer

















          • 1




            Really neat! I knew cos x could be written in this form, but I never saw that the equation could be manipulated like this!
            – Aditya
            Nov 26 at 8:38
















          5














          $$begin{align}
          e^{-ix}left(1+e^{-ix}right)=-1 &quadtoquad1 + e^{-ix} = -e^{ix} \[6pt]
          &quadtoquad e^{ix}+e^{-ix}=-1 \[6pt]
          &quadtoquad frac12left(e^{ix}+e^{-ix}right) = -frac12 \[6pt]
          &quadtoquad cos x = -frac12
          end{align}$$






          share|cite|improve this answer

















          • 1




            Really neat! I knew cos x could be written in this form, but I never saw that the equation could be manipulated like this!
            – Aditya
            Nov 26 at 8:38














          5












          5








          5






          $$begin{align}
          e^{-ix}left(1+e^{-ix}right)=-1 &quadtoquad1 + e^{-ix} = -e^{ix} \[6pt]
          &quadtoquad e^{ix}+e^{-ix}=-1 \[6pt]
          &quadtoquad frac12left(e^{ix}+e^{-ix}right) = -frac12 \[6pt]
          &quadtoquad cos x = -frac12
          end{align}$$






          share|cite|improve this answer












          $$begin{align}
          e^{-ix}left(1+e^{-ix}right)=-1 &quadtoquad1 + e^{-ix} = -e^{ix} \[6pt]
          &quadtoquad e^{ix}+e^{-ix}=-1 \[6pt]
          &quadtoquad frac12left(e^{ix}+e^{-ix}right) = -frac12 \[6pt]
          &quadtoquad cos x = -frac12
          end{align}$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 26 at 8:22









          Blue

          47.5k870151




          47.5k870151








          • 1




            Really neat! I knew cos x could be written in this form, but I never saw that the equation could be manipulated like this!
            – Aditya
            Nov 26 at 8:38














          • 1




            Really neat! I knew cos x could be written in this form, but I never saw that the equation could be manipulated like this!
            – Aditya
            Nov 26 at 8:38








          1




          1




          Really neat! I knew cos x could be written in this form, but I never saw that the equation could be manipulated like this!
          – Aditya
          Nov 26 at 8:38




          Really neat! I knew cos x could be written in this form, but I never saw that the equation could be manipulated like this!
          – Aditya
          Nov 26 at 8:38











          3














          Just take modulus and square the equation. You will get $cos, x=-0.5$ immediately.



          $$|e^{-ix}|^{2} |1+e^{-ix}|^{2}= |-1|^{2}$$



          $$1 times left ( (1+cos x)^2 +(-sin x)^2 right ) = 1$$



          $$ 2 + 2 cos x = 1 $$



          So $cos, x=frac{1-2}{2}$!!






          share|cite|improve this answer



















          • 2




            Taking the modulus might give false solutions. For example, $e^{ix}=1$ has only one point on the unit circle as a solution, but $|e^{ix}|=1$ is satisfied by every real $x$.
            – A.Γ.
            Nov 26 at 8:27






          • 1




            What I am saying is that taking modulus tells you that $cos , x$ must be $-0.5$ if it all there is a solution $x$. This proves rigorously that there is no solution with $cos , x neq -0.5$ and then yo u have to go back to the equation and verify that this indeed gives a solution.
            – Kavi Rama Murthy
            Nov 26 at 8:33


















          3














          Just take modulus and square the equation. You will get $cos, x=-0.5$ immediately.



          $$|e^{-ix}|^{2} |1+e^{-ix}|^{2}= |-1|^{2}$$



          $$1 times left ( (1+cos x)^2 +(-sin x)^2 right ) = 1$$



          $$ 2 + 2 cos x = 1 $$



          So $cos, x=frac{1-2}{2}$!!






          share|cite|improve this answer



















          • 2




            Taking the modulus might give false solutions. For example, $e^{ix}=1$ has only one point on the unit circle as a solution, but $|e^{ix}|=1$ is satisfied by every real $x$.
            – A.Γ.
            Nov 26 at 8:27






          • 1




            What I am saying is that taking modulus tells you that $cos , x$ must be $-0.5$ if it all there is a solution $x$. This proves rigorously that there is no solution with $cos , x neq -0.5$ and then yo u have to go back to the equation and verify that this indeed gives a solution.
            – Kavi Rama Murthy
            Nov 26 at 8:33
















          3












          3








          3






          Just take modulus and square the equation. You will get $cos, x=-0.5$ immediately.



          $$|e^{-ix}|^{2} |1+e^{-ix}|^{2}= |-1|^{2}$$



          $$1 times left ( (1+cos x)^2 +(-sin x)^2 right ) = 1$$



          $$ 2 + 2 cos x = 1 $$



          So $cos, x=frac{1-2}{2}$!!






          share|cite|improve this answer














          Just take modulus and square the equation. You will get $cos, x=-0.5$ immediately.



          $$|e^{-ix}|^{2} |1+e^{-ix}|^{2}= |-1|^{2}$$



          $$1 times left ( (1+cos x)^2 +(-sin x)^2 right ) = 1$$



          $$ 2 + 2 cos x = 1 $$



          So $cos, x=frac{1-2}{2}$!!







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 26 at 8:34









          Aditya

          268314




          268314










          answered Nov 26 at 8:17









          Kavi Rama Murthy

          49.1k31854




          49.1k31854








          • 2




            Taking the modulus might give false solutions. For example, $e^{ix}=1$ has only one point on the unit circle as a solution, but $|e^{ix}|=1$ is satisfied by every real $x$.
            – A.Γ.
            Nov 26 at 8:27






          • 1




            What I am saying is that taking modulus tells you that $cos , x$ must be $-0.5$ if it all there is a solution $x$. This proves rigorously that there is no solution with $cos , x neq -0.5$ and then yo u have to go back to the equation and verify that this indeed gives a solution.
            – Kavi Rama Murthy
            Nov 26 at 8:33
















          • 2




            Taking the modulus might give false solutions. For example, $e^{ix}=1$ has only one point on the unit circle as a solution, but $|e^{ix}|=1$ is satisfied by every real $x$.
            – A.Γ.
            Nov 26 at 8:27






          • 1




            What I am saying is that taking modulus tells you that $cos , x$ must be $-0.5$ if it all there is a solution $x$. This proves rigorously that there is no solution with $cos , x neq -0.5$ and then yo u have to go back to the equation and verify that this indeed gives a solution.
            – Kavi Rama Murthy
            Nov 26 at 8:33










          2




          2




          Taking the modulus might give false solutions. For example, $e^{ix}=1$ has only one point on the unit circle as a solution, but $|e^{ix}|=1$ is satisfied by every real $x$.
          – A.Γ.
          Nov 26 at 8:27




          Taking the modulus might give false solutions. For example, $e^{ix}=1$ has only one point on the unit circle as a solution, but $|e^{ix}|=1$ is satisfied by every real $x$.
          – A.Γ.
          Nov 26 at 8:27




          1




          1




          What I am saying is that taking modulus tells you that $cos , x$ must be $-0.5$ if it all there is a solution $x$. This proves rigorously that there is no solution with $cos , x neq -0.5$ and then yo u have to go back to the equation and verify that this indeed gives a solution.
          – Kavi Rama Murthy
          Nov 26 at 8:33






          What I am saying is that taking modulus tells you that $cos , x$ must be $-0.5$ if it all there is a solution $x$. This proves rigorously that there is no solution with $cos , x neq -0.5$ and then yo u have to go back to the equation and verify that this indeed gives a solution.
          – Kavi Rama Murthy
          Nov 26 at 8:33




















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