Taking complex solution on $frac{1}{x^2 + 1} = 0$












1












$begingroup$


Is $i$ a complex solution to $$ frac{1}{x^2 + 1} = 0 ;?$$

Knowing that it has no real solutions, substituting $i$ leaves $frac{1}{0}$ which for me is equal to zero. Is this right?






algebra-precalculus complex-numbers







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    Is $i$ a complex solution to $$ frac{1}{x^2 + 1} = 0 ;?$$

    Knowing that it has no real solutions, substituting $i$ leaves $frac{1}{0}$ which for me is equal to zero. Is this right?






    algebra-precalculus complex-numbers







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      Is $i$ a complex solution to $$ frac{1}{x^2 + 1} = 0 ;?$$

      Knowing that it has no real solutions, substituting $i$ leaves $frac{1}{0}$ which for me is equal to zero. Is this right?






      algebra-precalculus complex-numbers







      share|cite|improve this question











      $endgroup$




      Is $i$ a complex solution to $$ frac{1}{x^2 + 1} = 0 ;?$$

      Knowing that it has no real solutions, substituting $i$ leaves $frac{1}{0}$ which for me is equal to zero. Is this right?






      algebra-precalculus complex-numbers




      algebra-precalculus complex-numbers






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 19 '18 at 20:53









      Eric Wofsey

      187k14215344




      187k14215344










      asked Nov 10 '18 at 10:34









      MMJMMMJM

      3351111




      3351111






















          4 Answers
          4






          active

          oldest

          votes


















          7












          $begingroup$

          No, this is wrong. $1/0$ is definitely not equal to zero, it is rather an undefined expression. Note that in complex analysis functions are defined in similar ways. If you wanted to consider that expression over $mathbb C$, it would be :



          $$f(z) = frac{1}{z^2+1}, quad z in mathbb Csetminus{pm i}$$



          The denominator can never be zero no matter if you're studying strictly real numbers or complex numbers.



          The equation



          $$f(z) = 0 implies frac{1}{z^2+1} = 0$$



          is false, as it has no solutions at all.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            If we solve for $$frac{1}{x^2 + 1} = frac{1}{0}$$ will it have $i$ be a solution or not ?(assuming that $frac{1}{0}$ would not be simplified) since begin{align} frac{1}{x^2 + 1} &= frac{1}{0}\ frac{1}{(i)^2 + 1} &= frac{1}{0}\ frac{1}{0} &= frac{1}{0} end{align}
            $endgroup$
            – MMJM
            Nov 10 '18 at 11:13












          • $begingroup$
            @MMJM You cannot use $1/0$ as an expression in whatever you do, it is not a properly defined mathematical substance. It is an undefined expression and that's why you yield contradictions when you use it. Recall that a fraction $frac{a}{b}$ is defined iff $b neq 0$.
            $endgroup$
            – Rebellos
            Nov 10 '18 at 11:15












          • $begingroup$
            I misunderstood $frac{1}{0}$ at first when I posted the question. Now I get the point.
            $endgroup$
            – MMJM
            Nov 10 '18 at 11:20



















          2












          $begingroup$

          We have that



          $$frac{1}{x^2 + 1} = 0$$



          is an expression defined for $x^2+1 neq 0$ and it has not (real nor complex) solutions.



          Indeed for any $zin mathbb{C}$ we have



          $$frac{1}{z^2 + 1} = 0 implies left|frac{1}{z^2 + 1}right| = |0| $$



          but



          $$left|frac{1}{z^2 + 1}right|=frac{1}{|z^2 + 1|}neq 0quad forall z$$






          share|cite|improve this answer











          $endgroup$





















            1












            $begingroup$

            No, the complex field does not provide solution to $$frac {1}{x^2+1}=0$$



            That is equivalent to $$1=0times (x^2+1)$$ and as you know the product of $0$ with any number is $0$ not $1$.






            share|cite|improve this answer









            $endgroup$





















              0












              $begingroup$

              Note that for $xinmathbb{C}$ and polynomials $P$ and $Q$,



              $$ frac{P(x)}{Q(x)} = 0 Leftrightarrow P(x) = 0 text{ and } Q(x)neq 0 $$



              Since in this case $P(x) equiv 1 neq 0$, there are no $x$ such that $frac{P(x)}{Q(x)} = 0$.






              share|cite|improve this answer









              $endgroup$













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                4 Answers
                4






                active

                oldest

                votes








                4 Answers
                4






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes









                7












                $begingroup$

                No, this is wrong. $1/0$ is definitely not equal to zero, it is rather an undefined expression. Note that in complex analysis functions are defined in similar ways. If you wanted to consider that expression over $mathbb C$, it would be :



                $$f(z) = frac{1}{z^2+1}, quad z in mathbb Csetminus{pm i}$$



                The denominator can never be zero no matter if you're studying strictly real numbers or complex numbers.



                The equation



                $$f(z) = 0 implies frac{1}{z^2+1} = 0$$



                is false, as it has no solutions at all.






                share|cite|improve this answer









                $endgroup$













                • $begingroup$
                  If we solve for $$frac{1}{x^2 + 1} = frac{1}{0}$$ will it have $i$ be a solution or not ?(assuming that $frac{1}{0}$ would not be simplified) since begin{align} frac{1}{x^2 + 1} &= frac{1}{0}\ frac{1}{(i)^2 + 1} &= frac{1}{0}\ frac{1}{0} &= frac{1}{0} end{align}
                  $endgroup$
                  – MMJM
                  Nov 10 '18 at 11:13












                • $begingroup$
                  @MMJM You cannot use $1/0$ as an expression in whatever you do, it is not a properly defined mathematical substance. It is an undefined expression and that's why you yield contradictions when you use it. Recall that a fraction $frac{a}{b}$ is defined iff $b neq 0$.
                  $endgroup$
                  – Rebellos
                  Nov 10 '18 at 11:15












                • $begingroup$
                  I misunderstood $frac{1}{0}$ at first when I posted the question. Now I get the point.
                  $endgroup$
                  – MMJM
                  Nov 10 '18 at 11:20
















                7












                $begingroup$

                No, this is wrong. $1/0$ is definitely not equal to zero, it is rather an undefined expression. Note that in complex analysis functions are defined in similar ways. If you wanted to consider that expression over $mathbb C$, it would be :



                $$f(z) = frac{1}{z^2+1}, quad z in mathbb Csetminus{pm i}$$



                The denominator can never be zero no matter if you're studying strictly real numbers or complex numbers.



                The equation



                $$f(z) = 0 implies frac{1}{z^2+1} = 0$$



                is false, as it has no solutions at all.






                share|cite|improve this answer









                $endgroup$













                • $begingroup$
                  If we solve for $$frac{1}{x^2 + 1} = frac{1}{0}$$ will it have $i$ be a solution or not ?(assuming that $frac{1}{0}$ would not be simplified) since begin{align} frac{1}{x^2 + 1} &= frac{1}{0}\ frac{1}{(i)^2 + 1} &= frac{1}{0}\ frac{1}{0} &= frac{1}{0} end{align}
                  $endgroup$
                  – MMJM
                  Nov 10 '18 at 11:13












                • $begingroup$
                  @MMJM You cannot use $1/0$ as an expression in whatever you do, it is not a properly defined mathematical substance. It is an undefined expression and that's why you yield contradictions when you use it. Recall that a fraction $frac{a}{b}$ is defined iff $b neq 0$.
                  $endgroup$
                  – Rebellos
                  Nov 10 '18 at 11:15












                • $begingroup$
                  I misunderstood $frac{1}{0}$ at first when I posted the question. Now I get the point.
                  $endgroup$
                  – MMJM
                  Nov 10 '18 at 11:20














                7












                7








                7





                $begingroup$

                No, this is wrong. $1/0$ is definitely not equal to zero, it is rather an undefined expression. Note that in complex analysis functions are defined in similar ways. If you wanted to consider that expression over $mathbb C$, it would be :



                $$f(z) = frac{1}{z^2+1}, quad z in mathbb Csetminus{pm i}$$



                The denominator can never be zero no matter if you're studying strictly real numbers or complex numbers.



                The equation



                $$f(z) = 0 implies frac{1}{z^2+1} = 0$$



                is false, as it has no solutions at all.






                share|cite|improve this answer









                $endgroup$



                No, this is wrong. $1/0$ is definitely not equal to zero, it is rather an undefined expression. Note that in complex analysis functions are defined in similar ways. If you wanted to consider that expression over $mathbb C$, it would be :



                $$f(z) = frac{1}{z^2+1}, quad z in mathbb Csetminus{pm i}$$



                The denominator can never be zero no matter if you're studying strictly real numbers or complex numbers.



                The equation



                $$f(z) = 0 implies frac{1}{z^2+1} = 0$$



                is false, as it has no solutions at all.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 10 '18 at 10:38









                RebellosRebellos

                14.9k31248




                14.9k31248












                • $begingroup$
                  If we solve for $$frac{1}{x^2 + 1} = frac{1}{0}$$ will it have $i$ be a solution or not ?(assuming that $frac{1}{0}$ would not be simplified) since begin{align} frac{1}{x^2 + 1} &= frac{1}{0}\ frac{1}{(i)^2 + 1} &= frac{1}{0}\ frac{1}{0} &= frac{1}{0} end{align}
                  $endgroup$
                  – MMJM
                  Nov 10 '18 at 11:13












                • $begingroup$
                  @MMJM You cannot use $1/0$ as an expression in whatever you do, it is not a properly defined mathematical substance. It is an undefined expression and that's why you yield contradictions when you use it. Recall that a fraction $frac{a}{b}$ is defined iff $b neq 0$.
                  $endgroup$
                  – Rebellos
                  Nov 10 '18 at 11:15












                • $begingroup$
                  I misunderstood $frac{1}{0}$ at first when I posted the question. Now I get the point.
                  $endgroup$
                  – MMJM
                  Nov 10 '18 at 11:20


















                • $begingroup$
                  If we solve for $$frac{1}{x^2 + 1} = frac{1}{0}$$ will it have $i$ be a solution or not ?(assuming that $frac{1}{0}$ would not be simplified) since begin{align} frac{1}{x^2 + 1} &= frac{1}{0}\ frac{1}{(i)^2 + 1} &= frac{1}{0}\ frac{1}{0} &= frac{1}{0} end{align}
                  $endgroup$
                  – MMJM
                  Nov 10 '18 at 11:13












                • $begingroup$
                  @MMJM You cannot use $1/0$ as an expression in whatever you do, it is not a properly defined mathematical substance. It is an undefined expression and that's why you yield contradictions when you use it. Recall that a fraction $frac{a}{b}$ is defined iff $b neq 0$.
                  $endgroup$
                  – Rebellos
                  Nov 10 '18 at 11:15












                • $begingroup$
                  I misunderstood $frac{1}{0}$ at first when I posted the question. Now I get the point.
                  $endgroup$
                  – MMJM
                  Nov 10 '18 at 11:20
















                $begingroup$
                If we solve for $$frac{1}{x^2 + 1} = frac{1}{0}$$ will it have $i$ be a solution or not ?(assuming that $frac{1}{0}$ would not be simplified) since begin{align} frac{1}{x^2 + 1} &= frac{1}{0}\ frac{1}{(i)^2 + 1} &= frac{1}{0}\ frac{1}{0} &= frac{1}{0} end{align}
                $endgroup$
                – MMJM
                Nov 10 '18 at 11:13






                $begingroup$
                If we solve for $$frac{1}{x^2 + 1} = frac{1}{0}$$ will it have $i$ be a solution or not ?(assuming that $frac{1}{0}$ would not be simplified) since begin{align} frac{1}{x^2 + 1} &= frac{1}{0}\ frac{1}{(i)^2 + 1} &= frac{1}{0}\ frac{1}{0} &= frac{1}{0} end{align}
                $endgroup$
                – MMJM
                Nov 10 '18 at 11:13














                $begingroup$
                @MMJM You cannot use $1/0$ as an expression in whatever you do, it is not a properly defined mathematical substance. It is an undefined expression and that's why you yield contradictions when you use it. Recall that a fraction $frac{a}{b}$ is defined iff $b neq 0$.
                $endgroup$
                – Rebellos
                Nov 10 '18 at 11:15






                $begingroup$
                @MMJM You cannot use $1/0$ as an expression in whatever you do, it is not a properly defined mathematical substance. It is an undefined expression and that's why you yield contradictions when you use it. Recall that a fraction $frac{a}{b}$ is defined iff $b neq 0$.
                $endgroup$
                – Rebellos
                Nov 10 '18 at 11:15














                $begingroup$
                I misunderstood $frac{1}{0}$ at first when I posted the question. Now I get the point.
                $endgroup$
                – MMJM
                Nov 10 '18 at 11:20




                $begingroup$
                I misunderstood $frac{1}{0}$ at first when I posted the question. Now I get the point.
                $endgroup$
                – MMJM
                Nov 10 '18 at 11:20











                2












                $begingroup$

                We have that



                $$frac{1}{x^2 + 1} = 0$$



                is an expression defined for $x^2+1 neq 0$ and it has not (real nor complex) solutions.



                Indeed for any $zin mathbb{C}$ we have



                $$frac{1}{z^2 + 1} = 0 implies left|frac{1}{z^2 + 1}right| = |0| $$



                but



                $$left|frac{1}{z^2 + 1}right|=frac{1}{|z^2 + 1|}neq 0quad forall z$$






                share|cite|improve this answer











                $endgroup$


















                  2












                  $begingroup$

                  We have that



                  $$frac{1}{x^2 + 1} = 0$$



                  is an expression defined for $x^2+1 neq 0$ and it has not (real nor complex) solutions.



                  Indeed for any $zin mathbb{C}$ we have



                  $$frac{1}{z^2 + 1} = 0 implies left|frac{1}{z^2 + 1}right| = |0| $$



                  but



                  $$left|frac{1}{z^2 + 1}right|=frac{1}{|z^2 + 1|}neq 0quad forall z$$






                  share|cite|improve this answer











                  $endgroup$
















                    2












                    2








                    2





                    $begingroup$

                    We have that



                    $$frac{1}{x^2 + 1} = 0$$



                    is an expression defined for $x^2+1 neq 0$ and it has not (real nor complex) solutions.



                    Indeed for any $zin mathbb{C}$ we have



                    $$frac{1}{z^2 + 1} = 0 implies left|frac{1}{z^2 + 1}right| = |0| $$



                    but



                    $$left|frac{1}{z^2 + 1}right|=frac{1}{|z^2 + 1|}neq 0quad forall z$$






                    share|cite|improve this answer











                    $endgroup$



                    We have that



                    $$frac{1}{x^2 + 1} = 0$$



                    is an expression defined for $x^2+1 neq 0$ and it has not (real nor complex) solutions.



                    Indeed for any $zin mathbb{C}$ we have



                    $$frac{1}{z^2 + 1} = 0 implies left|frac{1}{z^2 + 1}right| = |0| $$



                    but



                    $$left|frac{1}{z^2 + 1}right|=frac{1}{|z^2 + 1|}neq 0quad forall z$$







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Nov 10 '18 at 10:52

























                    answered Nov 10 '18 at 10:38









                    gimusigimusi

                    92.9k84494




                    92.9k84494























                        1












                        $begingroup$

                        No, the complex field does not provide solution to $$frac {1}{x^2+1}=0$$



                        That is equivalent to $$1=0times (x^2+1)$$ and as you know the product of $0$ with any number is $0$ not $1$.






                        share|cite|improve this answer









                        $endgroup$


















                          1












                          $begingroup$

                          No, the complex field does not provide solution to $$frac {1}{x^2+1}=0$$



                          That is equivalent to $$1=0times (x^2+1)$$ and as you know the product of $0$ with any number is $0$ not $1$.






                          share|cite|improve this answer









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            No, the complex field does not provide solution to $$frac {1}{x^2+1}=0$$



                            That is equivalent to $$1=0times (x^2+1)$$ and as you know the product of $0$ with any number is $0$ not $1$.






                            share|cite|improve this answer









                            $endgroup$



                            No, the complex field does not provide solution to $$frac {1}{x^2+1}=0$$



                            That is equivalent to $$1=0times (x^2+1)$$ and as you know the product of $0$ with any number is $0$ not $1$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Nov 10 '18 at 10:42









                            Mohammad Riazi-KermaniMohammad Riazi-Kermani

                            41.6k42061




                            41.6k42061























                                0












                                $begingroup$

                                Note that for $xinmathbb{C}$ and polynomials $P$ and $Q$,



                                $$ frac{P(x)}{Q(x)} = 0 Leftrightarrow P(x) = 0 text{ and } Q(x)neq 0 $$



                                Since in this case $P(x) equiv 1 neq 0$, there are no $x$ such that $frac{P(x)}{Q(x)} = 0$.






                                share|cite|improve this answer









                                $endgroup$


















                                  0












                                  $begingroup$

                                  Note that for $xinmathbb{C}$ and polynomials $P$ and $Q$,



                                  $$ frac{P(x)}{Q(x)} = 0 Leftrightarrow P(x) = 0 text{ and } Q(x)neq 0 $$



                                  Since in this case $P(x) equiv 1 neq 0$, there are no $x$ such that $frac{P(x)}{Q(x)} = 0$.






                                  share|cite|improve this answer









                                  $endgroup$
















                                    0












                                    0








                                    0





                                    $begingroup$

                                    Note that for $xinmathbb{C}$ and polynomials $P$ and $Q$,



                                    $$ frac{P(x)}{Q(x)} = 0 Leftrightarrow P(x) = 0 text{ and } Q(x)neq 0 $$



                                    Since in this case $P(x) equiv 1 neq 0$, there are no $x$ such that $frac{P(x)}{Q(x)} = 0$.






                                    share|cite|improve this answer









                                    $endgroup$



                                    Note that for $xinmathbb{C}$ and polynomials $P$ and $Q$,



                                    $$ frac{P(x)}{Q(x)} = 0 Leftrightarrow P(x) = 0 text{ and } Q(x)neq 0 $$



                                    Since in this case $P(x) equiv 1 neq 0$, there are no $x$ such that $frac{P(x)}{Q(x)} = 0$.







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered Nov 10 '18 at 10:43









                                    AlexanderJ93AlexanderJ93

                                    6,173823




                                    6,173823






























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                                        Dieringhausen

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                                        Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
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