Finding the angle of a complex fraction












2












$begingroup$


If I have a complex fraction, say $$z=frac{a+bi}{c+di}$$, how can I find the phase angle of this? Could I just calculate it using $$arctan(frac{b}{a})-arctan(frac{d}{c})?$$



For example, my textbook says that for the fraction $$z=frac{jomega}{1+jomega},$$ the phase angle should be $$arctan(frac{1}{omega}).$$ I'm not really understanding what method/identities they're using to arrive at this answer.Is there a standard way to find the phase angle of a complex fraction?






trigonometry complex-numbers







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  • $begingroup$
    If $z=frac{z_1}{z_2}$ then $arg(z)=arg(z_1)-arg(z_2)$. It is known.
    $endgroup$
    – G-man
    Oct 19 '15 at 16:25


















2












$begingroup$


If I have a complex fraction, say $$z=frac{a+bi}{c+di}$$, how can I find the phase angle of this? Could I just calculate it using $$arctan(frac{b}{a})-arctan(frac{d}{c})?$$



For example, my textbook says that for the fraction $$z=frac{jomega}{1+jomega},$$ the phase angle should be $$arctan(frac{1}{omega}).$$ I'm not really understanding what method/identities they're using to arrive at this answer.Is there a standard way to find the phase angle of a complex fraction?






trigonometry complex-numbers







share|cite|improve this question









$endgroup$












  • $begingroup$
    If $z=frac{z_1}{z_2}$ then $arg(z)=arg(z_1)-arg(z_2)$. It is known.
    $endgroup$
    – G-man
    Oct 19 '15 at 16:25
















2












2








2


1



$begingroup$


If I have a complex fraction, say $$z=frac{a+bi}{c+di}$$, how can I find the phase angle of this? Could I just calculate it using $$arctan(frac{b}{a})-arctan(frac{d}{c})?$$



For example, my textbook says that for the fraction $$z=frac{jomega}{1+jomega},$$ the phase angle should be $$arctan(frac{1}{omega}).$$ I'm not really understanding what method/identities they're using to arrive at this answer.Is there a standard way to find the phase angle of a complex fraction?






trigonometry complex-numbers







share|cite|improve this question









$endgroup$




If I have a complex fraction, say $$z=frac{a+bi}{c+di}$$, how can I find the phase angle of this? Could I just calculate it using $$arctan(frac{b}{a})-arctan(frac{d}{c})?$$



For example, my textbook says that for the fraction $$z=frac{jomega}{1+jomega},$$ the phase angle should be $$arctan(frac{1}{omega}).$$ I'm not really understanding what method/identities they're using to arrive at this answer.Is there a standard way to find the phase angle of a complex fraction?






trigonometry complex-numbers




trigonometry complex-numbers






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asked Oct 19 '15 at 16:17









nichinichi

1313




1313












  • $begingroup$
    If $z=frac{z_1}{z_2}$ then $arg(z)=arg(z_1)-arg(z_2)$. It is known.
    $endgroup$
    – G-man
    Oct 19 '15 at 16:25




















  • $begingroup$
    If $z=frac{z_1}{z_2}$ then $arg(z)=arg(z_1)-arg(z_2)$. It is known.
    $endgroup$
    – G-man
    Oct 19 '15 at 16:25


















$begingroup$
If $z=frac{z_1}{z_2}$ then $arg(z)=arg(z_1)-arg(z_2)$. It is known.
$endgroup$
– G-man
Oct 19 '15 at 16:25






$begingroup$
If $z=frac{z_1}{z_2}$ then $arg(z)=arg(z_1)-arg(z_2)$. It is known.
$endgroup$
– G-man
Oct 19 '15 at 16:25












3 Answers
3






active

oldest

votes


















2












$begingroup$

If $z = a+bj$ where $a$ and $b$ are real, then you can determine the phase angle as $arctanleft(frac{b}{a}right)$ (as long as you are careful about the signs and quadrants), since $a+bj$ represents a point in the complex plane, and the phase angle is the angle this vector makes with the $x$-axis. So for the example from your textbook,
begin{align*}
frac{jomega}{1+jomega} = frac{jomega(1-jomega)}{(1+jomega)(1-jomega)}
= frac{omega^2 + jomega}{1+omega^2}.
end{align*}
Then the phase angle is $arctanleft(frac{omega}{omega^2}right) = arctanleft(frac{1}{omega}right)$, as the book says.



If you take the fraction you are given, $frac{a+bi}{c+di}$ and rationalize the denominator, you should be able to write the phase angle as an arctangent just as above.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you very much! This is the answer I was looking for. The book didn't go into much depth into how to find the phase angle for fractions, so I had to try to understand what they were saying.
    $endgroup$
    – nichi
    Oct 19 '15 at 16:38





















1












$begingroup$

The argument of the complex number $frac{a+bi}{c+di}$ is just the difference between the arguments of $a+bi$ and $c+di$.



So you really just need to know how to compute the argument of any complex number $a+bi$ (and then you simply compute a difference of two arguments).




  • If $a > 0$, this is indeed $arctan left(frac{b}{a}right)$.

  • If $a < 0$, this is $arctan left(frac{b}{a}right) pmpi$ depending on the sign of $b$ (I suggest you draw it to see).

  • If $a = 0$, this is $pm frac{pi}{2}$ depending depending on the sign of $b$.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$


    If I have a complex fraction, say $$z=frac{a+bi}{c+di}$$, how can I find the phase angle of this? Could I just calculate it using $$arctan(frac{b}{a})-arctan(frac{d}{c})?$$




    Yes, you can. In fact, I can show you that find phasor angle of the rationalized $z$ is the same as the phasor angle of the numerator minus that of the denominator.



    First the rationalization approach:



    $$angle frac{a+bi}{c+di} = angle frac{(a+bi)(c-di)}{(c+di)(c-di)} = anglefrac{(ac+bd)+(bc-ad)i}{c^2+d^2} = arctan(frac{bc-ad}{ac+bd})$$



    Now calculate the numerator and denominator separately and use Sum of Arctangents: $$arctan a + arctan b = arctan dfrac {a + b} {1 - a b}$$



    $$angle(a+bi)-angle(c+di) = arctan(frac{b}{a})-arctan(frac{d}{c}) = arctan(frac{frac{b}{a}-frac{d}{c}}{1+frac{bd}{ac}})=arctan(frac{bc-ad}{ac+bd})$$






    share|cite|improve this answer









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      3 Answers
      3






      active

      oldest

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      3 Answers
      3






      active

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      active

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      2












      $begingroup$

      If $z = a+bj$ where $a$ and $b$ are real, then you can determine the phase angle as $arctanleft(frac{b}{a}right)$ (as long as you are careful about the signs and quadrants), since $a+bj$ represents a point in the complex plane, and the phase angle is the angle this vector makes with the $x$-axis. So for the example from your textbook,
      begin{align*}
      frac{jomega}{1+jomega} = frac{jomega(1-jomega)}{(1+jomega)(1-jomega)}
      = frac{omega^2 + jomega}{1+omega^2}.
      end{align*}
      Then the phase angle is $arctanleft(frac{omega}{omega^2}right) = arctanleft(frac{1}{omega}right)$, as the book says.



      If you take the fraction you are given, $frac{a+bi}{c+di}$ and rationalize the denominator, you should be able to write the phase angle as an arctangent just as above.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Thank you very much! This is the answer I was looking for. The book didn't go into much depth into how to find the phase angle for fractions, so I had to try to understand what they were saying.
        $endgroup$
        – nichi
        Oct 19 '15 at 16:38


















      2












      $begingroup$

      If $z = a+bj$ where $a$ and $b$ are real, then you can determine the phase angle as $arctanleft(frac{b}{a}right)$ (as long as you are careful about the signs and quadrants), since $a+bj$ represents a point in the complex plane, and the phase angle is the angle this vector makes with the $x$-axis. So for the example from your textbook,
      begin{align*}
      frac{jomega}{1+jomega} = frac{jomega(1-jomega)}{(1+jomega)(1-jomega)}
      = frac{omega^2 + jomega}{1+omega^2}.
      end{align*}
      Then the phase angle is $arctanleft(frac{omega}{omega^2}right) = arctanleft(frac{1}{omega}right)$, as the book says.



      If you take the fraction you are given, $frac{a+bi}{c+di}$ and rationalize the denominator, you should be able to write the phase angle as an arctangent just as above.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Thank you very much! This is the answer I was looking for. The book didn't go into much depth into how to find the phase angle for fractions, so I had to try to understand what they were saying.
        $endgroup$
        – nichi
        Oct 19 '15 at 16:38
















      2












      2








      2





      $begingroup$

      If $z = a+bj$ where $a$ and $b$ are real, then you can determine the phase angle as $arctanleft(frac{b}{a}right)$ (as long as you are careful about the signs and quadrants), since $a+bj$ represents a point in the complex plane, and the phase angle is the angle this vector makes with the $x$-axis. So for the example from your textbook,
      begin{align*}
      frac{jomega}{1+jomega} = frac{jomega(1-jomega)}{(1+jomega)(1-jomega)}
      = frac{omega^2 + jomega}{1+omega^2}.
      end{align*}
      Then the phase angle is $arctanleft(frac{omega}{omega^2}right) = arctanleft(frac{1}{omega}right)$, as the book says.



      If you take the fraction you are given, $frac{a+bi}{c+di}$ and rationalize the denominator, you should be able to write the phase angle as an arctangent just as above.






      share|cite|improve this answer









      $endgroup$



      If $z = a+bj$ where $a$ and $b$ are real, then you can determine the phase angle as $arctanleft(frac{b}{a}right)$ (as long as you are careful about the signs and quadrants), since $a+bj$ represents a point in the complex plane, and the phase angle is the angle this vector makes with the $x$-axis. So for the example from your textbook,
      begin{align*}
      frac{jomega}{1+jomega} = frac{jomega(1-jomega)}{(1+jomega)(1-jomega)}
      = frac{omega^2 + jomega}{1+omega^2}.
      end{align*}
      Then the phase angle is $arctanleft(frac{omega}{omega^2}right) = arctanleft(frac{1}{omega}right)$, as the book says.



      If you take the fraction you are given, $frac{a+bi}{c+di}$ and rationalize the denominator, you should be able to write the phase angle as an arctangent just as above.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Oct 19 '15 at 16:27









      rogerlrogerl

      18k22747




      18k22747












      • $begingroup$
        Thank you very much! This is the answer I was looking for. The book didn't go into much depth into how to find the phase angle for fractions, so I had to try to understand what they were saying.
        $endgroup$
        – nichi
        Oct 19 '15 at 16:38




















      • $begingroup$
        Thank you very much! This is the answer I was looking for. The book didn't go into much depth into how to find the phase angle for fractions, so I had to try to understand what they were saying.
        $endgroup$
        – nichi
        Oct 19 '15 at 16:38


















      $begingroup$
      Thank you very much! This is the answer I was looking for. The book didn't go into much depth into how to find the phase angle for fractions, so I had to try to understand what they were saying.
      $endgroup$
      – nichi
      Oct 19 '15 at 16:38






      $begingroup$
      Thank you very much! This is the answer I was looking for. The book didn't go into much depth into how to find the phase angle for fractions, so I had to try to understand what they were saying.
      $endgroup$
      – nichi
      Oct 19 '15 at 16:38













      1












      $begingroup$

      The argument of the complex number $frac{a+bi}{c+di}$ is just the difference between the arguments of $a+bi$ and $c+di$.



      So you really just need to know how to compute the argument of any complex number $a+bi$ (and then you simply compute a difference of two arguments).




      • If $a > 0$, this is indeed $arctan left(frac{b}{a}right)$.

      • If $a < 0$, this is $arctan left(frac{b}{a}right) pmpi$ depending on the sign of $b$ (I suggest you draw it to see).

      • If $a = 0$, this is $pm frac{pi}{2}$ depending depending on the sign of $b$.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        The argument of the complex number $frac{a+bi}{c+di}$ is just the difference between the arguments of $a+bi$ and $c+di$.



        So you really just need to know how to compute the argument of any complex number $a+bi$ (and then you simply compute a difference of two arguments).




        • If $a > 0$, this is indeed $arctan left(frac{b}{a}right)$.

        • If $a < 0$, this is $arctan left(frac{b}{a}right) pmpi$ depending on the sign of $b$ (I suggest you draw it to see).

        • If $a = 0$, this is $pm frac{pi}{2}$ depending depending on the sign of $b$.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          The argument of the complex number $frac{a+bi}{c+di}$ is just the difference between the arguments of $a+bi$ and $c+di$.



          So you really just need to know how to compute the argument of any complex number $a+bi$ (and then you simply compute a difference of two arguments).




          • If $a > 0$, this is indeed $arctan left(frac{b}{a}right)$.

          • If $a < 0$, this is $arctan left(frac{b}{a}right) pmpi$ depending on the sign of $b$ (I suggest you draw it to see).

          • If $a = 0$, this is $pm frac{pi}{2}$ depending depending on the sign of $b$.






          share|cite|improve this answer









          $endgroup$



          The argument of the complex number $frac{a+bi}{c+di}$ is just the difference between the arguments of $a+bi$ and $c+di$.



          So you really just need to know how to compute the argument of any complex number $a+bi$ (and then you simply compute a difference of two arguments).




          • If $a > 0$, this is indeed $arctan left(frac{b}{a}right)$.

          • If $a < 0$, this is $arctan left(frac{b}{a}right) pmpi$ depending on the sign of $b$ (I suggest you draw it to see).

          • If $a = 0$, this is $pm frac{pi}{2}$ depending depending on the sign of $b$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Oct 19 '15 at 16:33









          Joel CohenJoel Cohen

          7,33412137




          7,33412137























              0












              $begingroup$


              If I have a complex fraction, say $$z=frac{a+bi}{c+di}$$, how can I find the phase angle of this? Could I just calculate it using $$arctan(frac{b}{a})-arctan(frac{d}{c})?$$




              Yes, you can. In fact, I can show you that find phasor angle of the rationalized $z$ is the same as the phasor angle of the numerator minus that of the denominator.



              First the rationalization approach:



              $$angle frac{a+bi}{c+di} = angle frac{(a+bi)(c-di)}{(c+di)(c-di)} = anglefrac{(ac+bd)+(bc-ad)i}{c^2+d^2} = arctan(frac{bc-ad}{ac+bd})$$



              Now calculate the numerator and denominator separately and use Sum of Arctangents: $$arctan a + arctan b = arctan dfrac {a + b} {1 - a b}$$



              $$angle(a+bi)-angle(c+di) = arctan(frac{b}{a})-arctan(frac{d}{c}) = arctan(frac{frac{b}{a}-frac{d}{c}}{1+frac{bd}{ac}})=arctan(frac{bc-ad}{ac+bd})$$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$


                If I have a complex fraction, say $$z=frac{a+bi}{c+di}$$, how can I find the phase angle of this? Could I just calculate it using $$arctan(frac{b}{a})-arctan(frac{d}{c})?$$




                Yes, you can. In fact, I can show you that find phasor angle of the rationalized $z$ is the same as the phasor angle of the numerator minus that of the denominator.



                First the rationalization approach:



                $$angle frac{a+bi}{c+di} = angle frac{(a+bi)(c-di)}{(c+di)(c-di)} = anglefrac{(ac+bd)+(bc-ad)i}{c^2+d^2} = arctan(frac{bc-ad}{ac+bd})$$



                Now calculate the numerator and denominator separately and use Sum of Arctangents: $$arctan a + arctan b = arctan dfrac {a + b} {1 - a b}$$



                $$angle(a+bi)-angle(c+di) = arctan(frac{b}{a})-arctan(frac{d}{c}) = arctan(frac{frac{b}{a}-frac{d}{c}}{1+frac{bd}{ac}})=arctan(frac{bc-ad}{ac+bd})$$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$


                  If I have a complex fraction, say $$z=frac{a+bi}{c+di}$$, how can I find the phase angle of this? Could I just calculate it using $$arctan(frac{b}{a})-arctan(frac{d}{c})?$$




                  Yes, you can. In fact, I can show you that find phasor angle of the rationalized $z$ is the same as the phasor angle of the numerator minus that of the denominator.



                  First the rationalization approach:



                  $$angle frac{a+bi}{c+di} = angle frac{(a+bi)(c-di)}{(c+di)(c-di)} = anglefrac{(ac+bd)+(bc-ad)i}{c^2+d^2} = arctan(frac{bc-ad}{ac+bd})$$



                  Now calculate the numerator and denominator separately and use Sum of Arctangents: $$arctan a + arctan b = arctan dfrac {a + b} {1 - a b}$$



                  $$angle(a+bi)-angle(c+di) = arctan(frac{b}{a})-arctan(frac{d}{c}) = arctan(frac{frac{b}{a}-frac{d}{c}}{1+frac{bd}{ac}})=arctan(frac{bc-ad}{ac+bd})$$






                  share|cite|improve this answer









                  $endgroup$




                  If I have a complex fraction, say $$z=frac{a+bi}{c+di}$$, how can I find the phase angle of this? Could I just calculate it using $$arctan(frac{b}{a})-arctan(frac{d}{c})?$$




                  Yes, you can. In fact, I can show you that find phasor angle of the rationalized $z$ is the same as the phasor angle of the numerator minus that of the denominator.



                  First the rationalization approach:



                  $$angle frac{a+bi}{c+di} = angle frac{(a+bi)(c-di)}{(c+di)(c-di)} = anglefrac{(ac+bd)+(bc-ad)i}{c^2+d^2} = arctan(frac{bc-ad}{ac+bd})$$



                  Now calculate the numerator and denominator separately and use Sum of Arctangents: $$arctan a + arctan b = arctan dfrac {a + b} {1 - a b}$$



                  $$angle(a+bi)-angle(c+di) = arctan(frac{b}{a})-arctan(frac{d}{c}) = arctan(frac{frac{b}{a}-frac{d}{c}}{1+frac{bd}{ac}})=arctan(frac{bc-ad}{ac+bd})$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 21 '18 at 13:41









                  johanjohan

                  11




                  11






























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                      Guatemaltekische Davis-Cup-Mannschaft

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                      Guatemala Kapitän Anthony Vásquez Aktuelles ITF-Ranking 73 Statistik Erste Teilnahme 1990 Davis-Cup-Teilnahmen 22 Bestes Ergebnis Amerika-Gruppenzone II HF (1995, 2000, 2001) Ewige Bilanz 42:30 Erfolgreichste Spieler Meiste Siege gesamt Daniel Chávez Morales (29) Meiste Einzelsiege Cristian Paiz (21) Meiste Doppelsiege Daniel Chávez Morales (17) Bestes Doppel Jacobo Chávez / Daniel Chávez Morales (9) Meiste Teilnahmen Cristian Paiz (32) Meiste Jahre Daniel Chávez Morales (10) Letzte Aktualisierung der Infobox: 3. Juni 2012 Die guatemaltekische Davis-Cup-Mannschaft ist die Tennisnationalmannschaft Guatemalas. Geschichte | 1990 nahm Guatemala erstmals am Davis Cup teil. Das beste Resultat erreichte die Mannschaft jeweils 1995, 2000 und 2001 mit dem Halbfinale in der Amerika-Gruppenzone II. Bester Spieler ist Daniel Chávez Morales mit 29 Siegen bei insgesamt 31 Teilnahmen. Rekord...
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