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Complex numbers z=-3-4i polar form



Unicorn Meta Zoo #1: Why another podcast?
Announcing the arrival of Valued Associate #679: Cesar ManaraExpressing $e^z$ where $z=a+bi$ in polar form.Polar form of Complex numbersConverting to polar formComplex number polar form equationWhy is e used for polar form of complex numbers?Show $-27$ in polar form.writing in polar form (complex numbers)Are all complex exponentials in polar form?Complex numbers polar form changeDifferentiating polar functions using complex numbers












4












$begingroup$


$z=-3-4i$ . In polar form this becomes $[5, 233° ]$. The question then asks for $z^2$, so the polar form becomes $[25,466]$ However in the solution they did $466° -360° $ and I am unsure why they did this. Is it a rule that needs to be applied.






complex-numbers polar-coordinates







share|cite|improve this question











$endgroup$

















    4












    $begingroup$


    $z=-3-4i$ . In polar form this becomes $[5, 233° ]$. The question then asks for $z^2$, so the polar form becomes $[25,466]$ However in the solution they did $466° -360° $ and I am unsure why they did this. Is it a rule that needs to be applied.






    complex-numbers polar-coordinates







    share|cite|improve this question











    $endgroup$















      4












      4








      4





      $begingroup$


      $z=-3-4i$ . In polar form this becomes $[5, 233° ]$. The question then asks for $z^2$, so the polar form becomes $[25,466]$ However in the solution they did $466° -360° $ and I am unsure why they did this. Is it a rule that needs to be applied.






      complex-numbers polar-coordinates







      share|cite|improve this question











      $endgroup$




      $z=-3-4i$ . In polar form this becomes $[5, 233° ]$. The question then asks for $z^2$, so the polar form becomes $[25,466]$ However in the solution they did $466° -360° $ and I am unsure why they did this. Is it a rule that needs to be applied.






      complex-numbers polar-coordinates




      complex-numbers polar-coordinates






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 1 hour ago









      José Carlos Santos

      178k24139251




      178k24139251










      asked 1 hour ago









      user221435user221435

      285




      285






















          2 Answers
          2






          active

          oldest

          votes


















          3












          $begingroup$

          It's because$$cos(x-360^circ)=cos(x)text{ and }sin(x-360^circ)=sin(x).$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Usually I don't do this when I get the degree, why is it that in this case I needed to do this?
            $endgroup$
            – user221435
            1 hour ago






          • 2




            $begingroup$
            We don't need to do this, but we usually try to work with angles in the range $[0^circ,360^circ)$.
            $endgroup$
            – José Carlos Santos
            1 hour ago










          • $begingroup$
            Great, thank you
            $endgroup$
            – user221435
            1 hour ago



















          2












          $begingroup$

          It is just to get the principal value of the angle, since if you rotate by an angle $466^{circ}$ you'll get to the same position as rotating $106^{circ}$ so we usually take the smallest angle that is needed to arrive at the desired position.



          The principal angle is an angle between $-180^{circ}$ and $+180^{circ}$






          share|cite|improve this answer











          $endgroup$














            Your Answer








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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            3












            $begingroup$

            It's because$$cos(x-360^circ)=cos(x)text{ and }sin(x-360^circ)=sin(x).$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Usually I don't do this when I get the degree, why is it that in this case I needed to do this?
              $endgroup$
              – user221435
              1 hour ago






            • 2




              $begingroup$
              We don't need to do this, but we usually try to work with angles in the range $[0^circ,360^circ)$.
              $endgroup$
              – José Carlos Santos
              1 hour ago










            • $begingroup$
              Great, thank you
              $endgroup$
              – user221435
              1 hour ago
















            3












            $begingroup$

            It's because$$cos(x-360^circ)=cos(x)text{ and }sin(x-360^circ)=sin(x).$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Usually I don't do this when I get the degree, why is it that in this case I needed to do this?
              $endgroup$
              – user221435
              1 hour ago






            • 2




              $begingroup$
              We don't need to do this, but we usually try to work with angles in the range $[0^circ,360^circ)$.
              $endgroup$
              – José Carlos Santos
              1 hour ago










            • $begingroup$
              Great, thank you
              $endgroup$
              – user221435
              1 hour ago














            3












            3








            3





            $begingroup$

            It's because$$cos(x-360^circ)=cos(x)text{ and }sin(x-360^circ)=sin(x).$$






            share|cite|improve this answer









            $endgroup$



            It's because$$cos(x-360^circ)=cos(x)text{ and }sin(x-360^circ)=sin(x).$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 1 hour ago









            José Carlos SantosJosé Carlos Santos

            178k24139251




            178k24139251












            • $begingroup$
              Usually I don't do this when I get the degree, why is it that in this case I needed to do this?
              $endgroup$
              – user221435
              1 hour ago






            • 2




              $begingroup$
              We don't need to do this, but we usually try to work with angles in the range $[0^circ,360^circ)$.
              $endgroup$
              – José Carlos Santos
              1 hour ago










            • $begingroup$
              Great, thank you
              $endgroup$
              – user221435
              1 hour ago


















            • $begingroup$
              Usually I don't do this when I get the degree, why is it that in this case I needed to do this?
              $endgroup$
              – user221435
              1 hour ago






            • 2




              $begingroup$
              We don't need to do this, but we usually try to work with angles in the range $[0^circ,360^circ)$.
              $endgroup$
              – José Carlos Santos
              1 hour ago










            • $begingroup$
              Great, thank you
              $endgroup$
              – user221435
              1 hour ago
















            $begingroup$
            Usually I don't do this when I get the degree, why is it that in this case I needed to do this?
            $endgroup$
            – user221435
            1 hour ago




            $begingroup$
            Usually I don't do this when I get the degree, why is it that in this case I needed to do this?
            $endgroup$
            – user221435
            1 hour ago




            2




            2




            $begingroup$
            We don't need to do this, but we usually try to work with angles in the range $[0^circ,360^circ)$.
            $endgroup$
            – José Carlos Santos
            1 hour ago




            $begingroup$
            We don't need to do this, but we usually try to work with angles in the range $[0^circ,360^circ)$.
            $endgroup$
            – José Carlos Santos
            1 hour ago












            $begingroup$
            Great, thank you
            $endgroup$
            – user221435
            1 hour ago




            $begingroup$
            Great, thank you
            $endgroup$
            – user221435
            1 hour ago











            2












            $begingroup$

            It is just to get the principal value of the angle, since if you rotate by an angle $466^{circ}$ you'll get to the same position as rotating $106^{circ}$ so we usually take the smallest angle that is needed to arrive at the desired position.



            The principal angle is an angle between $-180^{circ}$ and $+180^{circ}$






            share|cite|improve this answer











            $endgroup$


















              2












              $begingroup$

              It is just to get the principal value of the angle, since if you rotate by an angle $466^{circ}$ you'll get to the same position as rotating $106^{circ}$ so we usually take the smallest angle that is needed to arrive at the desired position.



              The principal angle is an angle between $-180^{circ}$ and $+180^{circ}$






              share|cite|improve this answer











              $endgroup$
















                2












                2








                2





                $begingroup$

                It is just to get the principal value of the angle, since if you rotate by an angle $466^{circ}$ you'll get to the same position as rotating $106^{circ}$ so we usually take the smallest angle that is needed to arrive at the desired position.



                The principal angle is an angle between $-180^{circ}$ and $+180^{circ}$






                share|cite|improve this answer











                $endgroup$



                It is just to get the principal value of the angle, since if you rotate by an angle $466^{circ}$ you'll get to the same position as rotating $106^{circ}$ so we usually take the smallest angle that is needed to arrive at the desired position.



                The principal angle is an angle between $-180^{circ}$ and $+180^{circ}$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited 1 hour ago









                J. W. Tanner

                5,2001520




                5,2001520










                answered 1 hour ago









                Fareed AFFareed AF

                1,025214




                1,025214






























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