Help on solving quadratic equations with complex coefficients












2














Please help me out here, I'm self-studying Complex Numbers and I've gotten to a point where I'm kinda stuck.



Given a general quadratic equation $az^2 + bz + c = 0$ with $a not= 0$.



Using the same algebraic manipulation as in the case of real coefficients, we obtain
$$aleft[left(z + frac{b} {2a}right)^2 - frac{Delta} {4a^2}right] = 0$$



This is equivalent to
$$left(z + frac{b} {2a}right)^2 = frac {Delta} {4a^2}quadtext{or}quad left({2az + b}right)^2 = Delta$$
where $Delta = b^2 - 4ac$ is called the discriminant of the quadratic equation, setting $y= 2az + b$, the expression is reduced to
$$y^2 = Delta = u + vi$$
where $u$ and $v$ are real numbers.



What I don't understand now is how the author obtained
$$y_{1,2} = pm left(sqrt { frac {r + u} {2} } + operatorname{sgn} (v)sqrt{ frac{r - u} {2} },iright)$$
where $r = |Delta|$ and $operatorname{sgn}(v)$ is the sign of the real number $v$.






complex-numbers







share|cite|improve this question





























    2














    Please help me out here, I'm self-studying Complex Numbers and I've gotten to a point where I'm kinda stuck.



    Given a general quadratic equation $az^2 + bz + c = 0$ with $a not= 0$.



    Using the same algebraic manipulation as in the case of real coefficients, we obtain
    $$aleft[left(z + frac{b} {2a}right)^2 - frac{Delta} {4a^2}right] = 0$$



    This is equivalent to
    $$left(z + frac{b} {2a}right)^2 = frac {Delta} {4a^2}quadtext{or}quad left({2az + b}right)^2 = Delta$$
    where $Delta = b^2 - 4ac$ is called the discriminant of the quadratic equation, setting $y= 2az + b$, the expression is reduced to
    $$y^2 = Delta = u + vi$$
    where $u$ and $v$ are real numbers.



    What I don't understand now is how the author obtained
    $$y_{1,2} = pm left(sqrt { frac {r + u} {2} } + operatorname{sgn} (v)sqrt{ frac{r - u} {2} },iright)$$
    where $r = |Delta|$ and $operatorname{sgn}(v)$ is the sign of the real number $v$.






    complex-numbers







    share|cite|improve this question



























      2












      2








      2







      Please help me out here, I'm self-studying Complex Numbers and I've gotten to a point where I'm kinda stuck.



      Given a general quadratic equation $az^2 + bz + c = 0$ with $a not= 0$.



      Using the same algebraic manipulation as in the case of real coefficients, we obtain
      $$aleft[left(z + frac{b} {2a}right)^2 - frac{Delta} {4a^2}right] = 0$$



      This is equivalent to
      $$left(z + frac{b} {2a}right)^2 = frac {Delta} {4a^2}quadtext{or}quad left({2az + b}right)^2 = Delta$$
      where $Delta = b^2 - 4ac$ is called the discriminant of the quadratic equation, setting $y= 2az + b$, the expression is reduced to
      $$y^2 = Delta = u + vi$$
      where $u$ and $v$ are real numbers.



      What I don't understand now is how the author obtained
      $$y_{1,2} = pm left(sqrt { frac {r + u} {2} } + operatorname{sgn} (v)sqrt{ frac{r - u} {2} },iright)$$
      where $r = |Delta|$ and $operatorname{sgn}(v)$ is the sign of the real number $v$.






      complex-numbers







      share|cite|improve this question















      Please help me out here, I'm self-studying Complex Numbers and I've gotten to a point where I'm kinda stuck.



      Given a general quadratic equation $az^2 + bz + c = 0$ with $a not= 0$.



      Using the same algebraic manipulation as in the case of real coefficients, we obtain
      $$aleft[left(z + frac{b} {2a}right)^2 - frac{Delta} {4a^2}right] = 0$$



      This is equivalent to
      $$left(z + frac{b} {2a}right)^2 = frac {Delta} {4a^2}quadtext{or}quad left({2az + b}right)^2 = Delta$$
      where $Delta = b^2 - 4ac$ is called the discriminant of the quadratic equation, setting $y= 2az + b$, the expression is reduced to
      $$y^2 = Delta = u + vi$$
      where $u$ and $v$ are real numbers.



      What I don't understand now is how the author obtained
      $$y_{1,2} = pm left(sqrt { frac {r + u} {2} } + operatorname{sgn} (v)sqrt{ frac{r - u} {2} },iright)$$
      where $r = |Delta|$ and $operatorname{sgn}(v)$ is the sign of the real number $v$.






      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 Jun 26 '17 at 9:38









      Robert Z

      93.2k1061132




      93.2k1061132










      asked Jun 26 '17 at 9:07









      Icosahedron

      331114




      331114






















          2 Answers
          2






          active

          oldest

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          2














          Note that if $y=s+it$ and $y^2=u+iv$ then, by considering separately the real and imaginary parts, we obtain
          $$s^2-t^2=u,quad 2st=v.$$
          Now we solve this system with respect to $s$ and $t$.
          From the second $t=v/(2s)$ and plugging it in the first we get
          $$s^2-frac{v^2}{4s^2}=u Leftrightarrow 4s^4-4us^2-v^2=0$$
          and we obtain
          $$s^2=frac{u+sqrt{u^2+v^2}}{2}=frac{|Delta|+u}{2}$$
          Note that we dropped the part with the minus sign because $s^2geq 0$.
          Finally
          $$t^2=frac{v^2}{4s^2}=frac{v^2}{2(|Delta|+u)}=frac{v^2(|Delta|-u)}{2(|Delta|^2-u^2)}=frac{|Delta|-u}{2}.$$
          The we take the square roots keeping in mind that $mbox{sgn}(scdot t)=mbox{sgn}(v).$






          share|cite|improve this answer























          • I understand your solution and thanks for the response but what I don't understand is this. If $y = s + it$ and $y^2 = u + iv$ then $s^2 - t^2 = u$ and $2st = v$ how did you arrive at this system? Please do explain your steps.
            – Icosahedron
            Jun 26 '17 at 10:03












          • Thanks your your response @Robert
            – Icosahedron
            Jun 26 '17 at 10:08










          • @Icosahedron You are welcome. BTW if you are new here please take a few minutes for a tour math.stackexchange.com/tour
            – Robert Z
            Jun 26 '17 at 10:11










          • Oops, never mind. I finally figured it out. You equated the real and imaginary part. THANKS a bunch.
            – Icosahedron
            Jun 26 '17 at 10:11



















          0














          Discriminant $Delta = b^2 - 4ac = alpha + ibeta$



          We need to find the square root of $Delta$
          $sqrt{alpha + ibeta} = p + iq$
          $implies alpha + ibeta = p^2 - q^2 + 2ipq$
          $implies p^2 - q^2 = alpha$ and $ 2pq = beta$
          $implies p^2 + q^2 = sqrt{alpha^2 + beta^2}$
          $implies p = sqrt{dfrac{alpha + |Delta|}{2}}$ and $ q = sqrt{dfrac{|Delta| - alpha}{2}}$



          $implies sqrt{Delta} = sqrt{dfrac{alpha + |Delta|}{2}} + dfrac{beta}{|beta|} sqrt{dfrac{|Delta| - alpha}{2}}i$



          Now you can use the quadratic formula.






          share|cite|improve this answer





















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            2 Answers
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            active

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            2 Answers
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            active

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            2














            Note that if $y=s+it$ and $y^2=u+iv$ then, by considering separately the real and imaginary parts, we obtain
            $$s^2-t^2=u,quad 2st=v.$$
            Now we solve this system with respect to $s$ and $t$.
            From the second $t=v/(2s)$ and plugging it in the first we get
            $$s^2-frac{v^2}{4s^2}=u Leftrightarrow 4s^4-4us^2-v^2=0$$
            and we obtain
            $$s^2=frac{u+sqrt{u^2+v^2}}{2}=frac{|Delta|+u}{2}$$
            Note that we dropped the part with the minus sign because $s^2geq 0$.
            Finally
            $$t^2=frac{v^2}{4s^2}=frac{v^2}{2(|Delta|+u)}=frac{v^2(|Delta|-u)}{2(|Delta|^2-u^2)}=frac{|Delta|-u}{2}.$$
            The we take the square roots keeping in mind that $mbox{sgn}(scdot t)=mbox{sgn}(v).$






            share|cite|improve this answer























            • I understand your solution and thanks for the response but what I don't understand is this. If $y = s + it$ and $y^2 = u + iv$ then $s^2 - t^2 = u$ and $2st = v$ how did you arrive at this system? Please do explain your steps.
              – Icosahedron
              Jun 26 '17 at 10:03












            • Thanks your your response @Robert
              – Icosahedron
              Jun 26 '17 at 10:08










            • @Icosahedron You are welcome. BTW if you are new here please take a few minutes for a tour math.stackexchange.com/tour
              – Robert Z
              Jun 26 '17 at 10:11










            • Oops, never mind. I finally figured it out. You equated the real and imaginary part. THANKS a bunch.
              – Icosahedron
              Jun 26 '17 at 10:11
















            2














            Note that if $y=s+it$ and $y^2=u+iv$ then, by considering separately the real and imaginary parts, we obtain
            $$s^2-t^2=u,quad 2st=v.$$
            Now we solve this system with respect to $s$ and $t$.
            From the second $t=v/(2s)$ and plugging it in the first we get
            $$s^2-frac{v^2}{4s^2}=u Leftrightarrow 4s^4-4us^2-v^2=0$$
            and we obtain
            $$s^2=frac{u+sqrt{u^2+v^2}}{2}=frac{|Delta|+u}{2}$$
            Note that we dropped the part with the minus sign because $s^2geq 0$.
            Finally
            $$t^2=frac{v^2}{4s^2}=frac{v^2}{2(|Delta|+u)}=frac{v^2(|Delta|-u)}{2(|Delta|^2-u^2)}=frac{|Delta|-u}{2}.$$
            The we take the square roots keeping in mind that $mbox{sgn}(scdot t)=mbox{sgn}(v).$






            share|cite|improve this answer























            • I understand your solution and thanks for the response but what I don't understand is this. If $y = s + it$ and $y^2 = u + iv$ then $s^2 - t^2 = u$ and $2st = v$ how did you arrive at this system? Please do explain your steps.
              – Icosahedron
              Jun 26 '17 at 10:03












            • Thanks your your response @Robert
              – Icosahedron
              Jun 26 '17 at 10:08










            • @Icosahedron You are welcome. BTW if you are new here please take a few minutes for a tour math.stackexchange.com/tour
              – Robert Z
              Jun 26 '17 at 10:11










            • Oops, never mind. I finally figured it out. You equated the real and imaginary part. THANKS a bunch.
              – Icosahedron
              Jun 26 '17 at 10:11














            2












            2








            2






            Note that if $y=s+it$ and $y^2=u+iv$ then, by considering separately the real and imaginary parts, we obtain
            $$s^2-t^2=u,quad 2st=v.$$
            Now we solve this system with respect to $s$ and $t$.
            From the second $t=v/(2s)$ and plugging it in the first we get
            $$s^2-frac{v^2}{4s^2}=u Leftrightarrow 4s^4-4us^2-v^2=0$$
            and we obtain
            $$s^2=frac{u+sqrt{u^2+v^2}}{2}=frac{|Delta|+u}{2}$$
            Note that we dropped the part with the minus sign because $s^2geq 0$.
            Finally
            $$t^2=frac{v^2}{4s^2}=frac{v^2}{2(|Delta|+u)}=frac{v^2(|Delta|-u)}{2(|Delta|^2-u^2)}=frac{|Delta|-u}{2}.$$
            The we take the square roots keeping in mind that $mbox{sgn}(scdot t)=mbox{sgn}(v).$






            share|cite|improve this answer














            Note that if $y=s+it$ and $y^2=u+iv$ then, by considering separately the real and imaginary parts, we obtain
            $$s^2-t^2=u,quad 2st=v.$$
            Now we solve this system with respect to $s$ and $t$.
            From the second $t=v/(2s)$ and plugging it in the first we get
            $$s^2-frac{v^2}{4s^2}=u Leftrightarrow 4s^4-4us^2-v^2=0$$
            and we obtain
            $$s^2=frac{u+sqrt{u^2+v^2}}{2}=frac{|Delta|+u}{2}$$
            Note that we dropped the part with the minus sign because $s^2geq 0$.
            Finally
            $$t^2=frac{v^2}{4s^2}=frac{v^2}{2(|Delta|+u)}=frac{v^2(|Delta|-u)}{2(|Delta|^2-u^2)}=frac{|Delta|-u}{2}.$$
            The we take the square roots keeping in mind that $mbox{sgn}(scdot t)=mbox{sgn}(v).$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jun 26 '17 at 10:14

























            answered Jun 26 '17 at 9:18









            Robert Z

            93.2k1061132




            93.2k1061132












            • I understand your solution and thanks for the response but what I don't understand is this. If $y = s + it$ and $y^2 = u + iv$ then $s^2 - t^2 = u$ and $2st = v$ how did you arrive at this system? Please do explain your steps.
              – Icosahedron
              Jun 26 '17 at 10:03












            • Thanks your your response @Robert
              – Icosahedron
              Jun 26 '17 at 10:08










            • @Icosahedron You are welcome. BTW if you are new here please take a few minutes for a tour math.stackexchange.com/tour
              – Robert Z
              Jun 26 '17 at 10:11










            • Oops, never mind. I finally figured it out. You equated the real and imaginary part. THANKS a bunch.
              – Icosahedron
              Jun 26 '17 at 10:11


















            • I understand your solution and thanks for the response but what I don't understand is this. If $y = s + it$ and $y^2 = u + iv$ then $s^2 - t^2 = u$ and $2st = v$ how did you arrive at this system? Please do explain your steps.
              – Icosahedron
              Jun 26 '17 at 10:03












            • Thanks your your response @Robert
              – Icosahedron
              Jun 26 '17 at 10:08










            • @Icosahedron You are welcome. BTW if you are new here please take a few minutes for a tour math.stackexchange.com/tour
              – Robert Z
              Jun 26 '17 at 10:11










            • Oops, never mind. I finally figured it out. You equated the real and imaginary part. THANKS a bunch.
              – Icosahedron
              Jun 26 '17 at 10:11
















            I understand your solution and thanks for the response but what I don't understand is this. If $y = s + it$ and $y^2 = u + iv$ then $s^2 - t^2 = u$ and $2st = v$ how did you arrive at this system? Please do explain your steps.
            – Icosahedron
            Jun 26 '17 at 10:03






            I understand your solution and thanks for the response but what I don't understand is this. If $y = s + it$ and $y^2 = u + iv$ then $s^2 - t^2 = u$ and $2st = v$ how did you arrive at this system? Please do explain your steps.
            – Icosahedron
            Jun 26 '17 at 10:03














            Thanks your your response @Robert
            – Icosahedron
            Jun 26 '17 at 10:08




            Thanks your your response @Robert
            – Icosahedron
            Jun 26 '17 at 10:08












            @Icosahedron You are welcome. BTW if you are new here please take a few minutes for a tour math.stackexchange.com/tour
            – Robert Z
            Jun 26 '17 at 10:11




            @Icosahedron You are welcome. BTW if you are new here please take a few minutes for a tour math.stackexchange.com/tour
            – Robert Z
            Jun 26 '17 at 10:11












            Oops, never mind. I finally figured it out. You equated the real and imaginary part. THANKS a bunch.
            – Icosahedron
            Jun 26 '17 at 10:11




            Oops, never mind. I finally figured it out. You equated the real and imaginary part. THANKS a bunch.
            – Icosahedron
            Jun 26 '17 at 10:11











            0














            Discriminant $Delta = b^2 - 4ac = alpha + ibeta$



            We need to find the square root of $Delta$
            $sqrt{alpha + ibeta} = p + iq$
            $implies alpha + ibeta = p^2 - q^2 + 2ipq$
            $implies p^2 - q^2 = alpha$ and $ 2pq = beta$
            $implies p^2 + q^2 = sqrt{alpha^2 + beta^2}$
            $implies p = sqrt{dfrac{alpha + |Delta|}{2}}$ and $ q = sqrt{dfrac{|Delta| - alpha}{2}}$



            $implies sqrt{Delta} = sqrt{dfrac{alpha + |Delta|}{2}} + dfrac{beta}{|beta|} sqrt{dfrac{|Delta| - alpha}{2}}i$



            Now you can use the quadratic formula.






            share|cite|improve this answer


























              0














              Discriminant $Delta = b^2 - 4ac = alpha + ibeta$



              We need to find the square root of $Delta$
              $sqrt{alpha + ibeta} = p + iq$
              $implies alpha + ibeta = p^2 - q^2 + 2ipq$
              $implies p^2 - q^2 = alpha$ and $ 2pq = beta$
              $implies p^2 + q^2 = sqrt{alpha^2 + beta^2}$
              $implies p = sqrt{dfrac{alpha + |Delta|}{2}}$ and $ q = sqrt{dfrac{|Delta| - alpha}{2}}$



              $implies sqrt{Delta} = sqrt{dfrac{alpha + |Delta|}{2}} + dfrac{beta}{|beta|} sqrt{dfrac{|Delta| - alpha}{2}}i$



              Now you can use the quadratic formula.






              share|cite|improve this answer
























                0












                0








                0






                Discriminant $Delta = b^2 - 4ac = alpha + ibeta$



                We need to find the square root of $Delta$
                $sqrt{alpha + ibeta} = p + iq$
                $implies alpha + ibeta = p^2 - q^2 + 2ipq$
                $implies p^2 - q^2 = alpha$ and $ 2pq = beta$
                $implies p^2 + q^2 = sqrt{alpha^2 + beta^2}$
                $implies p = sqrt{dfrac{alpha + |Delta|}{2}}$ and $ q = sqrt{dfrac{|Delta| - alpha}{2}}$



                $implies sqrt{Delta} = sqrt{dfrac{alpha + |Delta|}{2}} + dfrac{beta}{|beta|} sqrt{dfrac{|Delta| - alpha}{2}}i$



                Now you can use the quadratic formula.






                share|cite|improve this answer












                Discriminant $Delta = b^2 - 4ac = alpha + ibeta$



                We need to find the square root of $Delta$
                $sqrt{alpha + ibeta} = p + iq$
                $implies alpha + ibeta = p^2 - q^2 + 2ipq$
                $implies p^2 - q^2 = alpha$ and $ 2pq = beta$
                $implies p^2 + q^2 = sqrt{alpha^2 + beta^2}$
                $implies p = sqrt{dfrac{alpha + |Delta|}{2}}$ and $ q = sqrt{dfrac{|Delta| - alpha}{2}}$



                $implies sqrt{Delta} = sqrt{dfrac{alpha + |Delta|}{2}} + dfrac{beta}{|beta|} sqrt{dfrac{|Delta| - alpha}{2}}i$



                Now you can use the quadratic formula.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 26 at 9:43









                Rahul B.

                12




                12






























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