Help on solving quadratic equations with complex coefficients
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
edited Jun 26 '17 at 9:38
Robert Z
93.2k1061132
asked Jun 26 '17 at 9:07
Icosahedron
331114
add a comment |
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
edited Jun 26 '17 at 9:38
Robert Z
93.2k1061132
asked Jun 26 '17 at 9:07
Icosahedron
331114
add a comment |
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
edited Jun 26 '17 at 9:38
Robert Z
93.2k1061132
asked Jun 26 '17 at 9:07
Icosahedron
331114
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
edited Jun 26 '17 at 9:38
Robert Z
93.2k1061132
asked Jun 26 '17 at 9:07
Icosahedron
331114
edited Jun 26 '17 at 9:38
Robert Z
93.2k1061132
asked Jun 26 '17 at 9:07
Icosahedron
331114
edited Jun 26 '17 at 9:38
Robert Z
93.2k1061132
edited Jun 26 '17 at 9:38
Robert Z
93.2k1061132
edited Jun 26 '17 at 9:38
Robert Z
93.2k1061132
93.2k1061132
asked Jun 26 '17 at 9:07
Icosahedron
331114
asked Jun 26 '17 at 9:07
Icosahedron
331114
asked Jun 26 '17 at 9:07
Icosahedron
331114
331114
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
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 $mbox{sgn}(scdot t)=mbox{sgn}(v).$
answered Jun 26 '17 at 9:18
Robert Z
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
add a comment |
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.
answered Nov 26 at 9:43
Rahul B.
12
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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votes
active
oldest
votes
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 $mbox{sgn}(scdot t)=mbox{sgn}(v).$
answered Jun 26 '17 at 9:18
Robert Z
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
add a comment |
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 $mbox{sgn}(scdot t)=mbox{sgn}(v).$
answered Jun 26 '17 at 9:18
Robert Z
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
add a comment |
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 $mbox{sgn}(scdot t)=mbox{sgn}(v).$
answered Jun 26 '17 at 9:18
Robert Z
93.2k1061132
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 $mbox{sgn}(scdot t)=mbox{sgn}(v).$
answered Jun 26 '17 at 9:18
Robert Z
93.2k1061132
edited Jun 26 '17 at 10:14
answered Jun 26 '17 at 9:18
Robert Z
93.2k1061132
answered Jun 26 '17 at 9:18
Robert Z
93.2k1061132
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
add a comment |
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
add a comment |
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.
answered Nov 26 at 9:43
Rahul B.
12
add a comment |
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.
answered Nov 26 at 9:43
Rahul B.
12
add a comment |
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.
answered Nov 26 at 9:43
Rahul B.
12
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.
answered Nov 26 at 9:43
Rahul B.
12
answered Nov 26 at 9:43
Rahul B.
12
answered Nov 26 at 9:43
Rahul B.
12
answered Nov 26 at 9:43
Rahul B.
12
12
add a comment |
add a comment |
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