Find all the solutions of $z^2-(1+3i)z-8-i=0$












3












$begingroup$


I am stuck on a problem and I was hoping someone could tell me what I am doing wrong.




I want to find all the roots of:



$$z^2-(1+3i)z-8-i=0$$




There are two ways I tried to approach this.




  1. Quadratic formula:


$$begin{align} z_1,z_2 &=-frac{p}{2}pmsqrt{left( frac{p}{2}right)^2-q} \ & implies z_1,z_2=frac{1+3i}{2}pmsqrt{left( frac{(-1-3i)}{2}right)^2-(-8-i)}\ &= frac{1+3i}{2} pmsqrt{frac{24+10i}{4} }=frac{1+3i pmsqrt{24+10i}}{2}end{align}$$




  1. Completing the square:


$$begin{aligned} z^2-(1+3i)z-8-i &=0 \ & iff left(z-left( frac{1+3i}{2}right) right)^2-8-i=left( frac{1+3i}{2}right)^2 \ & iff u^2=frac{24+10i}{4} \ & iff u=pmfrac{sqrt{24+10i}}{2} \ & iff z_{1,2}=frac{1+3i pmsqrt{24+10i}}{2}end{aligned} $$



Using both methods I arrive at the same result. However, wolframalpha tells me the roots are:



$$z_1=-2+i \ z_2=3+2i$$



I have tried everything (writing it in in exponential form, graphing, etc.) but I have no idea how I can arrive at those two roots. What am I doing wrong?






polynomials complex-numbers roots quadratics







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Possible duplicate of How do I get the square root of a complex number?
    $endgroup$
    – Jyrki Lahtonen
    Dec 16 '18 at 10:36
















3












$begingroup$


I am stuck on a problem and I was hoping someone could tell me what I am doing wrong.




I want to find all the roots of:



$$z^2-(1+3i)z-8-i=0$$




There are two ways I tried to approach this.




  1. Quadratic formula:


$$begin{align} z_1,z_2 &=-frac{p}{2}pmsqrt{left( frac{p}{2}right)^2-q} \ & implies z_1,z_2=frac{1+3i}{2}pmsqrt{left( frac{(-1-3i)}{2}right)^2-(-8-i)}\ &= frac{1+3i}{2} pmsqrt{frac{24+10i}{4} }=frac{1+3i pmsqrt{24+10i}}{2}end{align}$$




  1. Completing the square:


$$begin{aligned} z^2-(1+3i)z-8-i &=0 \ & iff left(z-left( frac{1+3i}{2}right) right)^2-8-i=left( frac{1+3i}{2}right)^2 \ & iff u^2=frac{24+10i}{4} \ & iff u=pmfrac{sqrt{24+10i}}{2} \ & iff z_{1,2}=frac{1+3i pmsqrt{24+10i}}{2}end{aligned} $$



Using both methods I arrive at the same result. However, wolframalpha tells me the roots are:



$$z_1=-2+i \ z_2=3+2i$$



I have tried everything (writing it in in exponential form, graphing, etc.) but I have no idea how I can arrive at those two roots. What am I doing wrong?






polynomials complex-numbers roots quadratics







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Possible duplicate of How do I get the square root of a complex number?
    $endgroup$
    – Jyrki Lahtonen
    Dec 16 '18 at 10:36














3












3








3


1



$begingroup$


I am stuck on a problem and I was hoping someone could tell me what I am doing wrong.




I want to find all the roots of:



$$z^2-(1+3i)z-8-i=0$$




There are two ways I tried to approach this.




  1. Quadratic formula:


$$begin{align} z_1,z_2 &=-frac{p}{2}pmsqrt{left( frac{p}{2}right)^2-q} \ & implies z_1,z_2=frac{1+3i}{2}pmsqrt{left( frac{(-1-3i)}{2}right)^2-(-8-i)}\ &= frac{1+3i}{2} pmsqrt{frac{24+10i}{4} }=frac{1+3i pmsqrt{24+10i}}{2}end{align}$$




  1. Completing the square:


$$begin{aligned} z^2-(1+3i)z-8-i &=0 \ & iff left(z-left( frac{1+3i}{2}right) right)^2-8-i=left( frac{1+3i}{2}right)^2 \ & iff u^2=frac{24+10i}{4} \ & iff u=pmfrac{sqrt{24+10i}}{2} \ & iff z_{1,2}=frac{1+3i pmsqrt{24+10i}}{2}end{aligned} $$



Using both methods I arrive at the same result. However, wolframalpha tells me the roots are:



$$z_1=-2+i \ z_2=3+2i$$



I have tried everything (writing it in in exponential form, graphing, etc.) but I have no idea how I can arrive at those two roots. What am I doing wrong?






polynomials complex-numbers roots quadratics







share|cite|improve this question











$endgroup$




I am stuck on a problem and I was hoping someone could tell me what I am doing wrong.




I want to find all the roots of:



$$z^2-(1+3i)z-8-i=0$$




There are two ways I tried to approach this.




  1. Quadratic formula:


$$begin{align} z_1,z_2 &=-frac{p}{2}pmsqrt{left( frac{p}{2}right)^2-q} \ & implies z_1,z_2=frac{1+3i}{2}pmsqrt{left( frac{(-1-3i)}{2}right)^2-(-8-i)}\ &= frac{1+3i}{2} pmsqrt{frac{24+10i}{4} }=frac{1+3i pmsqrt{24+10i}}{2}end{align}$$




  1. Completing the square:


$$begin{aligned} z^2-(1+3i)z-8-i &=0 \ & iff left(z-left( frac{1+3i}{2}right) right)^2-8-i=left( frac{1+3i}{2}right)^2 \ & iff u^2=frac{24+10i}{4} \ & iff u=pmfrac{sqrt{24+10i}}{2} \ & iff z_{1,2}=frac{1+3i pmsqrt{24+10i}}{2}end{aligned} $$



Using both methods I arrive at the same result. However, wolframalpha tells me the roots are:



$$z_1=-2+i \ z_2=3+2i$$



I have tried everything (writing it in in exponential form, graphing, etc.) but I have no idea how I can arrive at those two roots. What am I doing wrong?






polynomials complex-numbers roots quadratics




polynomials complex-numbers roots quadratics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 16 '18 at 10:28









greedoid

41.9k1152105




41.9k1152105










asked Dec 16 '18 at 10:21









NullspaceNullspace

372110




372110








  • 2




    $begingroup$
    Possible duplicate of How do I get the square root of a complex number?
    $endgroup$
    – Jyrki Lahtonen
    Dec 16 '18 at 10:36














  • 2




    $begingroup$
    Possible duplicate of How do I get the square root of a complex number?
    $endgroup$
    – Jyrki Lahtonen
    Dec 16 '18 at 10:36








2




2




$begingroup$
Possible duplicate of How do I get the square root of a complex number?
$endgroup$
– Jyrki Lahtonen
Dec 16 '18 at 10:36




$begingroup$
Possible duplicate of How do I get the square root of a complex number?
$endgroup$
– Jyrki Lahtonen
Dec 16 '18 at 10:36










5 Answers
5






active

oldest

votes


















2












$begingroup$

To find square roots $;pm(a+ ib);$ of $;24+10i,;$ solve the equation
$$(a+ib)^2=24+10i.$$
Real and imaginary parts of RHS and LHS are equal, and also absolute values:



$$begin{aligned}a^2-b^2&=24\ 2ab&=10\a^2+b^2&=26end{aligned}$$
We get $a^2=25$ and $ab=5$ hence $b$ has the sign of $a.$ This gives the two square roots $pm(5+i).$






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Thanks for explaining that. I am new to complex roots and complex polynomials so I didn't immediately think of doing this. Thank you!
    $endgroup$
    – Nullspace
    Dec 16 '18 at 10:46










  • $begingroup$
    You're welcome :)
    $endgroup$
    – user376343
    Dec 16 '18 at 10:55



















1












$begingroup$

That's just because the square roots of $24+10i$ are $pm(5+i)$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I should have seen that. Thank you!
    $endgroup$
    – Nullspace
    Dec 16 '18 at 10:28



















1












$begingroup$

Hint: Use the fact: $$ 24+10i = (5+i)^2$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I should have seen that. Thank you!
    $endgroup$
    – Nullspace
    Dec 16 '18 at 10:28



















1












$begingroup$

Hint: Expanding $$(5+i)^2=25+10i-1$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I should have seen that. Thank you!
    $endgroup$
    – Nullspace
    Dec 16 '18 at 10:28



















1












$begingroup$

See the answer to this question for how to extract complex square roots. Note carefully: when you have an odd prime root of a complex number you can't extract it except with trigonometry, but square roots can be rendered by algebra alone.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Ah I see. I would have do use de moivres formula then right?
    $endgroup$
    – Nullspace
    Dec 16 '18 at 10:29






  • 1




    $begingroup$
    For odd prime roots yes. For square roots no.
    $endgroup$
    – Oscar Lanzi
    Dec 16 '18 at 10:36










  • $begingroup$
    So for example: $z^8=-3+sqrt{3}i$ I could solve without de moivres formula?
    $endgroup$
    – Nullspace
    Dec 16 '18 at 10:44






  • 1




    $begingroup$
    Yup, you just do three square root extractions. The algebra gets complicated but the concept works.
    $endgroup$
    – Oscar Lanzi
    Dec 16 '18 at 10:57













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5 Answers
5






active

oldest

votes








5 Answers
5






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

To find square roots $;pm(a+ ib);$ of $;24+10i,;$ solve the equation
$$(a+ib)^2=24+10i.$$
Real and imaginary parts of RHS and LHS are equal, and also absolute values:



$$begin{aligned}a^2-b^2&=24\ 2ab&=10\a^2+b^2&=26end{aligned}$$
We get $a^2=25$ and $ab=5$ hence $b$ has the sign of $a.$ This gives the two square roots $pm(5+i).$






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Thanks for explaining that. I am new to complex roots and complex polynomials so I didn't immediately think of doing this. Thank you!
    $endgroup$
    – Nullspace
    Dec 16 '18 at 10:46










  • $begingroup$
    You're welcome :)
    $endgroup$
    – user376343
    Dec 16 '18 at 10:55
















2












$begingroup$

To find square roots $;pm(a+ ib);$ of $;24+10i,;$ solve the equation
$$(a+ib)^2=24+10i.$$
Real and imaginary parts of RHS and LHS are equal, and also absolute values:



$$begin{aligned}a^2-b^2&=24\ 2ab&=10\a^2+b^2&=26end{aligned}$$
We get $a^2=25$ and $ab=5$ hence $b$ has the sign of $a.$ This gives the two square roots $pm(5+i).$






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Thanks for explaining that. I am new to complex roots and complex polynomials so I didn't immediately think of doing this. Thank you!
    $endgroup$
    – Nullspace
    Dec 16 '18 at 10:46










  • $begingroup$
    You're welcome :)
    $endgroup$
    – user376343
    Dec 16 '18 at 10:55














2












2








2





$begingroup$

To find square roots $;pm(a+ ib);$ of $;24+10i,;$ solve the equation
$$(a+ib)^2=24+10i.$$
Real and imaginary parts of RHS and LHS are equal, and also absolute values:



$$begin{aligned}a^2-b^2&=24\ 2ab&=10\a^2+b^2&=26end{aligned}$$
We get $a^2=25$ and $ab=5$ hence $b$ has the sign of $a.$ This gives the two square roots $pm(5+i).$






share|cite|improve this answer











$endgroup$



To find square roots $;pm(a+ ib);$ of $;24+10i,;$ solve the equation
$$(a+ib)^2=24+10i.$$
Real and imaginary parts of RHS and LHS are equal, and also absolute values:



$$begin{aligned}a^2-b^2&=24\ 2ab&=10\a^2+b^2&=26end{aligned}$$
We get $a^2=25$ and $ab=5$ hence $b$ has the sign of $a.$ This gives the two square roots $pm(5+i).$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 16 '18 at 10:49

























answered Dec 16 '18 at 10:35









user376343user376343

3,7883827




3,7883827








  • 1




    $begingroup$
    Thanks for explaining that. I am new to complex roots and complex polynomials so I didn't immediately think of doing this. Thank you!
    $endgroup$
    – Nullspace
    Dec 16 '18 at 10:46










  • $begingroup$
    You're welcome :)
    $endgroup$
    – user376343
    Dec 16 '18 at 10:55














  • 1




    $begingroup$
    Thanks for explaining that. I am new to complex roots and complex polynomials so I didn't immediately think of doing this. Thank you!
    $endgroup$
    – Nullspace
    Dec 16 '18 at 10:46










  • $begingroup$
    You're welcome :)
    $endgroup$
    – user376343
    Dec 16 '18 at 10:55








1




1




$begingroup$
Thanks for explaining that. I am new to complex roots and complex polynomials so I didn't immediately think of doing this. Thank you!
$endgroup$
– Nullspace
Dec 16 '18 at 10:46




$begingroup$
Thanks for explaining that. I am new to complex roots and complex polynomials so I didn't immediately think of doing this. Thank you!
$endgroup$
– Nullspace
Dec 16 '18 at 10:46












$begingroup$
You're welcome :)
$endgroup$
– user376343
Dec 16 '18 at 10:55




$begingroup$
You're welcome :)
$endgroup$
– user376343
Dec 16 '18 at 10:55











1












$begingroup$

That's just because the square roots of $24+10i$ are $pm(5+i)$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I should have seen that. Thank you!
    $endgroup$
    – Nullspace
    Dec 16 '18 at 10:28
















1












$begingroup$

That's just because the square roots of $24+10i$ are $pm(5+i)$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I should have seen that. Thank you!
    $endgroup$
    – Nullspace
    Dec 16 '18 at 10:28














1












1








1





$begingroup$

That's just because the square roots of $24+10i$ are $pm(5+i)$.






share|cite|improve this answer









$endgroup$



That's just because the square roots of $24+10i$ are $pm(5+i)$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 16 '18 at 10:25









José Carlos SantosJosé Carlos Santos

160k22127232




160k22127232












  • $begingroup$
    I should have seen that. Thank you!
    $endgroup$
    – Nullspace
    Dec 16 '18 at 10:28


















  • $begingroup$
    I should have seen that. Thank you!
    $endgroup$
    – Nullspace
    Dec 16 '18 at 10:28
















$begingroup$
I should have seen that. Thank you!
$endgroup$
– Nullspace
Dec 16 '18 at 10:28




$begingroup$
I should have seen that. Thank you!
$endgroup$
– Nullspace
Dec 16 '18 at 10:28











1












$begingroup$

Hint: Use the fact: $$ 24+10i = (5+i)^2$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I should have seen that. Thank you!
    $endgroup$
    – Nullspace
    Dec 16 '18 at 10:28
















1












$begingroup$

Hint: Use the fact: $$ 24+10i = (5+i)^2$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I should have seen that. Thank you!
    $endgroup$
    – Nullspace
    Dec 16 '18 at 10:28














1












1








1





$begingroup$

Hint: Use the fact: $$ 24+10i = (5+i)^2$$






share|cite|improve this answer









$endgroup$



Hint: Use the fact: $$ 24+10i = (5+i)^2$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 16 '18 at 10:26









greedoidgreedoid

41.9k1152105




41.9k1152105












  • $begingroup$
    I should have seen that. Thank you!
    $endgroup$
    – Nullspace
    Dec 16 '18 at 10:28


















  • $begingroup$
    I should have seen that. Thank you!
    $endgroup$
    – Nullspace
    Dec 16 '18 at 10:28
















$begingroup$
I should have seen that. Thank you!
$endgroup$
– Nullspace
Dec 16 '18 at 10:28




$begingroup$
I should have seen that. Thank you!
$endgroup$
– Nullspace
Dec 16 '18 at 10:28











1












$begingroup$

Hint: Expanding $$(5+i)^2=25+10i-1$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I should have seen that. Thank you!
    $endgroup$
    – Nullspace
    Dec 16 '18 at 10:28
















1












$begingroup$

Hint: Expanding $$(5+i)^2=25+10i-1$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I should have seen that. Thank you!
    $endgroup$
    – Nullspace
    Dec 16 '18 at 10:28














1












1








1





$begingroup$

Hint: Expanding $$(5+i)^2=25+10i-1$$






share|cite|improve this answer









$endgroup$



Hint: Expanding $$(5+i)^2=25+10i-1$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 16 '18 at 10:26









Dr. Sonnhard GraubnerDr. Sonnhard Graubner

75.2k42865




75.2k42865












  • $begingroup$
    I should have seen that. Thank you!
    $endgroup$
    – Nullspace
    Dec 16 '18 at 10:28


















  • $begingroup$
    I should have seen that. Thank you!
    $endgroup$
    – Nullspace
    Dec 16 '18 at 10:28
















$begingroup$
I should have seen that. Thank you!
$endgroup$
– Nullspace
Dec 16 '18 at 10:28




$begingroup$
I should have seen that. Thank you!
$endgroup$
– Nullspace
Dec 16 '18 at 10:28











1












$begingroup$

See the answer to this question for how to extract complex square roots. Note carefully: when you have an odd prime root of a complex number you can't extract it except with trigonometry, but square roots can be rendered by algebra alone.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Ah I see. I would have do use de moivres formula then right?
    $endgroup$
    – Nullspace
    Dec 16 '18 at 10:29






  • 1




    $begingroup$
    For odd prime roots yes. For square roots no.
    $endgroup$
    – Oscar Lanzi
    Dec 16 '18 at 10:36










  • $begingroup$
    So for example: $z^8=-3+sqrt{3}i$ I could solve without de moivres formula?
    $endgroup$
    – Nullspace
    Dec 16 '18 at 10:44






  • 1




    $begingroup$
    Yup, you just do three square root extractions. The algebra gets complicated but the concept works.
    $endgroup$
    – Oscar Lanzi
    Dec 16 '18 at 10:57


















1












$begingroup$

See the answer to this question for how to extract complex square roots. Note carefully: when you have an odd prime root of a complex number you can't extract it except with trigonometry, but square roots can be rendered by algebra alone.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Ah I see. I would have do use de moivres formula then right?
    $endgroup$
    – Nullspace
    Dec 16 '18 at 10:29






  • 1




    $begingroup$
    For odd prime roots yes. For square roots no.
    $endgroup$
    – Oscar Lanzi
    Dec 16 '18 at 10:36










  • $begingroup$
    So for example: $z^8=-3+sqrt{3}i$ I could solve without de moivres formula?
    $endgroup$
    – Nullspace
    Dec 16 '18 at 10:44






  • 1




    $begingroup$
    Yup, you just do three square root extractions. The algebra gets complicated but the concept works.
    $endgroup$
    – Oscar Lanzi
    Dec 16 '18 at 10:57
















1












1








1





$begingroup$

See the answer to this question for how to extract complex square roots. Note carefully: when you have an odd prime root of a complex number you can't extract it except with trigonometry, but square roots can be rendered by algebra alone.






share|cite|improve this answer









$endgroup$



See the answer to this question for how to extract complex square roots. Note carefully: when you have an odd prime root of a complex number you can't extract it except with trigonometry, but square roots can be rendered by algebra alone.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 16 '18 at 10:28









Oscar LanziOscar Lanzi

12.7k12136




12.7k12136












  • $begingroup$
    Ah I see. I would have do use de moivres formula then right?
    $endgroup$
    – Nullspace
    Dec 16 '18 at 10:29






  • 1




    $begingroup$
    For odd prime roots yes. For square roots no.
    $endgroup$
    – Oscar Lanzi
    Dec 16 '18 at 10:36










  • $begingroup$
    So for example: $z^8=-3+sqrt{3}i$ I could solve without de moivres formula?
    $endgroup$
    – Nullspace
    Dec 16 '18 at 10:44






  • 1




    $begingroup$
    Yup, you just do three square root extractions. The algebra gets complicated but the concept works.
    $endgroup$
    – Oscar Lanzi
    Dec 16 '18 at 10:57




















  • $begingroup$
    Ah I see. I would have do use de moivres formula then right?
    $endgroup$
    – Nullspace
    Dec 16 '18 at 10:29






  • 1




    $begingroup$
    For odd prime roots yes. For square roots no.
    $endgroup$
    – Oscar Lanzi
    Dec 16 '18 at 10:36










  • $begingroup$
    So for example: $z^8=-3+sqrt{3}i$ I could solve without de moivres formula?
    $endgroup$
    – Nullspace
    Dec 16 '18 at 10:44






  • 1




    $begingroup$
    Yup, you just do three square root extractions. The algebra gets complicated but the concept works.
    $endgroup$
    – Oscar Lanzi
    Dec 16 '18 at 10:57


















$begingroup$
Ah I see. I would have do use de moivres formula then right?
$endgroup$
– Nullspace
Dec 16 '18 at 10:29




$begingroup$
Ah I see. I would have do use de moivres formula then right?
$endgroup$
– Nullspace
Dec 16 '18 at 10:29




1




1




$begingroup$
For odd prime roots yes. For square roots no.
$endgroup$
– Oscar Lanzi
Dec 16 '18 at 10:36




$begingroup$
For odd prime roots yes. For square roots no.
$endgroup$
– Oscar Lanzi
Dec 16 '18 at 10:36












$begingroup$
So for example: $z^8=-3+sqrt{3}i$ I could solve without de moivres formula?
$endgroup$
– Nullspace
Dec 16 '18 at 10:44




$begingroup$
So for example: $z^8=-3+sqrt{3}i$ I could solve without de moivres formula?
$endgroup$
– Nullspace
Dec 16 '18 at 10:44




1




1




$begingroup$
Yup, you just do three square root extractions. The algebra gets complicated but the concept works.
$endgroup$
– Oscar Lanzi
Dec 16 '18 at 10:57






$begingroup$
Yup, you just do three square root extractions. The algebra gets complicated but the concept works.
$endgroup$
– Oscar Lanzi
Dec 16 '18 at 10:57




















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