Find the Solution to $a^2-b^2$ with $a+bi$ a Root of $z(z+i)(z+3i)=2002i$

[Albert1]

$a+bi$ is a root of :$z(z+i)(z+3i)=2002i$
here :$a,b>0$ and $i=\sqrt {-1}$
please find :$a^2-b^2=?$

 

[Opalg]

$a+bi$ is a root of :$z(z+i)(z+3i)=2002i$
here :$a,b>0$ and $i=\sqrt {-1}$
please find :$a^2-b^2=?$

[sp]$$0 = z(z+i)(z+3i) - 2002i = z^3 + 4iz^2 - 3z - 2002i = (z+14i)(z^2 - 10iz - 143).$$ One solution is $z = a+bi = -14i.$ But in that case, $b = -14 <0$ and we want a solution with $b>0$. So we must look for one of the solutions of $z^2 - 10iz - 143 = 0.$ These are $z = 5i \pm\sqrt{118}$, and the required solution is $z = \sqrt{118} + 5i$. So $a = \sqrt{118}$, $b = 5$, and $a^2-b^2 = 118 - 25 = 93.$[/sp]

 

[Albert1]

[sp]$$0 = z(z+i)(z+3i) - 2002i = z^3 + 4iz^2 - 3z - 2002i = (z+14i)(z^2 - 10iz - 143).$$ One solution is $z = a+bi = -14i.$ But in that case, $b = -14 <0$ and we want a solution with $b>0$. So we must look for one of the solutions of $z^2 - 10iz - 143 = 0.$ These are $z = 5i \pm\sqrt{118}$, and the required solution is $z = \sqrt{118} + 5i$. So $a = \sqrt{118}$, $b = 5$, and $a^2-b^2 = 118 - 25 = 93.$[/sp]

very good slution !

 

[Nono713]

Spoiler

As an addendum, an efficient way to notice the $z = -14i$ root is to notice that $2002 = 11 \cdot 13 \cdot 14$.

 

[Opalg]

Spoiler

As an addendum, an efficient way to notice the $z = -14i$ root is to notice that $2002 = 11 \cdot 13 \cdot 14$.

[sp]What I actually did was to notice that if you put $z=iw$ then the equation $z(z+i)(z+3i) - 2002i = 0$ becomes $w(w+1)(w+3) + 2002 = 0$, with all the coefficients real. You can then easily find the solution $w = -14$ by factorising $2002 = 11 \cdot 13 \cdot 14$.[/sp]