[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]Albert said:$a+bi$ is a root of :$z(z+i)(z+3i)=2002i$
here :$a,b>0$ and $i=\sqrt {-1}$
please find :$a^2-b^2=?$
very good slution !Opalg said:[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]
[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]Bacterius said:As an addendum, an efficient way to notice the $z = -14i$ root is to notice that $2002 = 11 \cdot 13 \cdot 14$.
The solution to $a^2-b^2$ can be found by solving the given equation using the root $a+bi$. This will result in two values for $a^2-b^2$, since complex roots always come in conjugate pairs.
The easiest way to find the roots of this equation is by using the quadratic formula. You can also use the factoring method if possible.
Yes, since complex roots always come in conjugate pairs, there will be two solutions to $a^2-b^2$ for a given root $a+bi$.
Complex roots will result in two solutions for $a^2-b^2$, which will be complex conjugates of each other. This means that the real parts will be the same, but the imaginary parts will have opposite signs.
Yes, you can use the difference of squares formula to simplify $a^2-b^2$ when dealing with complex roots. This will result in a real number as the solution.