Let z_1,z_2 C be the distinct solutions of the equation z^2+4z-(1+12i)=0.

Question

Let ${z}_{1},{z}_{2}\in C$ be the distinct solutions of the equation ${z}^{2}+4z-(1+12i)=0$.

Then ${|{z}_{1}|}^{2}+{|{z}_{2}|}^{2}$ is equal to :

Accepted Answer

Given the quadratic equation

${z}^{2}+4z-(1+12i)=0$

let its roots be ${z}_{1}$ and ${z}_{2}$.

We need to find

$|{z}_{1}{|}^{2}+|{z}_{2}{|}^{2}$

Using Vieta's formulas for

$a{z}^{2}+bz+c=0,$

we have

${z}_{1}+{z}_{2}=-\frac{b}{a}=-4$

and

${z}_{1}{z}_{2}=\frac{c}{a}=-(1+12i)=-1-12i$

Now solve the equation to identify the roots.

${z}^{2}+4z-(1+12i)=0$

Using the quadratic formula,

$z=\frac{-4\pm \sqrt{16+4(1+12i)}}{2}$

$z=\frac{-4\pm \sqrt{20+48i}}{2}$

Now observe that

$(6+4i{)}^{2}=36+48i-16=20+48i$

So,

$\sqrt{20+48i}=6+4i$

Hence the roots are

${z}_{1}=\frac{-4+(6+4i)}{2}=1+2i$

and

${z}_{2}=\frac{-4-(6+4i)}{2}=-5-2i$

Now compute their moduli squared:

$|{z}_{1}{|}^{2}=|1+2i{|}^{2}={1}^{2}+{2}^{2}=5$

$|{z}_{2}{|}^{2}=|-5-2i{|}^{2}=(-5{)}^{2}+(-2{)}^{2}=25+4=29$

Therefore,

$|{z}_{1}{|}^{2}+|{z}_{2}{|}^{2}=5+29=34$

So the correct answer is

$34$

Hence, Option D is correct.

Solutions · 1