If |z₁| = 2 , |z₂| =3 , |z₃| = 4 and |2z₁ +3z₂ +4z₃| =9 ,then value of |8z₂z₃ +27z?

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4-1.Complex numbers

normal

If $|z_1| = 2 , |z_2| =3 , |z_3| = 4$ and $|2z_1 +3z_2 +4z_3| =9$ ,then value of $|8z_2z_3 +27z_3z_1 +64z_1z_2|$ is equal to:-

$216$

$18$

$64$

None

Solution

$\left| {{z_1}{z_2}{z_3}\left( {\frac{{2{{\left| {{z_1}} \right|}^2}}}{{{z_1}}} + \frac{{3{{\left| {{z_2}} \right|}^2}}}{{{z_2}}} + \frac{{4{{\left| {{z_3}} \right|}^2}}}{{{z_3}}}} \right)} \right|$

$\left|z_{1}\left\|z_{2}\right\| z_{3} \| 2 \bar{z}_{1}+3 \bar{z}_{2}+4 \bar{z}_{3}\right|$

$2 \times 3 \times 4 \times 9=216$

Standard 11

Mathematics

If $|{z_1}|\, = \,|{z_2}|$ and $amp\,{z_1} + amp\,\,{z_2} = 0$, then

normal

Let $z$ satisfy $\left| z \right| = 1$ and $z = 1 – \vec z$.

Statement $1$ : $z$ is a real number

Statement $2$ : Principal argument of $z$ is $\frac{\pi }{3}$

hard

(JEE MAIN-2013)

Let $\mathrm{z}$ be a complex number such that $|\mathrm{z}+2|=1$ and $\operatorname{Im}\left(\frac{z+1}{z+2}\right)=\frac{1}{5}$. Then the value of $|\operatorname{Re}(\overline{z+2})|$ is :

medium

(JEE MAIN-2024)

Argument and modulus of $\frac{{1 + i}}{{1 – i}}$ are respectively

easy

$arg\left( {\frac{{3 + i}}{{2 – i}} + \frac{{3 – i}}{{2 + i}}} \right)$ is equal to