JEE Main 2019 — Complex Number Question with Solution

Let $arg\left(\frac{3z_1}{2z_2}\right)=\theta ,$ and we know that, if the $arg\left(z\right)=\alpha ,$ then $arg\left(\frac{1}{z}\right)=-\alpha .$

$\therefore arg\left(\frac{2z_2}{3z_1}\right)=-\theta$

Also, we know that a complex number $w$ can be expressed as $w=\left|w\right|\left(\cos \alpha +i\sin \alpha \right),$ where $\alpha$ is the argument of the complex number.

$\therefore z=\frac{3}{2}\left|\frac{z_1}{z_2}\right|\left(\cos \theta +i\sin \theta \right)+\frac{2}{3}\left|\frac{z_2}{z_1}\right|\left(\cos \theta -i\sin \theta \right)$

Given, $3\left|z_1\right|=4\left|z_2\right|, \Rightarrow \left|\frac{z_1}{z_2}\right|=\frac{4}{3},$

$\Rightarrow z=\frac{3}{2}\times \frac{4}{3}\left(\cos \theta +i\sin \theta \right)+\frac{2}{3}\times \frac{3}{4}\left(\cos \theta -i\sin \theta \right)$

$\Rightarrow z=2\left(\cos \theta +i\sin \theta \right)+\frac{1}{2}\left(\cos \theta -i\sin \theta \right)$

$\Rightarrow z=\frac{5}{2}\cos \theta +\left(\frac{3}{2}\sin \theta \right)i$

$\therefore \left|z\right|=\sqrt{\frac{25}{4}{\cos }^2\theta +\frac{9}{4}{\sin }^2\theta }$

Now, using ${\sin }^2\theta +{\cos }^2\theta =1,$ we get

$\left|z\right|=\sqrt{\frac{25}{4}{\cos }^2\theta +\frac{9}{4}\left(1-{\cos }^2\theta \right)}$

$\Rightarrow \left|z\right|=\sqrt{4{\cos }^2\theta +\frac{9}{4}}$

And, we know that the maximum value of $\cos \theta$ is $1.$
$\therefore {\left|z\right|}_{\mathrm{m}\mathrm{a}\mathrm{x}}=\sqrt{4+\frac{9}{4}}=\frac{5}{2}.$