Let z _ 1 and z _ 2 be any two non-zero complex numbers such that...

Let $arg\left(\dfrac{3{z}_{1}}{2{z}_{2}}\right) = \theta,$ and we know that, if the $arg\left(z\right) = \alpha,$ then $arg$\left(\dfrac{1}{z}\right) = - \alpha .$<br /><br />$∴ arg\left(\dfrac{2{z}_{2}}{3{z}_{1}}\right) = - \theta$</p><p>Also, we know that a complex number$w$can be expressed as$w = \left|w\right|\left(cos\alpha + isin\alpha\right),$where$\alpha$is the argument of the complex number.</p><p>$∴ z = \dfrac{3}{2}\left|\dfrac{{z}_{1}}{{z}_{2}}\right|\left(cos\theta + isin\theta\right) + \dfrac{2}{3}\left|\dfrac{{z}_{2}}{{z}_{1}}\right|\left(cos\theta - isin\theta\right)$</p><p>Given,$3\left|{z}_{1}\right| = 4\left|{z}_{2}\right|, ⇒ \left|\dfrac{{z}_{1}}{{z}_{2}}\right| = \dfrac{4}{3},$</p><p>$⇒ z = \dfrac{3}{2} \times \dfrac{4}{3}\left(cos\theta + isin\theta\right) + \dfrac{2}{3} \times \dfrac{3}{4}\left(cos\theta - isin\theta\right)$</p><p>$⇒ z = 2\left(cos\theta + isin\theta\right) + \dfrac{1}{2}\left(cos\theta - isin\theta\right)$</p><p>$⇒ z = \dfrac{5}{2}cos\theta + \left(\dfrac{3}{2}sin\theta\right)i$</p><p>$∴ \left|z\right| = \sqrt{\dfrac{25}{4}\cos^{2}\theta + \dfrac{9}{4}\sin^{2}\theta}$</p><p>Now, using$\sin^{2}\theta + \cos^{2}\theta = 1,$we get</p><p>\left|z\right| = \sqrt{\dfrac{25}{4}\cos^{2}\theta + \dfrac{9}{4}\left(1 - \cos^{2}\theta\right)}$

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

And, we know that the maximum value of $cos\theta$ is $1 .$
∴ {\left|z\right|}_\max = \sqrt{4 + \dfrac{9}{4}} = \dfrac{5}{2} .