JEE Main 2019 — Mathematics Algebra

Let $arg(\frac{3{z}_{1}}{2{z}_{2}})=\theta ,$arg(2z2​3z1​​)=θ, and we know that, if the $arg(z)=\alpha ,$arg(z)=α, then $arg(\frac{1}{z})=-\alpha .$arg(z1​)=−α.

$$\therefore arg(\frac{2{z}_{2}}{3{z}_{1}})=-\theta$$∴arg(3z1​2z2​​)=−θ

Also, we know that a complex number $w$w can be expressed as $w=|w|(\mathrm{cos}\alpha +i\mathrm{sin}\alpha ),$w=∣w∣(cosα+isinα), where $\alpha$α is the argument of the complex number.

$$\therefore z=\frac{3}{2}|\frac{{z}_{1}}{{z}_{2}}|(\mathrm{cos}\theta +i\mathrm{sin}\theta )+\frac{2}{3}|\frac{{z}_{2}}{{z}_{1}}|(\mathrm{cos}\theta -i\mathrm{sin}\theta )$$∴z=23​∣z2​z1​​∣(cosθ+isinθ)+32​∣z1​z2​​∣(cosθ−isinθ)

Given, $3|{z}_{1}|=4|{z}_{2}|,\Rightarrow |\frac{{z}_{1}}{{z}_{2}}|=\frac{4}{3},$3∣z1​∣=4∣z2​∣,⇒∣z2​z1​​∣=34​,

$$\Rightarrow z=\frac{3}{2}\times \frac{4}{3}(\mathrm{cos}\theta +i\mathrm{sin}\theta )+\frac{2}{3}\times \frac{3}{4}(\mathrm{cos}\theta -i\mathrm{sin}\theta )$$⇒z=23​×34​(cosθ+isinθ)+32​×43​(cosθ−isinθ)

$$\Rightarrow z=2(\mathrm{cos}\theta +i\mathrm{sin}\theta )+\frac{1}{2}(\mathrm{cos}\theta -i\mathrm{sin}\theta )$$⇒z=2(cosθ+isinθ)+21​(cosθ−isinθ)

$$\Rightarrow z=\frac{5}{2}\mathrm{cos}\theta +(\frac{3}{2}\mathrm{sin}\theta )i$$⇒z=25​cosθ+(23​sinθ)i

$$\therefore |z|=\sqrt{\frac{25}{4}{\mathrm{cos}}^{2}\theta +\frac{9}{4}{\mathrm{sin}}^{2}\theta }$$∴∣z∣=425​cos2θ+49​sin2θ​

Now, using ${\mathrm{sin}}^{2}\theta +{\mathrm{cos}}^{2}\theta =1,$sin2θ+cos2θ=1, we get

$$|z|=\sqrt{\frac{25}{4}{\mathrm{cos}}^{2}\theta +\frac{9}{4}(1-{\mathrm{cos}}^{2}\theta )}$$∣z∣=425​cos2θ+49​(1−cos2θ)​

$$\Rightarrow |z|=\sqrt{4{\mathrm{cos}}^{2}\theta +\frac{9}{4}}$$⇒∣z∣=4cos2θ+49​​

And, we know that the maximum value of $\mathrm{cos}\theta$cosθ is $1.$1.

$$\therefore {|z|}_{max}=\sqrt{4+\frac{9}{4}}=\frac{5}{2}.$$∴∣z∣max​=4+49​​=25​.