Verified Solution • 2025–26 Session moderate

QJEEM24Ma85a8: JEE Main 2024 Mathematics - Class 11 JEEM24Ma85a8 Solution: Let and respectively be the modulus and amplitude of the…

Let r\mathrm{r} and θ\theta respectively be the modulus and amplitude of the complex number z=2−i(2tan⁡5π8)z=2-i\left(2 \tan \frac{5 \pi}{8}\right) , then (r,θ)(\mathrm{r}, \theta) is equal to

Multiple Choice Options

  • A (2sec⁡11π8,11π8)\left(2 \sec \frac{11 \pi}{8}, \frac{11 \pi}{8}\right)
  • B (2sec⁡3π8,3π8)\left(2 \sec \frac{3 \pi}{8}, \frac{3 \pi}{8}\right)
  • C (2sec⁡5π8,3π8)\left(2 \sec \frac{5 \pi}{8}, \frac{3 \pi}{8}\right)
  • D (2sec⁡3π8,5π8)\left(2 \sec \frac{3 \pi}{8}, \frac{5 \pi}{8}\right)

Step-by-Step Verified Solution

amp;z=2−i(2tan⁡5π8)=x+iy( let )amp;r=x2+y2 & θ=tan⁡−1yxamp;r=(2)2+(2tan⁡5π8)2amp;=∣2sec⁡5π8∣=∣2sec⁡(π−3π8)∣amp;=2sec⁡3π8amp;& θ=tan⁡−1(−2tan⁡5π82)amp;=tan⁡−1(tan⁡2(π−5π8))amp;=3π8\begin{aligned}& z=2-i\left(2 \tan \frac{5 \pi}{8}\right)=x+i y(\text { let }) \\& r=\sqrt{x^2+y^2} ~\& ~\theta=\tan ^{-1} \frac{y}{x} \\& r=\sqrt{(2)^2+\left(2 \tan \frac{5 \pi}{8}\right)^2} \\& =\left|2 \sec \frac{5 \pi}{8}\right|=\left|2 \sec \left(\pi-\frac{3 \pi}{8}\right)\right| \\& =2 \sec \frac{3 \pi}{8} \\& \& ~\theta = {\tan ^{ - 1}}\left( {{{ - 2\tan {{5\pi } \over 8}} \over 2}} \right) \\& =\tan ^{-1}\left(\tan ^2\left(\pi-\frac{5 \pi}{8}\right)\right) \\& =\frac{3 \pi}{8}\end{aligned}

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