Complex Numbers
Modulus and Argument
nta_pyq_2024_jan
Grade 11
Question:
Let $r$ and $\theta$ respectively be the modulus and amplitude of the complex number $z=2-i\left(2\tan\frac{5\pi}{8}\right)$, then $(r,\theta)$ is equal to:
$\left(2\sec\frac{3\pi}{8},\,\frac{3\pi}{8}\right)$
$\left(2\sec\frac{3\pi}{8},\,\frac{5\pi}{8}\right)$
$\left(2\sec\frac{5\pi}{8},\,\frac{3\pi}{8}\right)$
$\left(2\sec\frac{11\pi}{8},\,\frac{11\pi}{8}\right)$
Step-by-Step Solution
Key Concept: Compute $r=\sqrt{4+4\tan^2\frac{5\pi}{8}}=2|\sec\frac{5\pi}{8}|=2\sec\frac{3\pi}{8}$ (using $\sec(\pi-\theta)=-\sec\theta$ and taking absolute value). For $\theta$: $\tan^{-1}\left(\frac{-2\tan(5\pi/8)}{2}\right)=\tan^{-1}(\tan(\pi-5\pi/8))=3\pi/8$.
$r=\sqrt{4+4\tan^2\frac{5\pi}{8}}=2\sqrt{\sec^2\frac{5\pi}{8}}=2|\sec\frac{5\pi}{8}|=2\sec(\pi-\frac{5\pi}{8})=2\sec\frac{3\pi}{8}$.
$\theta=\tan^{-1}\left(\frac{-2\tan\frac{5\pi}{8}}{2}\right)=\tan^{-1}\left(\tan(\pi-\frac{5\pi}{8})\right)=\tan^{-1}\left(\tan\frac{3\pi}{8}\right)=\frac{3\pi}{8}$.
$(r,\theta)=\left(2\sec\frac{3\pi}{8},\frac{3\pi}{8}\right)$.
Correct Answer: 1