Let $\mathrm{r}$ and $\theta$ respectively be the modulus and amplitude of the complex number $z=2-i\left(2 \tan \frac{5 \pi}{8}\right)$, then $(\mathrm{r}, \theta)$ is equal to
Solution
<p>$$\begin{aligned}
& z=2-i\left(2 \tan \frac{5 \pi}{8}\right)=x+i y(\text { let }) \\
& r=\sqrt{x^2+y^2} ~\& ~\theta=\tan ^{-1} \frac{y}{x} \\
& r=\sqrt{(2)^2+\left(2 \tan \frac{5 \pi}{8}\right)^2} \\
& =\left|2 \sec \frac{5 \pi}{8}\right|=\left|2 \sec \left(\pi-\frac{3 \pi}{8}\right)\right| \\
& =2 \sec \frac{3 \pi}{8} \\
& \& ~\theta = {\tan ^{ - 1}}\left( {{{ - 2\tan {{5\pi } \over 8}} \over 2}} \right) \\
& =\tan ^{-1}\left(\tan ^2\left(\pi-\frac{5 \pi}{8}\right)\right) \\
& =\frac{3 \pi}{8}
\end{aligned}$$</p>
About this question
Subject: Mathematics · Chapter: Complex Numbers and Quadratic Equations · Topic: Complex Numbers and Argand Plane
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