- Home
- Standard 11
- Mathematics
4-1.Complex numbers
easy
The argument of the complex number $\frac{{13 - 5i}}{{4 - 9i}}$is
A
$\frac{\pi }{3}$
B
$\frac{\pi }{4}$
C
$\frac{\pi }{5}$
D
$\frac{\pi }{6}$
Solution
(b)$arg\left( {\frac{{13 – 5i}}{{4 – 9i}}} \right) = arg(13 – 5i) – arg(4 – 9i)$
$ = – {\tan ^{ – 1}}\left( {\frac{5}{{13}}} \right) + {\tan ^{ – 1}}\frac{9}{4} = \frac{\pi }{4}$
Standard 11
Mathematics
Similar Questions
Let $z_k=\cos \left(\frac{2 k \pi}{10}\right)+ i \sin \left(\frac{2 k \pi}{10}\right) ; k =1,2, \ldots 9$.
| List $I$ | List $II$ |
| $P.$ For each $z_k$ there exists a $z_j$ such that $z_k \cdot z_j=1$ | $1.$ True |
| $Q.$ There exists a $k \in\{1,2, \ldots ., 9\}$ such that $z_{1 .} . z=z_k$ has no solution $z$ in the set of complex numbers. | $2.$ False |
| $R.$ $\frac{\left|1-z_1\right|\left|1-z_2\right| \ldots . .\left|1-z_9\right|}{10}$ equals | $3.$ $1$ |
| $S.$ $1-\sum_{k=1}^9 \cos \left(\frac{2 k \pi}{10}\right)$ equals | $4.$ $2$ |
Codes: $ \quad P \quad Q \quad R \quad S$
