Complex Numbers and Quadratic Equations — MATHS STD 11 Science — Question

2. $\sqrt{2},\frac{3\pi}{4}$

Solution:

$\text{r}=\sqrt{\text{x}^2+\text{y}^2}=\sqrt{(-1)^2+\text{1}^2}=\sqrt{1+1}=\sqrt{2}$

$\text{r}\cos\theta = -1$ and $\text{r}\sin\theta = 1$ so, $\theta$ is in 2^nd quadrant since sin is positive and cos is negative.

$\tan\theta = -1\Rightarrow \tan\theta =\frac{-\tan\pi}{4}$

$\Rightarrow\tan\theta=\tan(\frac{\pi-\pi}{4})=\frac{\tan3\pi}{4}$

$\Rightarrow\theta=\frac{3\pi}{4}$