MCQ
Convert -1 + i into polar form:
- A$\sqrt{2},\frac{5\pi}{4}$
- B$\sqrt{2},\frac{3\pi}{4}$
- C$-\sqrt{2},\frac{\pi}{4}$
- D$\sqrt{2},\frac{\pi}{4}$
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 2nd quadrant since sin is positive and cos is negative.
$\tan\theta = -1\Rightarrow \tan\theta =\frac{-\tan\pi}{4}$
$\Rightarrow\tan\theta=\tan(\frac{\pi-\pi}{4})=\frac{\tan3\pi}{4}$
$\Rightarrow\theta=\frac{3\pi}{4}$
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