MCQ
The polar form of $(\text{i}^{25})^3$ is:
- A$\cos\frac{\pi}{2}+\text{i}\sin\frac{\pi}{2}$
- B$\cos\pi+\text{i}\sin\pi$
- C$\cos\pi-\text{i}\sin\pi$
- D$\cos\frac{\pi}{2}-\text{i}\sin\frac{\pi}{2}$
Solution:
$(\text{i}^{25})^3=(\text{i})^{75}$
$=(\text{i})^{4\times18+3}$
$=(\text{i})^3$
$=-\text{i} \ (\because\text{i}^4=1)$
Let $\text{z}=0-\text{i}$
Since, the point (0,−1) lies on the negative direction of imaginary axis.Therefore, $\text{arg(z)}=\frac{-\pi}{2}$
Modulus, $\text{r}=|\text{z}|=|1|=1$
$\therefore$ Polar form $=\text{r}(\cos\theta+\text{i}\sin\theta)$
$=\cos\Big(\frac{-\pi}{2}\Big)+\text{i}\sin\Big(\frac{-\pi}{2}\Big)$
$=\cos\frac{\pi}{2}-\text{i}\sin\frac{\pi}{2}$
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The lines 2x - 3y = 5 and 3x - 4y = 7 are two diameters of a circel of area 154sq. units. Then the equation of circle is:
$\text{x}=\frac{\text{n}\pi}{7}+\frac{\pi}{3},\ \text{n}\in\text{Z}$