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

4. $\cos\frac{\pi}{2}-\text{i}\sin\frac{\pi}{2}$

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 $=\text{r}(\cos\theta+\text{i}\sin\theta)$

$=\cos\Big(\frac{-\pi}{2}\Big)+\text{i}\sin\Big(\frac{-\pi}{2}\Big)$

$=\cos\frac{\pi}{2}-\text{i}\sin\frac{\pi}{2}$