# Write (I25)3 in Polar Form. - Mathematics

Write (i25)3 in polar form.

#### Solution

$\left( i^{25} \right)^3 = i^{75}$

$= i^{4 \times 18 + 3}$

$= \left( i^4 \right)^{18} . i^3$

$= i^3 [ \because i^4 = 1]$

$= - i [ \because i^3 = - i]$

Let $z = 0 - i$

Then,

$\left| z \right| = \sqrt{0^2 + \left( - 1 \right)^2} = 1$.

Let θ be the argument of z and α be the acute angle given by

$\tan\alpha = \frac{\left| Im\left( z \right) \right|}{\left| Re\left( z \right) \right|}$
Then,

$\tan\alpha = \frac{1}{0} = \infty$

$\Rightarrow \alpha = \frac{\pi}{2}$

Clearly, z lies in fourth quadrant. So, arg(z) =

$- \alpha = - \frac{\pi}{2}$.
∴ the polar form of is  $\left| z \right|\left( \cos\theta + i\sin\theta \right) = \cos\left( - \frac{\pi}{2} \right) + i\sin\left( - \frac{\pi}{2} \right)$.
Thus, the polar form of  (i25)is $\cos\left( \frac{\pi}{2} \right) - i\sin\left( \frac{\pi}{2} \right)$.
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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 13 Complex Numbers
Exercise 13.4 | Q 2 | Page 57