Question

The polar form of (i²⁵)³ is

Options

• \[\cos\frac{\pi}{2} + i \sin\frac{\pi}{2}\]

• cos π + i sin π

•  cos π − i sin π

• \[\cos\frac{\pi}{2} - i \sin\frac{\pi}{2}\]

Answer 1

\[\cos\frac{\pi}{2} - i \sin\frac{\pi}{2}\]

(i²⁵)³ = (i)⁷⁵

= (i)^{4 \[\times\] 18+ 3}

   = (i)^{3
    =\[-\] i                (\[\because\]}  i⁴=1)

\[\text { Let } z = 0 - i \]

\[\text { Since, the point (0, - 1) lies on the negative direction of imaginary axis }. \]

\[\text { Therefore,} \arg (z) = \frac{- \pi}{2}\]

Modulus, r =\[\left| z \right| = \left| 1 \right| = 1\]

\[\therefore\] Polar form = r (cos \[\theta\] + i sin \[\theta\])

= cos \[\left( \frac{- \pi}{2} \right)\] +i sin \[\left( \frac{- \pi}{2} \right)\]

= cos \[\frac{\pi}{2}\] \[-\] i sin \[\frac{\pi}{2}\]