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Mathematics Question on Complex Numbers and Quadratic Equations

What is the polar form of the complex number (i25)3\left(i^{25}\right)^{3} ?

A

cosπ3−isinπ3cos \frac{\pi}{3}-isin \frac{\pi}{3}

B

(cos(−π2)+i sin(−π2))\left(cos\left(-\frac{\pi}{2}\right)+i\, sin\left(-\frac{\pi}{2}\right)\right)

C

cosπ4+i sinπ4cos \frac{\pi}{4}+ i\, sin \frac{\pi}{4}

D

(cosπ2+i sinπ2)\left(cos\frac{\pi}{2}+i\, sin \frac{\pi}{2}\right)

Answer

(cos(−π2)+i sin(−π2))\left(cos\left(-\frac{\pi}{2}\right)+i\, sin\left(-\frac{\pi}{2}\right)\right)

Explanation

Solution

Let z=(i25)3=(i)75z =\left(i^{25}\right)^{3} =\left(i\right)^{75} =i4×18+3=(i4)18(i)3=i^{4\times18+3}=\left(i^{4}\right)^{18}\left(i\right)^{3} =i3=−i=0−i=i^{3}=-i=0-i Polar form of z=r(cos θ+isin θ)z = r \left(cos\,\theta+isin\,\theta\right) =1\left\\{cos\left(-\frac{\pi}{2}\right)+i\, sin \left(-\frac{\pi}{2}\right)\right\\} =cosπ2−i sinπ2=cos \frac{\pi}{2}-i \, sin \frac{\pi}{2}