Question

Simplify 1\(1-i√3)³ in polar form

Answer 1

Answer:

Step-by-step explanation:

We have,

$z=\dfrac{1}{\left(1-i\sqrt{3}\right)^3}$

$\implies\,z=\dfrac{1}{2^3\left(\dfrac{1}{2}-i\dfrac{\sqrt{3}}{2}\right)^3}$

$\implies\,z=\dfrac{1}{8\left\{\cos\left(\dfrac{\pi}{3}\right)-i\,\sin\left(\dfrac{\pi}{3}\right)\right\}^3}$

Since the point $(1,-\sqrt{3})$ lies in 4th quadrant, so,

$\implies\,z=\dfrac{1}{8\left\{\cos\left(-\dfrac{\pi}{3}\right)+i\,\sin\left(-\dfrac{\pi}{3}\right)\right\}^3}$

$\implies\,z=\dfrac{1}{8\left\{\,e^{\displaystyle\,i\,\dfrac{\pi}{3}}\right\}^3}$

$\implies\,z=\dfrac{1}{8\,e^{\displaystyle\,i\,\pi}}$

$\implies\,z=\dfrac{1}{8}\,e^{\displaystyle\,-i\,\pi}$

$\implies\,z=\dfrac{1}{8}\big\{\cos(-\pi)+i\,\sin(-\pi)\big\}$