1. **State the problem:** We want to express the roots of $\left(4-3i\right)^{-\frac{2}{3}}$ in polar form. 2. **Convert the base to polar form:** The complex number $4-3i$ has modulus $r=\sqrt{4^2+(-3)^2}=\sqrt{16+9}=\sqrt{25}=5$. 3. Find the argument $\theta$ using $\tan \theta = \frac{-3}{4}$, so $\theta = \arctan\left(-\frac{3}{4}\right)$, which is in the fourth quadrant, approximately $-0.6435$ radians. Thus, $4-3i = 5 \left(\cos(-0.6435) + i \sin(-0.6435)\right)$. 4. **Apply the exponent $-{2}/{3}$:** Using De Moivre's theorem, $$\left(r(\cos \theta + i \sin \theta)\right)^n = r^n \left(\cos(n\theta) + i \sin(n\theta)\right)$$ Here, $n = -\frac{2}{3}$, so $$\left(4-3i\right)^{-\frac{2}{3}} = 5^{-\frac{2}{3}} \left(\cos\left(-\frac{2}{3} \times -0.6435\right) + i \sin\left(-\frac{2}{3} \times -0.6435\right)\right)$$ 5. Calculate the modulus: $$5^{-\frac{2}{3}} = \frac{1}{5^{2/3}} = \frac{1}{\left(5^{1/3}\right)^2} = \frac{1}{\sqrt[3]{5}^2}$$ 6. Calculate the new argument: $$-\frac{2}{3} \times -0.6435 = 0.4290\, \text{radians}$$ 7. **Roots due to multi-valued nature:** For fractional powers, the argument shifts by multiples of $2\pi$ for roots $k=0,1,2$, so $$\theta_k = 0.4290 + \frac{2\pi k}{3}, \quad k=0,1,2$$ 8. **Final polar forms of the roots:** $$5^{-\frac{2}{3}} \left(\cos \theta_k + i \sin \theta_k\right), \quad k=0,1,2$$ Explicitly, - For $k=0$: $\theta_0 = 0.4290$ - For $k=1$: $\theta_1 = 0.4290 + \frac{2\pi}{3} \approx 2.5316$ - For $k=2$: $\theta_2 = 0.4290 + \frac{4\pi}{3} \approx 4.6341$ So the three roots in polar form are: $$\frac{1}{\sqrt[3]{5}^2} \left( \cos(\theta_k) + i \sin(\theta_k) \right), \quad k=0,1,2$$