1. **State the problem:** Simplify the expression $$\frac{2(2-3i)}{1+5i}$$ and then evaluate $$\left(\frac{2(2-3i)}{1+5i}\right)^8$$. 2. **Simplify the fraction:** Multiply numerator and denominator by the conjugate of the denominator to remove the imaginary part from the denominator. The conjugate of $$1+5i$$ is $$1-5i$$. So, $$\frac{2(2-3i)}{1+5i} \times \frac{1-5i}{1-5i} = \frac{2(2-3i)(1-5i)}{(1+5i)(1-5i)}$$ 3. **Calculate the denominator:** $$ (1+5i)(1-5i) = 1^2 - (5i)^2 = 1 - (-25) = 1 + 25 = 26 $$ 4. **Calculate the numerator:** First expand $$2(2-3i) = 4 - 6i$$. Now multiply by $$1-5i$$: $$ (4 - 6i)(1 - 5i) = 4(1) + 4(-5i) - 6i(1) - 6i(-5i) = 4 - 20i - 6i + 30i^2 $$ Recall $$i^2 = -1$$, so: $$ 4 - 20i - 6i + 30(-1) = 4 - 26i - 30 = (4 - 30) - 26i = -26 - 26i $$ 5. **Put numerator and denominator together:** $$ \frac{-26 - 26i}{26} $$ 6. **Simplify by dividing numerator and denominator by 26:** $$ \frac{\cancel{26}(-1 - i)}{\cancel{26}} = -1 - i $$ 7. **Evaluate the power:** We want to find: $$ (-1 - i)^8 $$ 8. **Convert to polar form:** Magnitude: $$ r = \sqrt{(-1)^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2} $$ Argument (angle): $$ \theta = \arctan\left(\frac{-1}{-1}\right) = \arctan(1) = \frac{\pi}{4} $$ Since both real and imaginary parts are negative, the point is in the third quadrant, so: $$ \theta = \pi + \frac{\pi}{4} = \frac{5\pi}{4} $$ 9. **Use De Moivre's theorem:** $$ (r(\cos \theta + i \sin \theta))^8 = r^8 (\cos 8\theta + i \sin 8\theta) $$ Calculate: $$ r^8 = (\sqrt{2})^8 = (2^{1/2})^8 = 2^{4} = 16 $$ $$ 8\theta = 8 \times \frac{5\pi}{4} = 10\pi $$ 10. **Evaluate trigonometric functions:** $$ \cos 10\pi = 1 $$ $$ \sin 10\pi = 0 $$ 11. **Final result:** $$ (-1 - i)^8 = 16 (1 + 0i) = 16 $$ **Answer:** $$ \boxed{16} $$