1. **State the problem:** Express $$\frac{2(2-3i)^i}{1+5i}$$ in polar form and then evaluate $$\left(\frac{2(2-3i)}{1+5i}\right)^8$$. 2. **Convert complex numbers to polar form:** - For $$2-3i$$, calculate modulus $$r = \sqrt{2^2 + (-3)^2} = \sqrt{4 + 9} = \sqrt{13}$$. - Calculate argument $$\theta = \tan^{-1}\left(\frac{-3}{2}\right)$$. 3. **Express $$2-3i$$ in polar form:** $$2-3i = \sqrt{13} \left(\cos \theta + i \sin \theta\right)$$ where $$\theta = \tan^{-1}\left(-\frac{3}{2}\right)$$. 4. **Calculate $$ (2-3i)^i $$ using Euler's formula:** $$ (2-3i)^i = \left(\sqrt{13} e^{i\theta}\right)^i = e^{i \ln \sqrt{13} + i^2 \theta} = e^{i \ln \sqrt{13} - \theta} = e^{-\theta} e^{i \ln \sqrt{13}} $$. 5. **Calculate $$\frac{2(2-3i)^i}{1+5i}$$:** - Express $$1+5i$$ in polar form: modulus $$r = \sqrt{1^2 + 5^2} = \sqrt{26}$$, argument $$\phi = \tan^{-1}(5)$$. - So $$\frac{2(2-3i)^i}{1+5i} = \frac{2 e^{-\theta} e^{i \ln \sqrt{13}}}{\sqrt{26} e^{i \phi}} = \frac{2 e^{-\theta}}{\sqrt{26}} e^{i (\ln \sqrt{13} - \phi)}$$. 6. **Evaluate $$\left(\frac{2(2-3i)}{1+5i}\right)^8$$:** - First express $$\frac{2(2-3i)}{1+5i}$$ in polar form: - Numerator modulus: $$2 \times \sqrt{13} = 2\sqrt{13}$$ - Numerator argument: $$\theta$$ - Denominator modulus: $$\sqrt{26}$$ - Denominator argument: $$\phi$$ - So modulus of fraction: $$\frac{2\sqrt{13}}{\sqrt{26}} = \frac{2\sqrt{13}}{\sqrt{26}}$$. - Simplify modulus: $$\frac{2\sqrt{13}}{\sqrt{26}} = \frac{2\sqrt{13}}{\sqrt{2 \times 13}} = \frac{2\sqrt{13}}{\sqrt{2} \sqrt{13}} = \frac{2}{\sqrt{2}} = \sqrt{2} \times \cancel{\frac{2}{\sqrt{2}}}$$ - Argument of fraction: $$\theta - \phi$$ - Raise to power 8: $$\left(\sqrt{2} e^{i(\theta - \phi)}\right)^8 = (\sqrt{2})^8 e^{i 8 (\theta - \phi)} = (2^{1/2})^8 e^{i 8 (\theta - \phi)} = 2^{4} e^{i 8 (\theta - \phi)} = 16 e^{i 8 (\theta - \phi)}$$. 7. **Final answers:** - $$\frac{2(2-3i)^i}{1+5i} = \frac{2 e^{-\theta}}{\sqrt{26}} e^{i (\ln \sqrt{13} - \phi)}$$ where $$\theta = \tan^{-1}\left(-\frac{3}{2}\right)$$ and $$\phi = \tan^{-1}(5)$$. - $$\left(\frac{2(2-3i)}{1+5i}\right)^8 = 16 e^{i 8 (\theta - \phi)}$$. This completes the problem.