1. **State the problem:** Evaluate the expression $$i \left( \frac{1+3i}{1-2i} \right)^n$$ where $i$ is the imaginary unit and $n$ is an integer. 2. **Simplify the fraction inside the parentheses:** To simplify $$\frac{1+3i}{1-2i}$$, multiply numerator and denominator by the conjugate of the denominator $$1+2i$$: $$\frac{1+3i}{1-2i} \times \frac{1+2i}{1+2i} = \frac{(1+3i)(1+2i)}{(1-2i)(1+2i)}$$ 3. **Calculate numerator:** $$(1+3i)(1+2i) = 1 \times 1 + 1 \times 2i + 3i \times 1 + 3i \times 2i = 1 + 2i + 3i + 6i^2$$ Since $i^2 = -1$, this becomes: $$1 + 5i + 6(-1) = 1 + 5i - 6 = -5 + 5i$$ 4. **Calculate denominator:** $$(1-2i)(1+2i) = 1^2 - (2i)^2 = 1 - 4i^2 = 1 - 4(-1) = 1 + 4 = 5$$ 5. **Simplify the fraction:** $$\frac{-5 + 5i}{5} = \frac{\cancel{5}(-1 + i)}{\cancel{5}} = -1 + i$$ 6. **Rewrite the original expression:** $$i \left( -1 + i \right)^n$$ 7. **Express $-1 + i$ in polar form:** Calculate modulus: $$r = \sqrt{(-1)^2 + 1^2} = \sqrt{1 + 1} = \sqrt{2}$$ Calculate argument (angle): $$\theta = \arctan\left( \frac{1}{-1} \right) = \arctan(-1)$$ Since the point $(-1,1)$ is in the second quadrant, angle is: $$\theta = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$$ So, $$-1 + i = \sqrt{2} \left( \cos \frac{3\pi}{4} + i \sin \frac{3\pi}{4} \right)$$ 8. **Apply De Moivre's theorem:** $$\left( -1 + i \right)^n = \left( \sqrt{2} \right)^n \left( \cos \frac{3n\pi}{4} + i \sin \frac{3n\pi}{4} \right) = 2^{\frac{n}{2}} \left( \cos \frac{3n\pi}{4} + i \sin \frac{3n\pi}{4} \right)$$ 9. **Multiply by $i$:** Recall that multiplying by $i$ rotates by $\frac{\pi}{2}$ radians: $$i \left( -1 + i \right)^n = 2^{\frac{n}{2}} i \left( \cos \frac{3n\pi}{4} + i \sin \frac{3n\pi}{4} \right) = 2^{\frac{n}{2}} \left( i \cos \frac{3n\pi}{4} + i^2 \sin \frac{3n\pi}{4} \right)$$ Since $i^2 = -1$: $$= 2^{\frac{n}{2}} \left( i \cos \frac{3n\pi}{4} - \sin \frac{3n\pi}{4} \right) = 2^{\frac{n}{2}} \left( - \sin \frac{3n\pi}{4} + i \cos \frac{3n\pi}{4} \right)$$ **Final answer:** $$i \left( \frac{1+3i}{1-2i} \right)^n = 2^{\frac{n}{2}} \left( - \sin \frac{3n\pi}{4} + i \cos \frac{3n\pi}{4} \right)$$