1. **State the problem:** Given the equation $$(1a + 2i + 152) \left( \frac{-11 + 21}{5} \right) = (-2 - 4i) \left( \frac{(-6 + 2i)^n}{5} - 1 \right)$$ Solve for $n$. 2. **Simplify constants:** Calculate the fraction $$\frac{-11 + 21}{5} = \frac{10}{5} = 2$$ So the equation becomes: $$(1a + 2i + 152) \times 2 = (-2 - 4i) \left( \frac{(-6 + 2i)^n}{5} - 1 \right)$$ 3. **Rewrite the left side:** $$2(1a + 2i + 152) = 2a + 4i + 304$$ 4. **Divide both sides by $-2 - 4i$ to isolate the right term:** $$\frac{2a + 4i + 304}{-2 - 4i} = \frac{(-6 + 2i)^n}{5} - 1$$ 5. **Add 1 to both sides:** $$\frac{2a + 4i + 304}{-2 - 4i} + 1 = \frac{(-6 + 2i)^n}{5}$$ 6. **Multiply both sides by 5:** $$5\left( \frac{2a + 4i + 304}{-2 - 4i} + 1 \right) = (-6 + 2i)^n$$ 7. **Simplify the left side:** Calculate the complex division and addition (requires knowing $a$) or treat $a$ as a constant. 8. **Express the right side in polar form:** Find magnitude and argument of $-6 + 2i$: $$| -6 + 2i | = \sqrt{(-6)^2 + 2^2} = \sqrt{36 + 4} = \sqrt{40} = 2\sqrt{10}$$ $$\theta = \arctan\left( \frac{2}{-6} \right) = \arctan(-\frac{1}{3}) = -0.32175 \text{ rad}$$ Since real part is negative and imaginary part is positive, the angle is in second quadrant: $$\theta = \pi - 0.32175 = 2.81984 \text{ rad}$$ 9. **Express $(-6 + 2i)^n$:** $$(-6 + 2i)^n = (2\sqrt{10})^n \left( \cos(n \times 2.81984) + i \sin(n \times 2.81984) \right)$$ 10. **Set this equal to the left side computed in step 6 and solve for $n$: (if desired, solve using logarithms on magnitude and arguments)**