Here are the solutions to the problems. Problem 1: Simplify $\frac{(1-i)^{13}}{64}$ Step 1: Rewrite the numerator. $$ \frac{(1-i)^{13}}{64} = \frac{(1-i)^{12}(1-i)}{64} $$ Step 2: Express $(1-i)^{12}$ as a power of $(1-i)^2$. $$ \frac{((1-i)^2)^6(1-i)}{64} $$ Step 3: Calculate $(1-i)^2$. $$ (1-i)^2 = 1^2 - 2(1)(i) + i^2 = 1 - 2i - 1 = -2i $$ Step 4: Substitute this back into the expression. $$ \frac{(-2i)^6(1-i)}{64} $$ Step 5: Calculate $(-2i)^6$. $$ (-2i)^6 = (-2)^6 i^6 = 64 i^6 $$ Step 6: Simplify $i^6$. $$ i^6 = i^4 \cdot i^2 = 1 \cdot (-1) = -1 $$ Step 7: Substitute $i^6 = -1$ back into the expression. $$ \frac{64(-1)(1-i)}{64} = \frac{-64(1-i)}{64} $$ Step 8: Cancel out the 64 in the numerator and denominator. $$ -(1-i) $$ Step 9: Distribute the negative sign. $$ -1+i $$ The simplified form is $\boxed{-1+i}$. --- Problem 2: Verify $\overline{\left(\frac{C_1}{C_2}\right)} = \frac{\overline{C_1}}{\overline{C_2}}$ given $C_1 = 7-4i$ and $C_2 = 2-3i$. Part A: Calculate the Left Hand Side (LHS) LHS: $\overline{\left(\frac{C_1}{C_2}\right)}$ Step 1: Calculate $\frac{C_1}{C_2}$. $$ \frac{C_1}{C_2} = \frac{7-4i}{2-3i} $$ Step 2: Multiply the numerator and denominator by the conjugate of the denominator, which is $2+3i$. $$ \frac{7-4i}{2-3i} \times \frac{2+3i}{2+3i} = \frac{(7-4i)(2+3i)}{(2-3i)(2+3i)} $$ Step 3: Expand the numerator and denominator. Numerator: $(7-4i)(2+3i) = 7(2) + 7(3i) - 4i(2) - 4i(3i) = 14 + 21i - 8i - 12i^2 = 14 + 13i - 12(-1) = 14 + 13i + 12 = 26 + 13i$. Denominator: $(2-3i)(2+3i) = 2^2 - (3i)^2 = 4 - 9i^2 = 4 - 9(-1) = 4 + 9 = 13$. Step 4: Simplify the fraction. $$ \frac{26+13i}{13} = \frac{26}{13} + \frac{13i}{13} = 2+i $$ Step 5: Find the conjugate of the result. $$ \overline{\left(\frac{C_1}{C_2}\right)} = \overline{(2+i)} = 2-i $$ So, LHS = $2-i$. Part B: Calculate the Right Hand Side (RHS) RHS: $\frac{\overline{C_1}}{\overline{C_2}}$ Step 1: Find the conjugates of $C_1$ and $C_2$. $$ \overline{C_1} = \overline{7-4i} = 7+4i $$ $$ \overline{C_2} = \overline{2-3i} = 2+3i $$ Step 2: Calculate $\frac{\overline{C_1}}{\overline{C_2}}$. $$ \frac{\overline{C_1}}{\overline{C_2}} = \frac{7+4i}{2+3i} $$ Step 3: Multiply the numerator and denominator by the conjugate of the denominator, which is $2-3i$. $$ \frac{7+4i}{2+3i} \times \frac{2-3i}{2-3i} = \frac{(7+4i)(2-3i)}{(2+3i)(2-3i)} $$ Step 4: Expand the numerator and denominator. Numerator: $(7+4i)(2-3i) = 7(2) - 7(3i) + 4i(2) - 4i(3i) = 14 - 21i + 8i - 12i^2 = 14 - 13i - 12(-1) = 14 - 13i + 12 = 26 - 13i$. Denominator: $(2+3i)(2-3i) = 2^2 - (3i)^2 = 4 - 9i^2 = 4 - 9(-1) = 4 + 9 = 13$. Step 5: Simplify the fraction. $$ \frac{26-13i}{13} = \frac{26}{13} - \frac{13i}{13} = 2-i $$ So, RHS = $2-i$. Conclusion: Since LHS = $2-i$ and RHS = $2-i$, we have verified that $\overline{\left(\frac{C_1}{C_2}\right)} = \frac{\overline{C_1}}{\overline{C_2}}$. Send me the next one 📸