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Here are the solutions to the problems.

: Simplify (1−i)1364\frac{(1-i)^{13}}{64}

Step 1: Rewrite the numerator.

(1−i)1364=(1−i)12(1−i)64\frac{(1-i)^{13}}{64} = \frac{(1-i)^{12}(1-i)}{64}

Step 2: Express (1−i)12(1-i)^{12} as a power of (1−i)2(1-i)^2.

((1−i)2)6(1−i)64\frac{((1-i)^2)^6(1-i)}{64}

Step 3: Calculate (1−i)2(1-i)^2.

(1−i)2=12−2(1)(i)+i2=1−2i−1=−2i(1-i)^2 = 1^2 - 2(1)(i) + i^2 = 1 - 2i - 1 = -2i

Step 4: Substitute this back into the expression.

(−2i)6(1−i)64\frac{(-2i)^6(1-i)}{64}

Step 5: Calculate (−2i)6(-2i)^6.

(−2i)6=(−2)6i6=64i6(-2i)^6 = (-2)^6 i^6 = 64 i^6

Step 6: Simplify i6i^6.

i6=i4⋅i2=1⋅(−1)=−1i^6 = i^4 \cdot i^2 = 1 \cdot (-1) = -1

Step 7: Substitute i6=−1i^6 = -1 back into the expression.

64(−1)(1−i)64=−64(1−i)64\frac{64(-1)(1-i)}{64} = \frac{-64(1-i)}{64}

Step 8: Cancel out the 64 in the numerator and denominator.

−(1−i)-(1-i)

Step 9: Distribute the negative sign.

−1+i-1+i

The simplified form is −1+i\boxed{-1+i}.

: Verify (C1C2)‾=C1‾C2‾\overline{\left(\frac{C_1}{C_2}\right)} = \frac{\overline{C_1}}{\overline{C_2}} given C1=7−4iC_1 = 7-4i and C2=2−3iC_2 = 2-3i.

Part a· 5 steps

Calculate the Left Hand Side (LHS)

LHS: (C1C2)‾\overline{\left(\frac{C_1}{C_2}\right)}

  1. Calculate C1C2\frac{C_1}{C_2}.

    C1C2=7−4i2−3i\frac{C_1}{C_2} = \frac{7-4i}{2-3i}

  2. Multiply the numerator and denominator by the conjugate of the denominator, which is 2+3i2+3i.

    7−4i2−3i×2+3i2+3i=(7−4i)(2+3i)(2−3i)(2+3i)\frac{7-4i}{2-3i} \times \frac{2+3i}{2+3i} = \frac{(7-4i)(2+3i)}{(2-3i)(2+3i)}

  3. Expand the numerator and denominator.

    Numerator: (7−4i)(2+3i)=7(2)+7(3i)−4i(2)−4i(3i)=14+21i−8i−12i2=14+13i−12(−1)=14+13i+12=26+13i(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)=22−(3i)2=4−9i2=4−9(−1)=4+9=13(2-3i)(2+3i) = 2^2 - (3i)^2 = 4 - 9i^2 = 4 - 9(-1) = 4 + 9 = 13.

  4. Simplify the fraction.

    26+13i13=2613+13i13=2+i\frac{26+13i}{13} = \frac{26}{13} + \frac{13i}{13} = 2+i

  5. Find the conjugate of the result.

    (C1C2)‾=(2+i)‾=2−i\overline{\left(\frac{C_1}{C_2}\right)} = \overline{(2+i)} = 2-i

    So, LHS = 2−i2-i.

Part b· 5 steps

Calculate the Right Hand Side (RHS)

RHS: C1‾C2‾\frac{\overline{C_1}}{\overline{C_2}}

  1. Find the conjugates of C1C_1 and C2C_2.

    C1‾=7−4i‾=7+4i\overline{C_1} = \overline{7-4i} = 7+4i C2‾=2−3i‾=2+3i\overline{C_2} = \overline{2-3i} = 2+3i

  2. Calculate C1‾C2‾\frac{\overline{C_1}}{\overline{C_2}}.

    C1‾C2‾=7+4i2+3i\frac{\overline{C_1}}{\overline{C_2}} = \frac{7+4i}{2+3i}

  3. Multiply the numerator and denominator by the conjugate of the denominator, which is 2−3i2-3i.

    7+4i2+3i×2−3i2−3i=(7+4i)(2−3i)(2+3i)(2−3i)\frac{7+4i}{2+3i} \times \frac{2-3i}{2-3i} = \frac{(7+4i)(2-3i)}{(2+3i)(2-3i)}

  4. Expand the numerator and denominator.

    Numerator: (7+4i)(2−3i)=7(2)−7(3i)+4i(2)−4i(3i)=14−21i+8i−12i2=14−13i−12(−1)=14−13i+12=26−13i(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)=22−(3i)2=4−9i2=4−9(−1)=4+9=13(2+3i)(2-3i) = 2^2 - (3i)^2 = 4 - 9i^2 = 4 - 9(-1) = 4 + 9 = 13.

  5. Simplify the fraction.

    26−13i13=2613−13i13=2−i\frac{26-13i}{13} = \frac{26}{13} - \frac{13i}{13} = 2-i

    So, RHS = 2−i2-i.

    Conclusion:
    Since LHS = 2−i2-i and RHS = 2−i2-i, we have verified that (C1C2)‾=C1‾C2‾\overline{\left(\frac{C_1}{C_2}\right)} = \frac{\overline{C_1}}{\overline{C_2}}.

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