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Given Z1 = 2 − 5i and Z2 = 2 + 3i.

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Here's the solution for question 1, part (i) and (ii) verification.

Given Z1=2−5iZ_1 = 2 - 5i and Z2=2+3iZ_2 = 2 + 3i.

Part i· 5 steps

Verify Z1+Z2‾=Z1‾+Z2‾\overline{Z_1 + Z_2} = \overline{Z_1} + \overline{Z_2}

  1. Calculate Z1+Z2Z_1 + Z_2.

    Z1+Z2=(2−5i)+(2+3i)=(2+2)+(−5+3)i=4−2iZ_1 + Z_2 = (2 - 5i) + (2 + 3i) = (2 + 2) + (-5 + 3)i = 4 - 2i

  2. Find the conjugate of Z1+Z2Z_1 + Z_2.

    Z1+Z2‾=4−2i‾=4+2i\overline{Z_1 + Z_2} = \overline{4 - 2i} = 4 + 2i

  3. Find Z1‾\overline{Z_1} and Z2‾\overline{Z_2}.

    Z1‾=2−5i‾=2+5i\overline{Z_1} = \overline{2 - 5i} = 2 + 5i Z2‾=2+3i‾=2−3i\overline{Z_2} = \overline{2 + 3i} = 2 - 3i

  4. Calculate Z1‾+Z2‾\overline{Z_1} + \overline{Z_2}.

    Z1‾+Z2‾=(2+5i)+(2−3i)=(2+2)+(5−3)i=4+2i\overline{Z_1} + \overline{Z_2} = (2 + 5i) + (2 - 3i) = (2 + 2) + (5 - 3)i = 4 + 2i

  5. Compare the results.

    Since Z1+Z2‾=4+2i\overline{Z_1 + Z_2} = 4 + 2i and Z1‾+Z2‾=4+2i\overline{Z_1} + \overline{Z_2} = 4 + 2i,

    Z1+Z2‾=Z1‾+Z2‾\overline{Z_1 + Z_2} = \overline{Z_1} + \overline{Z_2}

    This property is verified.

Part ii· 5 steps

Verify Z1⋅Z2‾=Z1‾⋅Z2‾\overline{Z_1 \cdot Z_2} = \overline{Z_1} \cdot \overline{Z_2}

  1. Calculate Z1⋅Z2Z_1 \cdot Z_2.

    Z1⋅Z2=(2−5i)(2+3i)Z_1 \cdot Z_2 = (2 - 5i)(2 + 3i) Z1⋅Z2=2(2)+2(3i)−5i(2)−5i(3i)Z_1 \cdot Z_2 = 2(2) + 2(3i) - 5i(2) - 5i(3i) Z1⋅Z2=4+6i−10i−15i2Z_1 \cdot Z_2 = 4 + 6i - 10i - 15i^2

    Since i2=−1i^2 = -1:

    Z1⋅Z2=4−4i−15(−1)Z_1 \cdot Z_2 = 4 - 4i - 15(-1) Z1⋅Z2=4−4i+15=19−4iZ_1 \cdot Z_2 = 4 - 4i + 15 = 19 - 4i

  2. Find the conjugate of Z1⋅Z2Z_1 \cdot Z_2.

    Z1⋅Z2‾=19−4i‾=19+4i\overline{Z_1 \cdot Z_2} = \overline{19 - 4i} = 19 + 4i

  3. Use Z1‾\overline{Z_1} and Z2‾\overline{Z_2} from the previous part.

    Z1‾=2+5i\overline{Z_1} = 2 + 5i Z2‾=2−3i\overline{Z_2} = 2 - 3i

  4. Calculate Z1‾⋅Z2‾\overline{Z_1} \cdot \overline{Z_2}.

    Z1‾⋅Z2‾=(2+5i)(2−3i)\overline{Z_1} \cdot \overline{Z_2} = (2 + 5i)(2 - 3i) Z1‾⋅Z2‾=2(2)+2(−3i)+5i(2)+5i(−3i)\overline{Z_1} \cdot \overline{Z_2} = 2(2) + 2(-3i) + 5i(2) + 5i(-3i) Z1‾⋅Z2‾=4−6i+10i−15i2\overline{Z_1} \cdot \overline{Z_2} = 4 - 6i + 10i - 15i^2

    Since i2=−1i^2 = -1:

    Z1‾⋅Z2‾=4+4i−15(−1)\overline{Z_1} \cdot \overline{Z_2} = 4 + 4i - 15(-1) Z1‾⋅Z2‾=4+4i+15=19+4i\overline{Z_1} \cdot \overline{Z_2} = 4 + 4i + 15 = 19 + 4i

  5. Compare the results.

    Since Z1⋅Z2‾=19+4i\overline{Z_1 \cdot Z_2} = 19 + 4i and Z1‾⋅Z2‾=19+4i\overline{Z_1} \cdot \overline{Z_2} = 19 + 4i,

    Z1⋅Z2‾=Z1‾⋅Z2‾\overline{Z_1 \cdot Z_2} = \overline{Z_1} \cdot \overline{Z_2}

    This property is verified.

    Both properties are verified.

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