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You have two numbers and need the conjugate of their product. Do you multiply first and then flip the sign of i? Or flip each one first and then multiply? Your book proves it does not matter.
The same is true for sums and quotients. Taking the conjugate twice brings you back to where you started.
1. The conjugate of z1 + z2 is z̄1 + z̄2.
2. The conjugate of z1z2 is z̄1 z̄2.
3. The conjugate of z1 ÷ z2 is z̄1 ÷ z̄2, when z2 ≠ 0.
4. The conjugate of z̄ is z.
Re(z) = (z + z̄) ÷ 2 and Im(z) = (z − z̄) ÷ 2i.
Step 1 / 7
You have two numbers and need the conjugate of their product. Do you multiply first and then flip the sign of i? Or flip each one first and then multiply? Your book proves it does not matter.
The same is true for sums and quotients. Taking the conjugate twice brings you back to where you started.
1. The conjugate of z1 + z2 is z̄1 + z̄2.
2. The conjugate of z1z2 is z̄1 z̄2.
3. The conjugate of z1 ÷ z2 is z̄1 ÷ z̄2, when z2 ≠ 0.
4. The conjugate of z̄ is z.
Re(z) = (z + z̄) ÷ 2 and Im(z) = (z − z̄) ÷ 2i.
Conjugating and doing the arithmetic can be done in either order. Conjugating twice gives back z.
Q1. Example 11 (p. 11): z1 = 4 + 3i, z2 = 5 + 2i. Show conj(z1 + z2) = z̄1 + z̄2.
Both sides are 9 − 5i.
Q2. Example 12 (p. 12): z1 = 3 + 4i, z2 = 2 + 3i. Show conj(z1z2) = z̄1 z̄2.
Both sides are −6 − 17i.
✗ conj(−6 + 17i) = 6 − 17i.
✓ Only the i part changes sign: −6 − 17i.
✗ conj(z1z2) = z̄1 z2.
✓ Both factors are conjugated: z̄1 z̄2.
✗ z + z̄ = 8 + 8i for z = 4 + 4i.
✓ The i parts cancel: z + z̄ = 2Re(z) = 8.
1. If z = 4 + 4i, then z + z̄ =
(a) Review Q1(ix): (4 + 4i) + (4 − 4i) = 8.
2. The conjugate of the conjugate of 7 + 3i is:
(c) Property 4, Example 14 (p. 13).
3. conj(z1 ÷ z2) equals:
(d) Property 3 (p. 12), with z2 ≠ 0.
4. For z1 = 5 + 4i and z2 = 3 + 2i, conj(z1 ÷ z2) is (Example 13):
(b) z1 ÷ z2 = (23 + 2i) ÷ 13, so its conjugate is (23 − 2i) ÷ 13.
5. Which formula gives Im(z)? (p. 17)
(c) z − z̄ = 2iy, so dividing by 2i leaves y.
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