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Worked solution

Use a conjugate to recover components and a real product

Let z=2+3iz=2+3i. State the values of (a) z+z∗z+z^*, (b) z−z∗z-z^* and (c) zz∗zz^*.

CJC · Self-Practice 3

(a) Sum

Answer

44.

z∗=2−3iz^*=2-3i, so z+z∗=(2+3i)+(2−3i)=4z+z^*=(2+3i)+(2-3i)=4. Equivalently, 2Re⁡(z)=2(2)=42\operatorname{Re}(z)=2(2)=4.

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(b) Difference

Answer

6i6i.

z−z∗=(2+3i)−(2−3i)=6iz-z^*=(2+3i)-(2-3i)=6i. This is 2iIm⁡(z)2i\operatorname{Im}(z), not just twice the imaginary coefficient.

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(c) Product

Answer

1313.

zz∗=(2+3i)(2−3i)=22−(3i)2=4+9=13zz^*=(2+3i)(2-3i)=2^2-(3i)^2=4+9=13. The imaginary terms cancel.

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Solution notes

Both copies print all three answers. The calculations here explain those results.