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Exercise 423Algebra › Complex numbersHigh schoolAdded 3 Nov 2026

Properties of the conjugate

#conjugate#properties#modulus

Statement

Prove that for all z,w∈Cz,w\in\mathbb{C}:
1. z+w‾=zˉ+wˉ\overline{z+w}=\bar z+\bar w;
2. zw‾=zˉwˉ\overline{zw}=\bar z\bar w;
3. zzˉ=∣z∣2z\bar z=|z|^2;
4. z+zˉ=2Re⁡(z)z+\bar z=2\operatorname{Re}(z) and z−zˉ=2iIm⁡(z)z-\bar z=2i\operatorname{Im}(z).

Hint

Write z=a+biz=a+bi, w=c+diw=c+di.

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Solution

1. (a+c)+(b+d)i‾=(a+c)−(b+d)i=(a−bi)+(c−di)=zˉ+wˉ\overline{(a+c)+(b+d)i}=(a+c)-(b+d)i=(a-bi)+(c-di)=\bar z+\bar w.

2. zw=(ac−bd)+(ad+bc)izw=(ac-bd)+(ad+bc)i, so zw‾=(ac−bd)−(ad+bc)i\overline{zw}=(ac-bd)-(ad+bc)i. And zˉwˉ=(a−bi)(c−di)=ac−adi−bci−bd=ac−bd−(ad+bc)i\bar z\bar w=(a-bi)(c-di)=ac-ad i-bc i-bd=ac-bd-(ad+bc)i. Equal.

3. zzˉ=(a+bi)(a−bi)=a2+b2=∣z∣2z\bar z=(a+bi)(a-bi)=a^2+b^2=|z|^2.

4. z+zˉ=2a=2Re⁡(z)z+\bar z=2a=2\operatorname{Re}(z); z−zˉ=2bi=2iIm⁡(z)z-\bar z=2bi=2i\operatorname{Im}(z).

Conclusion. All four properties hold.