Theory of Intelligence
11TH•Mathematics•Complex Numbers•Question 6

Solution

Show that for all $z_1,z_2\in\mathbb{C}$, $\overline{z_1z_2}=\bar{z_1}\,\bar{z_2}$.

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Solution

Let:

z1=a+bi,z2=c+di(a,b,c,d∈R)z_1=a+bi,\quad z_2=c+di\qquad (a,b,c,d\in\mathbb{R})

Then:

z1z2=(a+bi)(c+di)=ac+adi+bci+bdi2=(ac−bd)+i(ad+bc)\begin{aligned} z_1z_2&=(a+bi)(c+di) \\ &=ac+adi+bci+bdi^2 \\ &=(ac-bd)+i(ad+bc) \end{aligned}

So:

z1z2‾=(ac−bd)−i(ad+bc)\overline{z_1z_2}=(ac-bd)-i(ad+bc)

Also:

z1ˉ=a−bi,z2ˉ=c−di\bar{z_1}=a-bi,\quad \bar{z_2}=c-di z1ˉ z2ˉ=(a−bi)(c−di)=ac−adi−bci+bdi2=(ac−bd)−i(ad+bc)\begin{aligned} \bar{z_1}\,\bar{z_2}&=(a-bi)(c-di) \\ &=ac-adi-bci+bdi^2 \\ &=(ac-bd)-i(ad+bc) \end{aligned}

Hence:

z1z2‾=z1ˉ z2ˉ\boxed{\overline{z_1z_2}=\bar{z_1}\,\bar{z_2}}

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