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I want to rigorously verify that $$\overline{\left(\frac{z_1}{z_2}\right)}={\left(\frac{\overline{z_1}}{\overline{z_2}}\right)}$$

Here's my attempt at thinking:

$$\overline{\left(\frac{z_1}{z_2}\right)}=\overline{\left({z_1}\cdot \frac{1}{z_2}\right)}={\left(\overline{z_1}\cdot \overline{\left(\frac{1}{z_2}\right)}\right)}$$

Now, it appears that we are in a "vicious circle" because we don't know if $$\overline{\left(\frac{1}{z_2}\right)}=\frac{1}{\overline{z_2}}$$

One approach might be to prove that

$$\overline{\left(\frac{1}{z_2}\right)}=\overline{{z_2}^{-1}}=\left(\overline{z_2}\right)^{-1}=\frac{1}{\overline{z_2}}$$

But is there a better way?

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3 Answers

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You are right when you reduce the original problem to the problem of proving that$$\overline{\left ( \frac 1z \right )}=\frac 1 {\overline z}.\tag1$$Now, in order to prove this, all you have to do is to notice that$$\overline{\left ( \frac 1z \right )}.\overline z=\overline{{\left ( \frac 1z \right)}\times z}=\overline1=1.$$Therefore, $(1)$ holds.

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I suppose, that you should just prove $$ \overline{\left(\dfrac1{z_2}\right)} = \dfrac1{\overline{z_2}} $$ Simply multiply both sides by $\overline{z_2}$ and get $$ \overline{\left(\dfrac1{z_2}\right)}\cdot\overline{z_2} = \overline{\left(\dfrac1{z_2}\right)\cdot z_2} = \overline{1} = 1 = \dfrac1{\overline{z_2}}\cdot\overline{z_2} $$ Done!

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By multiplying numerator and denominator by $\bar{z}_2$, noting that $|z|^2=z\bar{z}$ we get $$\left(\frac{z_1}{z_2}\right)=\left(\frac{z_1\bar{z}_2}{|z_2|^2}\right)$$ taking the conjugate on both sides $$\overline{\left(\frac{z_1}{z_2}\right)}=\overline{\left(\frac{z_1\bar{z}_2}{|z_2|^2}\right)} = \frac{\overline{z_1\bar{z}_2}}{|z_2|^2} = \frac{\bar{z}_1z_2}{|z_2|^2} =\bar{z}_1\frac{z_2}{|z_2|^2}$$ now going back on the absolute value $$\bar{z}_1\frac{z_2}{z_2\bar{z}_2} = \bar{z}_1\frac{1}{\bar{z}_2}=\frac{\bar{z}_1}{\bar{z}_2}$$

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