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Let $z_1=a_1+b_1i$ and $z_2=a_2+b_2i$ be complex numbers. Let $\overline{z_1}$ and $\overline{z_2}$ be the conjugate of $z_1$ and $z_2$, respectively. Prove: $$\overline{\left(\frac {z_1}{z_2}\right)}=\frac {\overline{z_1}}{\overline{z_2}}$$

My Attempt:

I started with $\overline{\left(\frac {z_1}{z_2}\right)}$ and multiplied it by $\frac {z_1}{z_2}$ to get $\left|\frac {z_1}{z_2}\right|^2=\frac {a_1^2+b_1^2}{a_2^2+b_2^2}$, but now, I don't know where to continue afterwards.

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$$ \begin{align} \frac{z_1}{z_2}&=\frac{a_1+ib_1}{a_2+ib_2}\\ \\ &=\frac{a_1+ib_1}{a_2+ib_2}\cdot\frac{a_2-ib_2}{a_2-ib_2}\\ \\ &=\frac{(a_1+ib_1)(a_2-ib_2)}{(a_2+ib_2)(a_2-ib_2)}\\ \\ &=\frac{(a_1+ib_1)(a_2-ib_2)}{(a_2^2+b_2^2)}\\ \\ &=\frac{(a_1a_2+b_1b_2)+i(-a_1b_2+a_2b_1)}{(a_2^2+b_2^2)} \end{align} $$ The $\frac{(a_1a_2+b_1b_2)+i(-a_1b_2+a_2b_1)}{(a_2^2+b_2^2)}$ fraction is a Complex Number with Real denominator ($a_2^2+b_2^2$) and conjugation of this fraction only changes the Numarator $(a_1a_2+b_1b_2)+i(-a_1b_2+a_2b_1)$.

So

$$ \begin{align} \overline{\Big(\frac{z_1}{z_2}\Big)}&=\overline{\Big(\frac{a_1+ib_1}{a_2+ib_2}\Big)}\\ \\ &=\overline{ \Big(\frac{(a_1a_2+b_1b_2)+i(-a_1b_2+a_2b_1)}{(a_2^2+b_2^2)}\Big)} \\ \\ &=\frac{(a_1a_2+b_1b_2)-i(-a_1b_2+a_2b_1)}{(a_2^2+b_2^2)} \\ \\ &=\frac{(a_1a_2+b_1b_2)+i(a_1b_2-a_2b_1)}{(a_2^2+b_2^2)} \\ \\ &=\frac{(a_1-ib_1)(a_2+ib_2)}{(a_2^2+b_2^2)} \\ \\ &=\frac{(a_1-ib_1)(a_2+ib_2)}{(a_2-ib_2)(a_2+ib_2)} \\ \\ &=\frac{a_1-ib_1}{a_2-ib_2} \cdot\frac{a_2+ib_2}{a_2+ib_2} \\ \\ &=\frac{a_1-ib_1}{a_2-ib_2} \\ \\ &=\frac{\overline{z_1}}{\overline{z_2}} \end{align} $$

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  • $\begingroup$ Ah! Thanks for elaborating why each step is like that! $\endgroup$ – user332252 Sep 17 '16 at 19:13
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Hint:

You can complete your proof multiplying also the RHS by $\frac{z_1}{z_2}$ then show that $\left|\frac {z_1}{z_2}\right|=\frac{|z_1|}{|z_2|}$ ( easy in polar form).

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we get $$\frac{a_1+b_1i}{a_2+b_2i}=\frac{a_1a_2+b_1b_2+(a_2b_1-a_1b_2)i}{a_2^2+b_2^2}$$ and the complement is $$\overline{\frac{a_1+b_1i}{a_2+b_2i}}=\frac{a_1a_2+b_1b_2-(a_2b_1-a_1b_2)i}{a_2^2+b_2^2}$$ and $$\frac{a_1-b_1i}{a_2-b_2i}=\frac{(a_1-b_1i)(a_2+b_2i)}{(a_2-b_2i)(a_2+b_2i)}=...$$ can you proceed?

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If $w_1=re^{i\theta}$ and $w_2=\rho e^{i\phi}$ then $$\overline{w_1}\ \overline{w_2}=re^{-i\theta} \rho e^{-i\phi}=r\ \rho e^{-i(\theta+\phi)}=\overline{w_1\ w_2}$$ so $$\boxed{\overline{w_1}\ \overline{w_2}=\overline{w_1\ w_2}}$$ now let $w_1=\dfrac{z_1}{z_2}$ and $z_2$. Also straightforward shows $$\overline{\left(\frac {w_1}{w_2}\right)}=\dfrac{r}{\rho} e^{-i(\theta-\phi)}=\dfrac{re^{-i\theta}}{\rho e^{-i\phi}}=\dfrac{\overline{w_1}}{\overline{w_2}}$$

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