Proof of conjugate verification.

Question

Prove $\frac{z_1}{z_2} = \frac{\overline{z_1}}{\overline{z_2}}$ if $z_2\neq0$.

Proof:

Let $z_1=a_1+ib_1$ and let $z_2=a_2+ib_2$, where $a_1, b_1, a_2, b_2 \in \Re$. $$\frac{z_1}{z_2} = (\overline{\frac{a_1a_2+b_1b_2}{a_2^2+b_2^2}+i\frac{a_2b_1-a_1b_2}{a_2^2+b_2^2}})$$ $$\Downarrow$$ $$=\frac{a_1a_2+b_1b_2}{a_2^2+b_2^2}-i\frac{a_2b_1-a_1b_2}{a_2^2+b_2^2}$$ $$\Downarrow$$ $$= \frac{a_1a_2+(-b_1)(-b_2)}{a_2^2+(-b_2)^2}+i\frac{a_2(-b_1)-a_1(-b_2)}{a_2^2+(-b_2)^2}$$ $$\Downarrow$$ $$=\frac{a_1-ib_1}{a_2-ib_2}$$ $$\Downarrow$$ $$=\frac{\overline{z_1}}{\overline{z_2}}$$ when $z_2\neq 0$.

Do you have a specific question or did you want to share this with us? — Kushashwa Ravi Shrimali, Jun 27 '14 at 16:31
@KushashwaRaviShrimali: the specific question was to prove the above, I just want to make sure I have done that appropriately. — Michelle, Jun 27 '14 at 16:32
Let $z_2=1$ then it says $z_1=\overline{z_1}$ for all $z_1$. — Rene Schipperus, Jun 27 '14 at 16:33
Your result is wrong, as shown by the counter example $z_1=i$, $z_2=1$, as you'd be claiming that $i=-i$. Go through your proof with these values of $z_1$ and $z_2$ to see where you're wrong. Hint: the conjugate you have in the first step is wrong. — gniourf_gniourf, Jun 27 '14 at 16:37

score 1 · Answer 1 · answered Jun 27 '14 at 17:01

As noted in the comments if $z_2=1$ we clearly have a contradiction. I believe that you are possibly confusing this false statement with a similar true one:
$$\bar{ab}=\bar{a}\bar{b}$$

I'm going to replace the bar with a star: One can quickly apply this to get:

$$\frac{a}{b} \neq^{\star}({\frac{a}{b}})^* = {b^{-1}}^*{a}^*=\frac{a^*}{b^*}$$

$\star$ Note: the $\neq$ will become equality if and only if the number on the far left is real.

score 0 · Accepted Answer · answered Jun 27 '14 at 16:59

I think you meant to prove : $\bar{\left(\cfrac{z_1}{z_2}\right)} = \cfrac{\bar {z_1}}{\bar{z_2}}$

$$\overline{\cfrac{z_1}{z_2}} = \overline{ \cfrac{a + ib}{c + id} } \\ = \overline{\cfrac{a+ib}{c+id} \times \cfrac{c-id}{c-id}} \\ = \overline{\cfrac{ac + bd + i(bd - ac)}{c^2 +d^2}} \\ = \cfrac{ac + bd + i(ad - bc) }{c^2 + d^2} $$

Now, we will simplify $\cfrac{\overline{z_1}}{\overline{z_2}}$

$$\cfrac{\overline{z_1}}{\overline{z_2}} = \cfrac{a-ib}{c-id} \\ = \cfrac{a-ib}{c-id} \times \cfrac{c+id}{c+id} \\ = \cfrac{ac + bd + i(ad - bc) }{c^2 + d^2} \\ = \overline{\cfrac{z_1}{z_2}} $$

Proof of conjugate verification.

2 Answers2