Analysis 1 Homework Help with Complex Numbers

[lema21]

Homework Statement

If z,w are in C then prove that bar(z/w) = bar(z)/bar(w).

Relevant Equations

z = a+bi
w = c+di
Bar(z) = a-bi
Bar(w) = c-di

I need help actually creating the proof. I've done the scratch needed for the problem, it's just forming the proof that I need help in.
Bar(a+bi/c+di)= (a-bi) / (c-di)
Bar ((a+bi/c+di)*(c-di/c-di)) = ((a-bi/c-di)*(c+di/c+di))
Bar((ac+bd/c^2 +d^2)+(i(bc-ad)/c^2+d^2)) = (ac+bd/c^2+d^2)+(i(ad-bc)/(c^2+d^2))
(ac+bd/c^2+d^2) - (i(bc-ad)/c^2+d^2) = (ac+bd/c^2+d^2) + (i(ad-bc)/c^2+d^2)
ibc+iad/ c^2+d^2 = iad-ibc/ c^2+d^2
-ibc+iad=iad-ibc

 

[lema21]

 

[Mark44]

Homework Statement:: If z,w are in C then prove that bar(z/w) = bar(z)/bar(w).
Relevant Equations:: z = a+bi
w = c+di
Bar(z) = a-bi
Bar(w) = c-di

I need help actually creating the proof. I've done the scratch needed for the problem, it's just forming the proof that I need help in.
Bar(a+bi/c+di)= (a-bi) / (c-di)
Bar ((a+bi/c+di)*(c-di/c-di)) = ((a-bi/c-di)*(c+di/c+di))
Bar((ac+bd/c^2 +d^2)+(i(bc-ad)/c^2+d^2)) = (ac+bd/c^2+d^2)+(i(ad-bc)/(c^2+d^2))
(ac+bd/c^2+d^2) - (i(bc-ad)/c^2+d^2) = (ac+bd/c^2+d^2) + (i(ad-bc)/c^2+d^2)
ibc+iad/ c^2+d^2 = iad-ibc/ c^2+d^2
-ibc+iad=iad-ibc

This is not what you want to end with, since it's obviously true that $(-bc + ad)i = (ad - bc)i$
You've shown that $\overline{\left(\frac z w\right)} = \frac{ac + bd + (ad - bc)i}{c^2 + d^2}$ and that $\frac {\overline z}{\overline w}$ equals the same value.

If you have x = some value and y = the same value, what can you conclude?

 

[FactChecker]

If you are wondering how to state the proof, one very direct way is to go down the left side of your hypothesized equations and then up the right side. That will start with $\overline{(\frac{z}{w})}$ and end up with $\frac{\overline{z}}{\overline{w}}$, and you can do it so that "=" clearly and undeniably means equals. That is what a proof needs.

PS. Your "=" signs are very confusing. It looks like you are stating equality when you are really just hypothesizing equality and trying to prove that they are equal.

 

[lema21]

This is not what you want to end with, since it's obviously true that $(-bc + ad)i = (ad - bc)i$
You've shown that $\overline{\left(\frac z w\right)} = \frac{ac + bd + (ad - bc)i}{c^2 + d^2}$ and that $\frac {\overline z}{\overline w}$ equals the same value.

If you have x = some value and y = the same value, what can you conclude?

So I can just finish this by stating that since they equal the same value that they then equal each other?

 

[Mark44]

So I can just finish this by stating that since they equal the same value that they then equal each other?

Yes, but you should organize your work a lot better so that whoever is reading it can follow what you're doing.

Show that $\overline{\left(\frac z w\right)} = \overline{\left(\frac{a + bi}{c + di}\right) } = \frac{ac + bd + (ad - bc)i}{c^2 + d^2}$ and then show that $\frac {\overline z}{\overline w}= \frac{\overline{a + bi}}{\overline{c + di}} = \frac{ac + bd + (ad - bc)i}{c^2 + d^2}$.
From that work, you can conclude that $\overline{\left(\frac z w\right)} = \frac {\overline z}{\overline w}$

 

[lema21]

Okay, thank you so much :)

 

[WWGD]

Okay, thank you so much :)

Have you considered using polar coordinates for the proof?