Analysis 1 Homework Help with Complex Numbers

  • #1
lema21
18
9
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
 

Answers and Replies

  • #2
lema21
18
9
1.1 - 1.2.jpg
 
  • #3
36,709
8,707
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?
 
  • #4
FactChecker
Science Advisor
Homework Helper
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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.
 
Last edited:
  • #5
lema21
18
9
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?
 
  • #6
36,709
8,707
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}##
 
  • #7
lema21
18
9
Okay, thank you so much :)
 
  • #8
WWGD
Science Advisor
Gold Member
6,291
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Okay, thank you so much :)
Have you considered using polar coordinates for the proof?
 

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