Analysis 1 Homework Help with Complex Numbers

In summary, if z,w are in C then prove that bar(z/w) = bar(z)/bar(w).In summary, if z,w are in C then prove that bar(z/w) = bar(z)/bar(w).
  • #1
lema21
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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
 
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  • #2
1.1 - 1.2.jpg
 
  • #3
lema21 said:
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?
 
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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.
 
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  • #5
Mark44 said:
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
lema21 said:
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}##
 
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  • #7
Okay, thank you so much :)
 
  • #8
lema21 said:
Okay, thank you so much :)
Have you considered using polar coordinates for the proof?
 

What are complex numbers?

Complex numbers are numbers that contain both a real and an imaginary part. They are typically written in the form a + bi, where a is the real part and bi is the imaginary part with i being the imaginary unit.

How do I add and subtract complex numbers?

To add or subtract complex numbers, you simply add or subtract the real parts and the imaginary parts separately. For example, (3 + 2i) + (1 + 4i) = (3 + 1) + (2i + 4i) = 4 + 6i.

How do I multiply and divide complex numbers?

To multiply complex numbers, you use the FOIL method (First, Outer, Inner, Last). For example, (3 + 2i)(1 + 4i) = 3 + 12i + 2i + 8i^2 = 3 + 14i - 8 = -5 + 14i. To divide complex numbers, you use the conjugate of the denominator to rationalize the fraction.

What is the complex conjugate?

The complex conjugate of a complex number a + bi is a - bi. It is used in dividing complex numbers and also in finding the modulus (absolute value) of a complex number, which is the square root of the sum of the squares of the real and imaginary parts.

How are complex numbers used in real life?

Complex numbers are used in many fields of science and engineering, including electrical engineering, quantum mechanics, and signal processing. They are also used in various applications such as computer graphics, fluid dynamics, and financial modeling.

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