what does this notation mean? "Verify (bar over)(z + w) = (bar over)(z) + (bar over)(w)" What are the bars?

allegansveritatem

Full Member
Joined
Jan 10, 2018
Messages
343
I came across this today in some problems at the end of a section and can find no reference to it in the text:
11194 What I don't understand is the symbols over the z+w and over the z and the w. What are they asking for here? Here is the solution the book gives but I can't figure out where they got any of this from:

11195
 

MarkFL

Super Moderator
Staff member
Joined
Nov 24, 2012
Messages
1,261
That overline notation refers to a complex conjugate. for example, if we are given:

\(\displaystyle z=x+yi\)

then:

\(\displaystyle \overline{z}=x-yi\)
 

allegansveritatem

Full Member
Joined
Jan 10, 2018
Messages
343
That overline notation refers to a complex conjugate. for example, if we are given:

\(\displaystyle z=x+yi\)

then:

\(\displaystyle \overline{z}=x-yi\)
So the line over the z refers to a complex conjugate. But I don't see here how they come to this solution. I mean how do they go from this:
11197

to:

11198

From z + w with the overhead line to a bunch of a's and b's....? I don't see how come by these terms. I can see conjugates in there but.... It astonishes me that this wasn't covered in the text but just thrown into the problems at the end of a section.
 
Last edited:

ksdhart2

Senior Member
Joined
Mar 25, 2016
Messages
1,039
So the line over the z refers to a complex conjugate. But I don't see here how they come to this solution. I mean how do they go from [...] z + w with the overhead line to a bunch of a's and b's....? I don't see how come by these terms. I can see conjugates in there but.... It astonishes me that this wasn't covered in the text but just thrown into the problems at the end of a section.
That first step there, where the a's and b's "magically" appear, is literally just the definition of a complex number. It doesn't make sense to talk about the complex conjugate of a real number, so z and w must be complex numbers. Let \(\displaystyle z = a + bi\) and \(\displaystyle w = c + di\). Then \(\displaystyle \overline{z + w} = \text{???}\)
 

pka

Elite Member
Joined
Jan 29, 2005
Messages
7,954
So the line over the z refers to a complex conjugate. But I don't see here how they come to this solution. I mean how do they go from this:
View attachment 11197

to:

View attachment 11198

From z + w with the overhead line to a bunch of a's and b's....? I don't see how come by these terms. I can see conjugates in there but.... It astonishes me that this wasn't covered in the text but just thrown into the problems at the end of a section.
First, do you understand or have you been taught that the notation \(\displaystyle \overline{3-2\bf{i}}\) stands for the conjugate \(\displaystyle 3+2\bf{i}~?\)

As for the \(\displaystyle a's~\&~b's\), in the theory of complex numbers are defined as ordered pairs of real numbers.
The pair (3,-2) is written as \(\displaystyle 3-2\bf{i}\). The real part \(\displaystyle \Re(3-2\bf{i})=3\) and imaginary part \(\displaystyle \Im(3-2\bf{i})=-2\)
So we define \(\displaystyle (a,b)+(c,d)=(a+c,b+d)\) , \(\displaystyle (a,b)\cdot(c,d)=(ac-bd,ad+bc)\), and \(\displaystyle \overline{(a,b)}=(a,-b)\)
 

allegansveritatem

Full Member
Joined
Jan 10, 2018
Messages
343
First, do you understand or have you been taught that the notation \(\displaystyle \overline{3-2\bf{i}}\) stands for the conjugate \(\displaystyle 3+2\bf{i}~?\)

As for the \(\displaystyle a's~\&~b's\), in the theory of complex numbers are defined as ordered pairs of real numbers.
The pair (3,-2) is written as \(\displaystyle 3-2\bf{i}\). The real part \(\displaystyle \Re(3-2\bf{i})=3\) and imaginary part \(\displaystyle \Im(3-2\bf{i})=-2\)
So we define \(\displaystyle (a,b)+(c,d)=(a+c,b+d)\) , \(\displaystyle (a,b)\cdot(c,d)=(ac-bd,ad+bc)\), and \(\displaystyle \overline{(a,b)}=(a,-b)\)
I know about complex numbers. I don't know about the sign, the overhead line that goes above them. I don't have a teacher but have been using a book by Robert Blixer. Recently I finished this book and got one by Earl Swokowski. I guess Swokowski uses this notation and Blixer does not. I have taken algebra courses in high school and a college course as well but that was many decades ago--still I don't recall this sign from those times either. Blixer's book is titled: College Algebra. Swokowski's is more advanced with the title: Precalculus, Equations and Graphs. But I begin to see my problem with this: One part of the problem is that I took the z and the w for single component terms, like a and b. I will study your reply later today and get back. Thanks.
 

allegansveritatem

Full Member
Joined
Jan 10, 2018
Messages
343
That first step there, where the a's and b's "magically" appear, is literally just the definition of a complex number. It doesn't make sense to talk about the complex conjugate of a real number, so z and w must be complex numbers. Let \(\displaystyle z = a + bi\) and \(\displaystyle w = c + di\). Then \(\displaystyle \overline{z + w} = \text{???}\)
I was having trouble translating z and w into anything other than terms with a single component. I wasn't making the leap from z or w to a+bi. Dumb but understandable. I have to go but I will get back to your post later and work it out. Thanks for reply.
 

Dr.Peterson

Elite Member
Joined
Nov 12, 2017
Messages
3,320
I checked the Blitzer book I've taught from, and see that, indeed, it doesn't use any notation for conjugates, because it doesn't do any more with them than use them to simplify expressions or point out that solutions to quadratic equations come in conjugate pairs, neither of which needs a notation.

Clearly your new book is at a higher level, so you may need to go back to where they introduced complex numbers and read carefully, to see their notation and conventions.

In the solution you quoted, they don't take the time, as one would in a proof, to define variables or justify steps, trusting that you know what to expect: When they replace z with a+bi, they are implicitly saying, "suppose that the complex number z has components a and b", and so on. So you have to build up your expectations to match theirs, by reading slowly and carefully, even when you think the topic is one you already know. Probably there has been an example that demonstrated these ways of thinking.
 

pka

Elite Member
Joined
Jan 29, 2005
Messages
7,954
(I) have been using a book by Robert Blixer. Recently I finished this book and got one by Earl Swokowski. I guess Swokowski uses this notation and Blixer does not. Blixer's book is titled: College Algebra. Swokowski's is more advanced with the title: Precalculus, Equations and Graphs. But I begin to see my problem with this: One part of the problem is that I took the z and the w for single component terms, like a and b. I will study your reply later today and get back.
Unlike Prof. Peterson, I do not have nor have I seen the text by Bilixer. I I knew Swokowski and his text is excellent. There are even solution manuals available on Amazon. I was typing a replay late last night when the LaTeX went haywire. So I simply stopped. I wish I had been able to save the code but alas its gone.
Here is the a bit of it. Any complex number is two parts: A real part & an imaginary part.
Say \(\displaystyle z=5-3\bf{i}\), the real part \(\displaystyle \Re(z)=5~\&~\)the imaginary part \(\displaystyle \Im{z}=-3\).
Thus \(\displaystyle z=a+b\bf{i}\), the real part \(\displaystyle \Re(z)=a\) & the imaginary part \(\displaystyle \Im{z}=b\) where both \(\displaystyle a~\&~b\) are real numbers.
Then we get Say \(\displaystyle |z|=\sqrt{a^2+b^2},\:\) \(\displaystyle \overline{\,z\,}=a-b\bf{i}\).
Moreover, \(\displaystyle \;z+\overline{\,z\,}=2\cdot\Re(z)\) and \(\displaystyle \;z-\overline{\,z\,}=2\cdot\Im(z)\)
\(\displaystyle z\cdot \overline{\,z\,}=|z|^2\)

 

allegansveritatem

Full Member
Joined
Jan 10, 2018
Messages
343
I checked the Blitzer book I've taught from, and see that, indeed, it doesn't use any notation for conjugates, because it doesn't do any more with them than use them to simplify expressions or point out that solutions to quadratic equations come in conjugate pairs, neither of which needs a notation.

Clearly your new book is at a higher level, so you may need to go back to where they introduced complex numbers and read carefully, to see their notation and conventions.

In the solution you quoted, they don't take the time, as one would in a proof, to define variables or justify steps, trusting that you know what to expect: When they replace z with a+bi, they are implicitly saying, "suppose that the complex number z has components a and b", and so on. So you have to build up your expectations to match theirs, by reading slowly and carefully, even when you think the topic is one you already know. Probably there has been an example that demonstrated these ways of thinking.
I think that is it. Blitzer introduces complex numbers and their conjugates and how to do operations with complex numbers. He demonstrates alos how to clear denominators of complex numbers by multiplying by conjugates. In the Swokowski book I skipped the first 65 pages because I couldn't see anything thee that wasn't some kind of review of things I have learned from Blitzer. Now I am going to check over those pages again and see if I missed something new. It is quite a nice book and pretty well illustrated.
 

allegansveritatem

Full Member
Joined
Jan 10, 2018
Messages
343
Unlike Prof. Peterson, I do not have nor have I seen the text by Bilixer. I I knew Swokowski and his text is excellent. There are even solution manuals available on Amazon. I was typing a replay late last night when the LaTeX went haywire. So I simply stopped. I wish I had been able to save the code but alas its gone.
Here is the a bit of it. Any complex number is two parts: A real part & an imaginary part.
Say \(\displaystyle z=5-3\bf{i}\), the real part \(\displaystyle \Re(z)=5~\&~\)the imaginary part \(\displaystyle \Im{z}=-3\).
Thus \(\displaystyle z=a+b\bf{i}\), the real part \(\displaystyle \Re(z)=a\) & the imaginary part \(\displaystyle \Im{z}=b\) where both \(\displaystyle a~\&~b\) are real numbers.
Then we get Say \(\displaystyle |z|=\sqrt{a^2+b^2},\:\) \(\displaystyle \overline{\,z\,}=a-b\bf{i}\).
Moreover, \(\displaystyle \;z+\overline{\,z\,}=2\cdot\Re(z)\) and \(\displaystyle \;z-\overline{\,z\,}=2\cdot\Im(z)\)
\(\displaystyle z\cdot \overline{\,z\,}=|z|^2\)

Yes, I think I follow what you are doing here. I worked something out on this problem today and substituted specific numbers for variables:
11209

And now that I think I understand the problem better I have another question: What is the difference between the two expressions, I mean between z+w with a continuous overhead line and z+w where each variable has its own separate overhead line?
 

pka

Elite Member
Joined
Jan 29, 2005
Messages
7,954
Yes all of that is correct. But it is easier to remember that the conjugate of a sum is the sum of the conjugates.

Here is a personal favorite of mine. \(\displaystyle \dfrac{1}{z}=\dfrac{\overline{\,z\,}}{|z|^2}\)
To see how useful this is, consider the following: \(\displaystyle \dfrac{2+\bf{i}}{3+4\bf{i}}= \dfrac{(2+\bf{i})(3-4\bf{i})}{5}\)
 

allegansveritatem

Full Member
Joined
Jan 10, 2018
Messages
343
Yes all of that is correct. But it is easier to remember that the conjugate of a sum is the sum of the conjugates.
^2
Here is a personal favorite of mine. \(\displaystyle \dfrac{1}{z}=\dfrac{\overline{\,z\,}}{|z|^2}\)
To see how useful this is, consider the following: \(\displaystyle \dfrac{2+\bf{i}}{3+4\bf{i}}= \dfrac{(2+\bf{i})(3-4\bf{i})}{5}\)
I don't get what you are driving at. The reciprocal of 1/z? Is the z a complex number?If si, that would mean that 1/2+i would equal 2-i/(2+i)^2. According to Wolfram Alpha, that is a false statement.
 

pka

Elite Member
Joined
Jan 29, 2005
Messages
7,954
I don't get what you are driving at. The reciprocal of 1/z? Is the z a complex number?If si, that would mean that 1/2+i would equal 2-i/(2+i)^2. According to Wolfram Alpha, that is a false statement.
I think you are missing the whole point. \(\displaystyle |2+\bf{i}|=\sqrt{2^2+1^2}=\sqrt5 \) so that \(\displaystyle |2+\bf{i}|^2=5 \).
So \(\displaystyle \dfrac{1}{2+\bf{i}}=\dfrac{2-\bf{i}}{5}=\frac{2}{5}-\frac{1}{5}\bf{i}\)
You need to learn absolute value.
Here is another example: \(\displaystyle \Re e[(4-3\bf{i})^{-1}]=\frac{4}{25}\;\&\;\Im m[(4-3\bf{i})^{-1}]=\frac{3}{25}\)
 

allegansveritatem

Full Member
Joined
Jan 10, 2018
Messages
343
I think you are missing the whole point. \(\displaystyle |2+\bf{i}|=\sqrt{2^2+1^2}=\sqrt5 \) so that \(\displaystyle |2+\bf{i}|^2=5 \).
So \(\displaystyle \dfrac{1}{2+\bf{i}}=\dfrac{2-\bf{i}}{5}=\frac{2}{5}-\frac{1}{5}\bf{i}\)
You need to learn absolute value.
Here is another example: \(\displaystyle \Re e[(4-3\bf{i})^{-1}]=\frac{4}{25}\;\&\;\Im m[(4-3\bf{i})^{-1}]=\frac{3}{25}\)
I know something about absolute value but not enough I guess to know how to make much use of it. I have copied this out and will try to work it out tomorrow. 1 over 4-3i is 1/4 +1/3i, no? Why do we need absolute value? Does that cancel out the i?
 

pka

Elite Member
Joined
Jan 29, 2005
Messages
7,954
1 over 4-3i is 1/4 +1/3i, no? Why do we need absolute value? Does that cancel out the i?
That is incorrect. Look at this: \(\displaystyle \frac{1}{7}=\frac{1}{3+4}\ne\frac{1}{3}+\frac{1}{4}\).
 

allegansveritatem

Full Member
Joined
Jan 10, 2018
Messages
343
I think you are missing the whole point. \(\displaystyle |2+\bf{i}|=\sqrt{2^2+1^2}=\sqrt5 \) so that \(\displaystyle |2+\bf{i}|^2=5 \).
So \(\displaystyle \dfrac{1}{2+\bf{i}}=\dfrac{2-\bf{i}}{5}=\frac{2}{5}-\frac{1}{5}\bf{i}\)
You need to learn absolute value.
Here is another example: \(\displaystyle \Re e[(4-3\bf{i})^{-1}]=\frac{4}{25}\;\&\;\Im m[(4-3\bf{i})^{-1}]=\frac{3}{25}\)
yes, I see what you did here: You took the reciprocal of 1/4-3i which is 4+3i/25. I have been studying this morning more about the concept of absolute value and have a clearer idea of what is going on now. I've also just watched several videos on solving equations containing absolute values. Thanks for pointing out the fact that I need to learn (or review?) the basics.
 

pka

Elite Member
Joined
Jan 29, 2005
Messages
7,954
I have been studying this morning more about the concept of absolute value and have a clearer idea of what is going on now. I've also just watched several videos on solving equations containing absolute values. Thanks for pointing out the fact that I need to learn (or review?) the basics.
In general \(\displaystyle |a+b\bf{i}|=\sqrt{a^2+b^2} \) Example \(\displaystyle |6-8\bf{i}|=\sqrt{(6)^2+(-8)^2}=10\)

Now lets look the connection to conjugates.
\(\displaystyle (a+b\bf{i})\overline{(a+b\bf{i})}=(a+b\bf{i})(a-b\bf{i})\)
\(\displaystyle (a+b\bf{i})(a-b\bf{i})=(a^2-b^2\bf{i}^2)+(ab\bf{i}-ba\bf{i})\)
\(\displaystyle (a+b\bf{i})(a-b\bf{i})=(a^2+b^2)\)
\(\displaystyle |a+b\bf{i}|^2=(a^2+b^2)\)

Thus we have shown that \(\displaystyle |z|^2=z\cdot\overline{\,z\,}\)
 

allegansveritatem

Full Member
Joined
Jan 10, 2018
Messages
343
here is my problem: When I multiply (a+bi)(a+bi) I get (a^+2bi-b^2). I don't see how you can say multiplying the first two terms = (a+bi)(a-bi).
 

ksdhart2

Senior Member
Joined
Mar 25, 2016
Messages
1,039
here is my problem: When I multiply (a+bi)(a+bi) I get (a^+2bi-b^2). I don't see how you can say multiplying the first two terms = (a+bi)(a-bi).
Well, that's mostly correct (accounting for a few typos). It is true that \(\displaystyle (a + bi)(a + bi) = (a + bi)^2 = a^2 + 2abi - b^2\) ... but that's not what we have here. Did you notice the line over the second \(\displaystyle (a + bi)\)? Do you recall that the overline means "complex conjugate?" What is the complex conjugate of \(\displaystyle a + bi\)? Can you see why that means that the statement \(\displaystyle (a + bi)\overline{(a +bi)} = (a + bi)(a - bi)\) is just a tautology?
 
Top