Hard Complex Numbers question

jasmeetcolumbia98

New member
Joined
May 15, 2020
Messages
33
I expanded Re((w+z)/(w-z))
lets say ((w+z)/(w-z)) = b
bcos(argb)
but I don't know where to go from here
 

Dr.Peterson

Elite Member
Joined
Nov 12, 2017
Messages
7,558
Have you tried for just the first equality, by multiplying numerator and denominator of \(\displaystyle \frac{w+z}{w-z}\) by the conjugate of \(\displaystyle w-z\)? Show your work on that step (not using the exponential form at all yet), and then we can move on from there.
 

jasmeetcolumbia98

New member
Joined
May 15, 2020
Messages
33
Ahhh so that would make it
((w+z)^2) / ((w^2)+(z^2))
 

Dr.Peterson

Elite Member
Joined
Nov 12, 2017
Messages
7,558
No. How did you get that? Did you do what I suggested, using conjugates?

What I said was to start with \(\displaystyle \frac{w+z}{w-z}\cdot\frac{\bar{w}-\bar{z}}{\bar{w}-\bar{z}}\).

Then you'll have to identify terms that are known to be real, and others that are pure imaginary. At some point you may want to identify \(\displaystyle z\) as \(\displaystyle x + iy\) and \(\displaystyle w\) as \(\displaystyle u+iv\), where \(\displaystyle x, y, u, v\) are real.
 

jasmeetcolumbia98

New member
Joined
May 15, 2020
Messages
33
No. How did you get that? Did you do what I suggested, using conjugates?

What I said was to start with \(\displaystyle \frac{w+z}{w-z}\cdot\frac{\bar{w}-\bar{z}}{\bar{w}-\bar{z}}\).

Then you'll have to identify terms that are known to be real, and others that are pure imaginary. At some point you may want to identify \(\displaystyle z\) as \(\displaystyle x + iy\) and \(\displaystyle w\) as \(\displaystyle u+iv\), where \(\displaystyle x, y, u, v\) are real.
I've almost solved it I have the right numerator and most of the denominator
I don't understand where you get cos(theta-rho) from
 

Dr.Peterson

Elite Member
Joined
Nov 12, 2017
Messages
7,558
Please take it one step at a time. Show me your work to get from \(\displaystyle Re\left(\frac{w+z}{w-z}\right)\) to \(\displaystyle \frac{|w|^2-|z|^2}{|w-z|^2}\). Then we can move on from there.
 

jasmeetcolumbia98

New member
Joined
May 15, 2020
Messages
33
Okay so I multiplied out the conjugates and (w+z) / (w-z)
that got me w¯w -z¯z for the numerator which is equivalent to |w|^2 - |z|^2
but then when I multiply out the denominator I get
w¯w +z¯z -z¯w-w¯z
then I took w= u+iv and z=x+iy

for the numerator
(u+iy)(u-iy)=u^2 if we ignore imaginary terms likewise we get x^2 for |z|^2
so Re(w+z)= u^2 -x^2 for the numerator essentially
but for the denominator
I immediately see w¯w +z¯z which from the numerator would be u^2 -x^2 in real terms
but then we have
-z¯w-w¯z which is like -(u+iv)(x-iy)-(x+iy)(u-iv)
for real terms we get -2ux
z=re^i*theta
w=Re^i*rho
would e^(i) be the imaginary part so we exclude this
u=re*theta and x=Re^rho
not sure at this point
 

Dr.Peterson

Elite Member
Joined
Nov 12, 2017
Messages
7,558
Okay so I multiplied out the conjugates and (w+z) / (w-z)
that got me w¯w -z¯z for the numerator which is equivalent to |w|^2 - |z|^2
but then when I multiply out the denominator I get
w¯w +z¯z -z¯w-w¯z
then I took w= u+iv and z=x+iy

for the numerator
(u+iy)(u-iy)=u^2 if we ignore imaginary terms likewise we get x^2 for |z|^2
so Re(w+z)= u^2 -x^2 for the numerator essentially
but for the denominator
I immediately see w¯w +z¯z which from the numerator would be u^2 -x^2 in real terms
but then we have
-z¯w-w¯z which is like -(u+iv)(x-iy)-(x+iy)(u-iv)
There are several errors here. Some things you wrote look like typos; for example, surely you didn't mean (u+iy), for example. Also, don't ignore imaginary terms too early.

I'll start at the beginning: \(\displaystyle \frac{w+z}{w-z}\cdot\frac{\bar{w}-\bar{z}}{\bar{w}-\bar{z}} = \frac{(w+z)(\bar{w}-\bar{z})}{(w-z)(\bar{w}-\bar{z})}\)

First, the denominator is easier than you are seeing; it's just \(\displaystyle (w-z)(\bar{w}-\bar{z}) = (w-z)\overline{(w-z)}\), which takes you directly to your goal.

As for the numerator, you have \(\displaystyle (w+z)(\bar{w}-\bar{z})\). Fully expand this as is, and one part will be obvious; split up the non-obvious terms as \(\displaystyle \bar{w}z-w\bar{z} = (u-iv)(x+iy) - (u+iv)(x-iy)\), fully expand that, and split it into real and imaginary parts. Then you can drop the imaginary part.
 

jasmeetcolumbia98

New member
Joined
May 15, 2020
Messages
33
(u−iv)(x+iy)−(u+iv)(x−iy) expanding alongside w and its conjunction also z and its conjunction this got me to
2iuy - 2ivx + u^2 + v^2 + x^2 + y^2
Real terms: u^2 + v^2 +x^2 + y^2
 

pka

Elite Member
Joined
Jan 29, 2005
Messages
9,809
(u−iv)(x+iy)−(u+iv)(x−iy) expanding alongside w and its conjunction also z and its conjunction this got me to 2iuy - 2ivx + u^2 + v^2 + x^2 + y^2
Real terms: u^2 + v^2 +x^2 + y^2
It clear that you are exhibiting some confusion here.
Do you understand that for all numbers \(z\cdot \overline{z}=|z|^2\) which is a real number.
Thus the denominator is \((w-z)(\overline{w}-\overline{z})=(w-z)\overline{(w-z)}=|w-z|^2\), which is real.
But your real confusion is in the numerator. I will use your notation.
Suppose that \(w=u+v{\bf{i}}~\&~z=x+y{\bf{i}}\) then we want
\([u+v{\bf{i}}]+[x+y{\bf{i}}])([u-v{\bf{i}}]-[x-y{\bf{i}}])\)
Please, please look into this LINK
Scroll down to the section labeled Expanded form and there see:
\(u^2 + v^2 - 2 i v x - x^2 + 2 i u y - y^2\) which is \((u^2+v^2)-(x^2+y^2)-2{\bf{i}}(ux+vy)\)
The real part of which is \(|w|^2-|z|^2\)
 

jasmeetcolumbia98

New member
Joined
May 15, 2020
Messages
33
It clear that you are exhibiting some confusion here.
Do you understand that for all numbers \(z\cdot \overline{z}=|z|^2\) which is a real number.
Thus the denominator is \((w-z)(\overline{w}-\overline{z})=(w-z)\overline{(w-z)}=|w-z|^2\), which is real.
But your real confusion is in the numerator. I will use your notation.
Suppose that \(w=u+v{\bf{i}}~\&~z=x+y{\bf{i}}\) then we want
\([u+v{\bf{i}}]+[x+y{\bf{i}}])([u-v{\bf{i}}]-[x-y{\bf{i}}])\)
Please, please look into this LINK
Scroll down to the section labeled Expanded form and there see:
\(u^2 + v^2 - 2 i v x - x^2 + 2 i u y - y^2\) which is \((u^2+v^2)-(x^2+y^2)-2{\bf{i}}(ux+vy)\)
The real part of which is \(|w|^2-|z|^2\)
Awesome I've just worked it out on paper
I have IwI^2 -IzI^2 on the numerator
Iw-zI^2 on the denominator
for the numerator w=Re^i*rho
w^2 = (R^2) e^2i*rho
would we only consider R^2 here because the nat log part is imaginary
if so
w^2 - z^2 for the numerator is R^2 - r^2
for the denominator
(Re^i*rho - re^I*theta)^2
R^2 + r^2 -2Rr
I basically don't understand where they get the (theta-rho bit from
 

pka

Elite Member
Joined
Jan 29, 2005
Messages
9,809
I basically don't understand where they get the (theta-rho bit from
Many, many times using the experiential form simplifies the workings.
However I doubt it helps here.
 

jasmeetcolumbia98

New member
Joined
May 15, 2020
Messages
33
Many, many times using the experiential form simplifies the workings.
However I doubt it helps here.
I don't really know what else to do
 

Dr.Peterson

Elite Member
Joined
Nov 12, 2017
Messages
7,558
Awesome I've just worked it out on paper
I have IwI^2 -IzI^2 on the numerator
Iw-zI^2 on the denominator
for the numerator w=Re^i*rho
w^2 = (R^2) e^2i*rho
would we only consider R^2 here because the nat log part is imaginary
if so
w^2 - z^2 for the numerator is R^2 - r^2
for the denominator
(Re^i*rho - re^I*theta)^2
R^2 + r^2 -2Rr
I basically don't understand where they get the (theta-rho bit from
What you are calling rho (\(\displaystyle \rho\)) is what they called phi (\(\displaystyle \phi\)).

The numerator is not \(\displaystyle w^2 - z^2\), but \(\displaystyle |w|^2 - |z|^2\). But that is exactly \(\displaystyle R^2 - r^2\), so maybe you didn't mean what you wrote. You do see that \(\displaystyle |w| = R\), right? Most of what you said about the numerator is unnecessary, if not wrong; in particular, "because the nat log part is imaginary" suggests you aren't thinking carefully (or else I'm just not seeing what you're thinking).

One way to handle the denominator is to revert to \(\displaystyle |(u+iv)-(x+iy)|^2 = |(u-x)+i(v-y)|^2 = (u-x)^2+(v-y)^2\). Expand this, and you'll find the \(\displaystyle R^2 + r^2\) part, together with a middle term. Express that in terms of sines and cosines, and you should recognize an angle-difference identity.
 

jasmeetcolumbia98

New member
Joined
May 15, 2020
Messages
33
What you are calling rho (\(\displaystyle \rho\)) is what they called phi (\(\displaystyle \phi\)).

The numerator is not \(\displaystyle w^2 - z^2\), but \(\displaystyle |w|^2 - |z|^2\). But that is exactly \(\displaystyle R^2 - r^2\), so maybe you didn't mean what you wrote. You do see that \(\displaystyle |w| = R\), right? Most of what you said about the numerator is unnecessary, if not wrong; in particular, "because the nat log part is imaginary" suggests you aren't thinking carefully (or else I'm just not seeing what you're thinking).

One way to handle the denominator is to revert to \(\displaystyle |(u+iv)-(x+iy)|^2 = |(u-x)+i(v-y)|^2 = (u-x)^2+(v-y)^2\). Expand this, and you'll find the \(\displaystyle R^2 + r^2\) part, together with a middle term. Express that in terms of sines and cosines, and you should recognize an angle-difference identity.
When we expand (u−x)^2+(v−y)^2
we get u^2 -2ux + x^2 +v^2 -2vy +y^2
u^2 + x^2 = R^2 + r^2
-2vy +y^2 +v^2 -2ux
Wouldn't I need to use the exponential form because
cos(theta) = (ei*theta + ei^-theta)/2
 

Dr.Peterson

Elite Member
Joined
Nov 12, 2017
Messages
7,558
When we expand (u−x)^2+(v−y)^2
we get u^2 -2ux + x^2 +v^2 -2vy +y^2
u^2 + x^2 = R^2 + r^2
-2vy +y^2 +v^2 -2ux
Wouldn't I need to use the exponential form because
cos(theta) = (ei*theta + ei^-theta)/2
At this point, as I said, you can use the fact that \(\displaystyle x = r\cos(\theta)\), \(\displaystyle y = r\sin(\theta)\), and so on. That is related to exponential form. In my mind, at this point my suggestion is more natural, but you can probably go explicitly to exponential form in some way. I haven't tried.

But the important thing is that it is not true that u^2 + x^2 = R^2 + r^2. Do you not see that?

What is R^2 equal to? What is r^2 equal to?
 

jasmeetcolumbia98

New member
Joined
May 15, 2020
Messages
33
Ah so can you say x^2 +y^2 = r^2 cos^2(theta) + r^2 sin^2 ( theta)
sin^2 + cos^2 =1
x^2 +y^2 = r^2
w=u+iv
so using the same principle
u^2 + v^2 = R^2
which sorts out the numerator
 

jasmeetcolumbia98

New member
Joined
May 15, 2020
Messages
33
Then for the denominator I listed
x=rcos(theta)
y=rsin(theta)
u=Rcos(thy)
v=Rsin(thy)
u^2 -2ux +x^2 +v^2 -2vy +y^2
u^2 +x^2 +v^2 + y^2 = r^2 cos^2(theta) + r^2 sin^2(theta) +R^2 cos^2(thy) + R^2 sin^2(thy) = R^2 + r^2
-2ux -2vy = -2(Rrcos(thy)*rcos(theta)) - 2(Rrsin(thy)*rsin(theta))
cos(theta-thy)=cos(theta)cos(thy)+sin(theta)sin(thy)
2Rrcos(theta-thy)=-2(Rrcos(thy)*rcos(theta)) + 2(Rrsin(thy)*rsin(theta))
 

Dr.Peterson

Elite Member
Joined
Nov 12, 2017
Messages
7,558
Ah so can you say x^2 +y^2 = r^2 cos^2(theta) + r^2 sin^2 ( theta)
sin^2 + cos^2 =1
x^2 +y^2 = r^2
w=u+iv
so using the same principle
u^2 + v^2 = R^2
which sorts out the numerator
Yes; or you can just recall that in exponential form, your r is the absolute value, so |z|^2 = r^2, with no intermediate work needed in the numerator.
Then for the denominator I listed
x=rcos(theta)
y=rsin(theta)
u=Rcos(thy)
v=Rsin(thy)
u^2 -2ux +x^2 +v^2 -2vy +y^2
u^2 +x^2 +v^2 + y^2 = r^2 cos^2(theta) + r^2 sin^2(theta) +R^2 cos^2(thy) + R^2 sin^2(thy) = R^2 + r^2
-2ux -2vy = -2(Rrcos(thy)*rcos(theta)) - 2(Rrsin(thy)*rsin(theta))
cos(theta-thy)=cos(theta)cos(thy)+sin(theta)sin(thy)
2Rrcos(theta-thy)=-2(Rrcos(thy)*rcos(theta)) + 2(Rrsin(thy)*rsin(theta))
You need to learn your Greek letters. It isn't "thy", but "phi". I'm guessing that you speak a language that doesn't strongly differentiate the "th" and "f" sounds ...

It looks like you got it now, so that quibble is all I can say about it ... except that you should check to see if you have too many r's in that last expression.
 
Top