Hard Complex Numbers question

Have you tried for just the first equality, by multiplying numerator and denominator of [MATH]\frac{w+z}{w-z}[/MATH] by the conjugate of [MATH]w-z[/MATH]? Show your work on that step (not using the exponential form at all yet), and then we can move on from there.
 
No. How did you get that? Did you do what I suggested, using conjugates?

What I said was to start with [MATH]\frac{w+z}{w-z}\cdot\frac{\bar{w}-\bar{z}}{\bar{w}-\bar{z}}[/MATH].

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 [MATH]z[/MATH] as [MATH]x + iy[/MATH] and [MATH]w[/MATH] as [MATH]u+iv[/MATH], where [MATH]x, y, u, v[/MATH] are real.
 
Dr.Peterson said:
No. How did you get that? Did you do what I suggested, using conjugates?

What I said was to start with [MATH]\frac{w+z}{w-z}\cdot\frac{\bar{w}-\bar{z}}{\bar{w}-\bar{z}}[/MATH].

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 [MATH]z[/MATH] as [MATH]x + iy[/MATH] and [MATH]w[/MATH] as [MATH]u+iv[/MATH], where [MATH]x, y, u, v[/MATH] are real.
Click to expand...
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
 
Please take it one step at a time. Show me your work to get from [MATH]Re\left(\frac{w+z}{w-z}\right)[/MATH] to [MATH]\frac{|w|^2-|z|^2}{|w-z|^2}[/MATH]. Then we can move on from there.
 
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
 
jasmeetcolumbia98 said:
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)
Click to expand...
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: [MATH]\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})}[/MATH]
First, the denominator is easier than you are seeing; it's just [MATH](w-z)(\bar{w}-\bar{z}) = (w-z)\overline{(w-z)}[/MATH], which takes you directly to your goal.

As for the numerator, you have [MATH](w+z)(\bar{w}-\bar{z})[/MATH]. Fully expand this as is, and one part will be obvious; split up the non-obvious terms as [MATH]\bar{w}z-w\bar{z} = (u-iv)(x+iy) - (u+iv)(x-iy)[/MATH], fully expand that, and split it into real and imaginary parts. Then you can drop the imaginary part.
 
(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
 
jasmeetcolumbia98 said:
(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
Click to expand...
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 \(|w|^2-|z|^2\)
 
pka said:
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 \(|w|^2-|z|^2\)
Click to expand...
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
 
jasmeetcolumbia98 said:
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
Click to expand...
What you are calling rho ([MATH]\rho[/MATH]) is what they called phi ([MATH]\phi[/MATH]).

The numerator is not [MATH]w^2 - z^2[/MATH], but [MATH]|w|^2 - |z|^2[/MATH]. But that is exactly [MATH]R^2 - r^2[/MATH], so maybe you didn't mean what you wrote. You do see that [MATH]|w| = R[/MATH], 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 [MATH]|(u+iv)-(x+iy)|^2 = |(u-x)+i(v-y)|^2 = (u-x)^2+(v-y)^2[/MATH]. Expand this, and you'll find the [MATH]R^2 + r^2[/MATH] part, together with a middle term. Express that in terms of sines and cosines, and you should recognize an angle-difference identity.
 
Dr.Peterson said:
What you are calling rho ([MATH]\rho[/MATH]) is what they called phi ([MATH]\phi[/MATH]).

The numerator is not [MATH]w^2 - z^2[/MATH], but [MATH]|w|^2 - |z|^2[/MATH]. But that is exactly [MATH]R^2 - r^2[/MATH], so maybe you didn't mean what you wrote. You do see that [MATH]|w| = R[/MATH], 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 [MATH]|(u+iv)-(x+iy)|^2 = |(u-x)+i(v-y)|^2 = (u-x)^2+(v-y)^2[/MATH]. Expand this, and you'll find the [MATH]R^2 + r^2[/MATH] part, together with a middle term. Express that in terms of sines and cosines, and you should recognize an angle-difference identity.
Click to expand...
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
 
jasmeetcolumbia98 said:
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
Click to expand...
At this point, as I said, you can use the fact that [MATH]x = r\cos(\theta)[/MATH], [MATH]y = r\sin(\theta)[/MATH], 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?
 
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
 
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))
 
jasmeetcolumbia98 said:
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
Click to expand...
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.
jasmeetcolumbia98 said:
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))
Click to expand...
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.
 
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