#### jasmeetcolumbia98

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lets say ((w+z)/(w-z)) = b

bcos(argb)

but I don't know where to go from here

- Thread starter jasmeetcolumbia98
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lets say ((w+z)/(w-z)) = b

bcos(argb)

but I don't know where to go from here

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Ahhh so that would make it

((w+z)^2) / ((w^2)+(z^2))

((w+z)^2) / ((w^2)+(z^2))

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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.

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I've almost solved it I have the right numerator and most of the denominator

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 don't understand where you get cos(theta-rho) from

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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

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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.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)

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.

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2iuy - 2ivx + u^2 + v^2 + x^2 + y^2

Real terms: u^2 + v^2 +x^2 + y^2

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It clear that you are exhibiting some confusion here.

Real terms: u^2 + v^2 +x^2 + y^2

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 \(|w|^2-|z|^2\)

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Awesome I've just worked it out on paperIt 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 \(|w|^2-|z|^2\)

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

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Many, many times using the experiential form simplifies the workings.I basically don't understand where they get the (theta-rho bit from

However I doubt it helps here.

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I don't really know what else to doMany, many times using the experiential form simplifies the workings.

However I doubt it helps here.

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What you are calling rho (\(\displaystyle \rho\)) is what they called phi (\(\displaystyle \phi\)).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

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.

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When we expand (u−x)^2+(v−y)^2What 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.

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

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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.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

But the important thing is that

What

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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

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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))

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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.

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

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 ...

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))

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.