So the shortest title I can come up with is: Complex Numbers and Real Solutions

[Dank2]

Homework Statement

z₁, z₂ are complex numbers.
If z₁z₂ =/= -1
and |z₁| = |z₂| = 1

then number :
z₁ + z₂
________
1 + z₁z₂

is real.

Homework Equations

The Attempt at a Solution

z1 = (a+bi), z2 = (c+di)[/B]
Should i use this extended form or is there a shorter path ? because it's pretty long.

 

[BvU]

Hi Dank,

Pity there are no relevant equations available to you. If you want a shorter path you might need something to work with. One thing I can think of is that for any complex number $\bf\alpha$ the form $\bf\alpha^*\alpha$ is real number. But in this case the given $|{\bf z}|$ seems to point in the direction of another kind of notation than the one you propose

(he said, mysterically, because he didn't want to give it all away so easily)

 

[BvU]

Does that signify you picked up the hint and found a fairly short path indeed ?

 

[Dank2]

Ohh

Hi Dank,

Pity there are no relevant equations available to you. If you want a shorter path you might need something to work with. One thing I can think of is that for any complex number $\bf\alpha$ the form $\bf\alpha^*\alpha$ is real number. But in this case the given $|{\bf z}|$ seems to point in the direction of another kind of notation than the one you propose

(he said, mysterically, because he didn't want to give it all away so easily)

Sorry it was z1 + z2 and not z1 + z1, and also i marked * as multiplication in mistake and edited.

So if i take the absolute value of the fraction i get : |z1+z2| over |1+z1z2|
, can't get much further here using the data

 

[Dank2]

Does that signify you picked up the hint and found a fairly short path indeed ?

nope, since i had a typo and now it's fixed. so i bet the hints will change ;D

 

[BvU]

It is z1 + z2 and not z1 + z1

That was clear. And the * in the denominator is indeed multiplication. The hints stay as they were.
But going for absolute values isn't productive: basically you don't care about them if you want to show something has imaginary part zero.

So I'll repeat the nudge: |z| = 1 should inspire a different notation : ${\bf z} = \alpha + i\beta$ has two unknowns. The notation I mean has only one

 

[Dank2]

That was clear. And the * in the denominator is indeed multiplication. The hints stay as they were.
But going for absolute values isn't productive: basically you don't care about them if you want to show something has imaginary part zero.

So I'll repeat the nudge: |z| = 1 should inspire a different notation : ${\bf z} = \alpha + i\beta$ has two unknowns. The notation I mean has only one

 

[Dank2]

Ok i think ill try somthing and post here in few min

 

[Dank2]

That was clear. And the * in the denominator is indeed multiplication. The hints stay as they were.
But going for absolute values isn't productive: basically you don't care about them if you want to show something has imaginary part zero.

So I'll repeat the nudge: |z| = 1 should inspire a different notation : ${\bf z} = \alpha + i\beta$ has two unknowns. The notation I mean has only one

Taking the conjugate of all the fraction might yield something meaningful? but if i then multiply it with the fraction ill get a real number, but that doesn't show me why z is real still...

 

[Dank2]

That was clear. And the * in the denominator is indeed multiplication. The hints stay as they were.
But going for absolute values isn't productive: basically you don't care about them if you want to show something has imaginary part zero.

So I'll repeat the nudge: |z| = 1 should inspire a different notation : ${\bf z} = \alpha + i\beta$ has two unknowns. The notation I mean has only one

I'm afraid i need another hint,

 

[BvU]

You loose a few brownie points, but what the heck: what is the locus of points $z_1$ and $z_2$ in the complex plane ?

 

[Dank2]

You loose a few brownie points, but what the heck: what is the locus of points $z_1$ and $z_2$ in the complex plane ?

Locus? maybe there is a different name for it, I'm studying linear algebra 1. hmm vectors you mean? or circle unit?

 

[BvU]

Unit circle is the name. And yes, locus is the collection of positions $z_i$ can occupy in the complex plane.
One way to pinpoint a complex number is by Re() and Im(), real and imaginary parts (your $a$ and $b$ ).
Another way is with Abs() and Arg(), absolute value and angle . Since in this exercise $Abs(z_i) = 1$ that's what you want to adopt !

 

[Dank2]

Unit circle is the name. And yes, locus is the collection of positions $z_i$ can occupy in the complex plane.
One way to pinpoint a complex number is by Re() and Im(), real and imaginary parts (your $a$ and $b$ ).
Another way is with Abs() and Arg(), absolute value and angle . Since in this exercise $Abs(z_i) = 1$ that's what you want to adopt !

Can write z1 = 1(cosx+isinx), and z2 = 1(cosy+isiny), where r = 1?
and then plug it in the fraction?

 

[BvU]

You can indeed. A short form for $\cos\phi + i\sin\phi$ is also available (saves writing effort ).

 

[Dank2]

You can indeed. A short form for $\cos\phi + i\sin\phi$ is also available (saves writing effort ).

cosx + isinx + cosy + isiny over 1 +cos(x+y) + isin(x+y) +1

 

[BvU]

Hehe, you reluctant to write $$
e^{i\phi_1} + e^{i\phi_2}\over 1+e^{i\left ( \phi_1 +\phi_2\right )} $$ or something ?

Now: what's the road ahead if you want to show this fraction is a real number ?

 

[Dank2]

Hehe, you reluctant to write $$
e^{i\phi_1} + e^{i\phi_2}\over 1+e^{i\left ( \phi_1 +\phi_2\right )} $$ or something ?

Now: what's the road ahead if you want to show this fraction is a real number ?

Haven't learned using e at all yet, my first course in linear algebra 1.

 

[Dank2]

Haven't learned using e at all yet, my first course in linear algebra 1.

well, i would want the imaginary part to be gone from the fraction

but sinx+siny =/= sin(x+y)i ;(

 

[BvU]

Oh, sorry. Well, in that case my respect that you got this far already !
We proceed with sin and cos. What's the way to ensure/make the denominator is real (remember post # 2)

 

[BvU]

Rapid fire postings ! We are at $$\cos\phi_1 +i\sin\phi_1 \ + \ \cos\phi_2 +i\sin\phi_2 \over 1 + \cos\left( \phi_1 + \phi_2 \right ) +i\sin\left( \phi_1 + \phi_2 \right ) $$ and we want to work the denominator around to something that is a real number (so we can forget it). Agreed ?

 

[BvU]

well, i would want the imaginary part to be gone from the fraction

but sinx+siny =/= sin(x+y)i ;(

Wouldn't help anyway: only would make the imaginary parts equal, but doesn't say anything about the imaginary part of the fraction!

 

[Dank2]

can i take conjugate of all the fraction?
on the bottom i'd get though sin(-x-y)i =/= sin(x+y)i

 

[Dank2]

Rapid fire postings ! We are at $$\cos\phi_1 +i\sin\phi_1 \ + \ \cos\phi_2 +i\sin\phi_2 \over 1 + \cos\left( \phi_1 + \phi_2 \right ) +i\sin\left( \phi_1 + \phi_2 \right ) $$ and we want to work the denominator around to something that is a real number (so we can forget it). Agreed ?

agreed

 

[BvU]

Not all the fraction ! Then you get something that is definitely real and non-negative, but is not equal to the fraction and has no information on the imaginary part of the fraction any more.

No, you multiply with 1 . Namely complex conjugate of denominator divided by complex conjugate of denominator !

And there would be a $\ -i\sin\left( \phi_1 + \phi_2 \right )$ in there indeed.

Note that you don't have to work ut the denominator any further: you know it's real, so that you can concentrate on the numerator.

 

[Dank2]

Not all the fraction ! Then you get something that is definitely real and non-negative, but is not equal to the fraction and has no information on the imaginary part of the fraction any more.

No, you multiply with 1 . Namely complex conjugate of denominator divided by complex conjugate of denominator !

And there would be a $\ -i\sin\left( \phi_1 + \phi_2 \right )$ in there indeed.

Note that you don't have to work ut the denominator any further: you know it's real, so that you can concentrate on the numerator.

Ok i didn't get any of what you just said hehe. If i take the whole fraction conjugate, tries to make equality between them(conjugate and non conjugate), then it's real number, i mean if there is equality , no ?
z* = z <==> z = real

 

[BvU]

I see what you mean. But I expect you'll have a hard time to show that $z^* = z$...

My proposal was to take the fraction, multiply with 1 in such a way that the denominator becomes a real number and then try to show that the numerator is a real number as well:$$ {\alpha\over \beta} = {\beta^*\alpha \over \beta^*\beta}$$ is real if ${\beta^*\alpha}$ is real. ($\ \ ^*$ means complex conjugate!)

 

[Dank2]

I see what you mean. But I expect you'll have a hard time to show that $z^* = z$...

My proposal was to take the fraction, multiply with 1 in such a way that the denominator becomes a real number and then try to show that the numerator is a real number as well:$$ {\alpha\over \beta} = {\beta^*\alpha \over \beta^*\beta}$$ is real if ${\beta^*\alpha}$ is real. ($\ \ ^*$ means complex conjugate!)

As for my suggestion, you need only to show that -six-siny = sinx + siny, and sin(x+y) = sin(-x-y)

As for your suggestion, you mean should i multiply Numerator and Denominator by 1+cos(x+y) - sin(x+y)i ? wouldn't it become mess in the numerator?

 

[BvU]

agreed

He, how did you so quickly agree to $$
\left (\cos\phi_1 +i\sin\phi_1 \right ) \left ( \cos\phi_2 +i\sin\phi_2 \right ) = \cos\left( \phi_1 + \phi_2 \right ) +i\sin\left( \phi_1 + \phi_2 \right ) \ \ \rm ?$$

As for my suggestion, you need only to show that -six-siny = sinx + siny, and sin(x+y) = sin(-x-y)

That's going to be difficult, I guess ?

wouldn't it become mess in the numerator

That's why I preferred the $e^{i\phi}$ notation -- it gave me a pretty quick result. Now it's a bit more involved (and I didn't write it out in full yet )

 

[Dank2]

the imaginary part : sinx + siny + cosxsin(x+y) -cosysin(x+y) + sinxcos(x+y) +sinycos(x+y) = sin (2(x+y)) -sin(0?) + sinx + siny.

sinx + siny should be add to 1 right ? since the length is 1. how is that looks?

seem like it might go to zero

 

[Sahil Kukreja]

is z1.z2 =1 or -1 ??
if z1.z2 = -1 then the number is not defined !

 

[Dank2]

is z1.z2 =1 or -1 ??
if z1.z2 = -1 then the number is not defined !

=/= not =

 

[Dank2]

Bump, problem still not solved.

 

[BvU]

Work to do, therefore ! It's almost more efficient to teach you this $e^{i\phi}$ than to drudge through all this cos sin and such...
But I can be hired to do your work if you pay better than my boss ... .

 

[Sahil Kukreja]

put z1= cosa + isina
z2= cosb + isinb
simplify

 

[Dank2]

I haven't learn using e yet, is it possible to do it with r(cosx+isinx) ?

 

[Sahil Kukreja]

I haven't learn using e yet, is it possible to do it with r(cosx+isinx) ?

In other questions e can be useful, but in this question you can use both the ways and get the answer in same time( actually same efficiency)

 

[Ray Vickson]

I have solved it

It is a violation of PF rules for you to post this. You are not supposed to present solutions!

 

[Dank2]

I haven't learn using e yet, is it possible to do it with r(cosx+isinx) ?[/QUOT

In other questions e can be useful, but in this question you can use both the ways and get the answer in same time( actually same efficiency)

Work to do, therefore ! It's almost more efficient to teach you this $e^{i\phi}$ than to drudge through all this cos sin and such...
But I can be hired to do your work if you pay better than my boss ... .

it's just 4 terms that needs to be organized, cause we don't matter about the real part

 

[Sahil Kukreja]

It is a violation of PF rules for you to post this. You are not supposed to present solutions!

Sorry. I have deleted the solution now

 

[Sahil Kukreja]

I have another method to solve and it is a lot easier:- (it just solves in three steps)

Hint:-
z' --> conjugate of z
if z-z' = 0 then z is purely real

also use that since |z1|=|z2|=1
then z1=1/z1' and z2=1/z2'

 

[Sahil Kukreja]

Now, I have a third method to solve :-
since |z1|=|z2|=1
then z1=1/z1' and z2=1/z2'

put this in the equation and use your previous knowledge to get the answer

 

[Dank2]

I have another method to solve and it is a lot easier:- (it just solves in three steps)

Hint:-
z' --> conjugate of z
if z-z' = 0 then z is purely real

also use that since |z1|=|z2|=1
then z1=1/z1' and z2=1/z2'

Very nice! it came so smooth, thank you sir, and also, for all the others that helped . i didn't even use z1 =1/z1'.

 

[Sahil Kukreja]

Very nice! it came so smooth, thank you sir, and also, for all the others that helped . i didn't even use z1 =1/z1'.

its good to remember that z.z' = $|z|^2$

 

[Dank2]

its good to remember that z.z' = $|z|^2$

yes i followed that, and all the terms in the numerator left was z1-z1' + z2-z2' which is ofc real, and denominator was 1+z1z2 + conjugate(z1z2), which is also real.

 

[BvU]

Very nice! it came so smooth, thank you sir, and also, for all the others that helped . i didn't even use z1 =1/z1'.

I learned from this too : even though you have found a way through (and I thought the $e^{i\phi}$ was pretty efficient ) , it's good to keep an eye open for alternatives, and Sahil had no trouble pointing out a very good one !

 

[Sahil Kukreja]

yes i followed that, and all the terms in the numerator left was z1-z1' + z2-z2' which is ofc real, and denominator was 1+z1z2 + conjugate(z1z2), which is also real.

z1-z1' + z2-z2' is not purely real, its purely imaginary.
z=(z1+z2)/(1+z1z2)
if you put z1=1/z1' and z2=1/z2' then you must have gotten z=z'
which implies z is purely real.

 

[Dank2]

z1-z1' + z2-z2' is not purely real, its purely imaginary.
z=(z1+z2)/(1+z1z2)
if you put z1=1/z1' and z2=1/z2' then you must have gotten z=z'
which implies z is purely real.

Ok , now i know that i had to use that z1=1/z1'

so it came up 1/z'+1/z+1/z2+1/z2', and i forgot to add |z1z2|^2 at the denominator in message #45