# Proof that the number is Real

• Dank2
In summary, Homework Equations z1, z2 are complex numbers. If z1z2 =/= -1 and |z1| = |z2| = 1, then number : z1 + z2.f

## Homework Statement

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

then number :
z1 + z2
________
1 + z1z2

is real.

## 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.

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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)

Dank2
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

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

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

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

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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...

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,

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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 ?

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?

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 !

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?

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

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

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 ?

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.

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 ;(

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)

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 ?

Dank2
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!

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

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

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

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?

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 )

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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

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

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

Bump, problem still not solved.

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 ... .

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

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Dank2