# 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