# Proof that the number is Real

## 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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BvU
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2019 Award
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
BvU
Homework Helper
2019 Award
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

BvU
Homework Helper
2019 Award
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

Last edited:
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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BvU
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2019 Award
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?

BvU
Homework Helper
2019 Award
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?

BvU
Homework Helper
2019 Award
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

BvU
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2019 Award
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 ;(

BvU
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2019 Award
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
BvU
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2019 Award
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
BvU
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2019 Award
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

BvU
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2019 Award
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