Complex numbers

[winged elm]

Find two complex numbers on the unit circle that their sum is equal to their multiply
Thank you for helping!!

[candid atlas]

asking for all possible pairs of complex numbers that satisfy this condition turned out to be a surprisingly difficult problem

that I couldn't brute force

[scarlet depot]

You seek two complex numbers, call them $z$ and $w$, such that

$|z| = |w| = 1$

$zw = z + w$

Use the second fact to write $w$ in terms of $z$,

then choose a suitable $z$ and find the corresponding $w$.

[vivid valley]

Jay

[candid atlas]

but if you're just looking for "a" pair then might I propose a third condition, that would make your math a lot easier:

$z_1 = \bar z_2$

[vivid valley]

HChan

[candid atlas]

if you're wondering how I got this, by the way, here's the justification:

notice that any number added to its conjugate is real, and any number multiplied by its conjugate is real.
However, since z1 and z2 are already on the unit circle to begin with, their multiple must also be on the unit circle and so must be +-1
Finally, notice that 1+1 = 2 and -1 + -1 = -2, so by continuity there must exist a complex number z between -1 and 1 (i.e. with phase between 0 and 180) such that z plus its conjugate is +-1

[winged elm]

Thank you so much!!

Ill try it now

Is there any algebrical way to prove that

$z_1 = \bar z_2$

[vivid valley]

IsrAlisa

[candid atlas]

you're only looking for one pair, right?

not all pairs?

[winged elm]

In the problem they wrote find "two" numbers
So I guess so
But if there is a way to prove it in general like for all the pairs it will be better I think

I don't know if my teacher will accept word explanation

[candid atlas]

I could give you my proof, but I ended up having to use a pretty geometrical argument

i.e. a looot of words

but that's what a proof looks like

[winged elm]

Maybe it will help me somehow thank you!

[winged elm]

Better just prove it cis or something😂

[candid atlas]

here goes:

If I have z1, z2 has to be within 180 degrees of z1 either to the left or to the right

so, without loss of generality, let's assume it's to the left

then, because of how complex addition works, notice that the points {0, z1, z2, z1 + z2} necessarily form the corners of a rhombus

if I add the requirement that z1 + z2 must be on the unit circle as well, then the rhombus becomes a diamond (i.e two equilateral triangles stuck together)

so what that tells us immediately, is that arg(z2) = 120 + arg(z1)

AND that arg(z1 + z2) = 60 + arg(z1)

however, because of how complex multiplication works, I also know that arg(z1 + z2) = arg(z1 * z2) = arg(z1) + arg (z2)

so now we can set up two simeltaneous equations, and solve for arg(z1) to find that it is equal to -60 degrees

and so arg(z2) = -60 + 120 = 60 degrees

and so that is the only possible solution pair

[winged elm]

Thank you very much! I'll try it now

[winged elm]

@candid atlas can you maybe help me prove arg(z1+z2)=60+arg(z1) algebraically?
I managed to prove algebraically arg(z1)=+-120+arg(z2) I can send you...
Thank you!

[candid atlas]

oh sure, send it here

but if you have that arg(z1) = +-120 + arg(z2) you should be able to derive the first one relatively easily

because x+y is always on the angle bisector of x and y for all complex numbers x and y IF x and y have the same magnitude

[winged elm]

I'll call z1 as cis(a) and z2 as cis(b)
So |cis(a)+cis(b)|=1 (since they gave to be on the unit circle)
(Cos(a)+cos(b))^2 + (sin(a)+sin(b))^2 =1
We know that cos()^2+sin()^2=1
Which leads as to cos(a)cos(b)+sin(a)sin(b)=-0.5
And cos(a)cos(b)+sin(a)sin(b) is cos(a-b)
So cos(a-b)=-0.5
Rcos(-0.5)=120
So a-b=+-120+360k

[candid atlas]

what do you mean by cis(a)

what is cis()

[winged elm]

Cis is cos + sin*i

Like cis(a) it's cos(a) + sin(a)i

e^ia

[winged elm]

@candid atlas I have proved it🥳
But if arg(60) and arg(-60) the conditions are not met

[candid atlas]

wdym?

[candid atlas]

also sorry I'll check this proof later

[winged elm]

OK no worries thank you!

[winged elm]

We have z1z2=z1+z2
But If arg(z2)=60, arg(z1)=-60
Isn't give as z1z2=z1+z2

[candid atlas]

it is the solution, though..?

[winged elm]

But isn't the solution have to work?

[candid atlas]

it does

can you show me what you got for z1*z2 and z1+z2

[winged elm]

So z1 is e^i60 and z2 is e^i-60

Z1+z2= 1+0

Z1*z2=0.25-0.75i

[candid atlas]

that's definitely not true

a number times its conjugate is real

and z1 is definitely the conjugate of z2

here's a hint, complex multiplication is a lot easier in polar form

and you already have z1 z2 in polar form

[winged elm]

Oh wait it should be 1

Omg I did a really stupid mistake
Thank you so much for helping!!

@candid atlas thank you very much! You helped me a lot!
Really hope I hope I didn't piss you off too much
And happy new year 🎉)

[candid atlas]

You’re good