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

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In summary, Homework Equations z1, z2 are complex numbers. If z1z2 =/= -1 and |z1| = |z2| = 1, then number : z1 + z2.
  • #36
I haven't learn using e yet, is it possible to do it with r(cosx+isinx) ?
 
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  • #37
Dank2 said:
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)
 
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  • #38
Sahil Kukreja said:
I have solved it
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It is a violation of PF rules for you to post this. You are not supposed to present solutions!
 
  • #39
Dank2 said:
I haven't learn using e yet, is it possible to do it with r(cosx+isinx) ?[/QUOT
Sahil Kukreja said:
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)
BvU said:
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 ... :smile:.
it's just 4 terms that needs to be organized, cause we don't matter about the real part
 
  • #40
Ray Vickson said:
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
 
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  • #41
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'
 
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  • #42
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
 
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  • #43
Sahil Kukreja said:
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'.
 
  • #44
Dank2 said:
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 ##
 
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  • #45
Sahil Kukreja said:
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.
 
  • #46
Dank2 said:
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 o:) ) , it's good to keep an eye open for alternatives, and Sahil had no trouble pointing out a very good one !
 
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  • #47
Dank2 said:
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.
 
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  • #48
Sahil Kukreja said:
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
 
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