Nonzero complex numbers

  • #1

Homework Statement


Consider 3 nonzero complex numbers $$z_1,z_2,z_3$$ each satisfying $$z^2=i \bar{z}$$. We are supposed to find $$z_1+z_2+z_3, z_1z_2z_3, z_1z_2+z_2z_3+z_3z_1$$.

The answers- 0, purely imaginary , purely real respectively.



Homework Equations




The Attempt at a Solution


I have no idea how to proceed. I tried to use the expansion for $$(a+b+c)^2$$ for them, but I am not getting anywhere. Please help. Thanks in advanced!
 

Answers and Replies

  • #2
BvU
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Asking you to show your work doesn't make sense for this expansion. Did you try to work out what ##z^2 = i\bar z ## means for the real and imaginary parts ##\alpha## and ##\beta## if you write ##z = \alpha + i\beta## ?
 
  • #3
Buzz Bloom
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Hi cr7:

I suggest that you work with the equation
z2 = i zbar .​
Try to understand the constraints on values of z that satisfy this equation. I suggest thinking about z in polar form. You also might find the following of some help:

Good luck.

Regards,
Buzz

PS. I have no idea why that PNG file got attached. Please ignore it.
 

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  • #4
BvU
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No typos in the problem statement ?
 
  • #5
haruspex
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z1+z2+z3, z1z2z3, z1z2+z2z3+z3z1
Do those three expressions remind you of any standard formulas?
 
  • #6
Yes @haruspex I got the answer .....the relation between coeffecients and roots....
 
  • #7
PeroK
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Homework Statement


Consider 3 nonzero complex numbers $$z_1,z_2,z_3$$ each satisfying $$z^2=i \bar{z}$$. We are supposed to find $$z_1+z_2+z_3, z_1z_2z_3, z_1z_2+z_2z_3+z_3z_1$$.

The answers- 0, purely imaginary , purely real respectively.

Two things. The question ought to say that ##z_1, z_2, z_3## are distinct. Also, can you not get much a much more specific answer than that?
 
  • #8
BvU
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Can someone help me out ? If I follow my own advice (#2) and solve, I get pure imaginary, 0, pure real, respectively (hence my #4, and indeed, the three roots must be distinct to get that).
 
  • #9
PeroK
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Can someone help me out ? If I follow my own advice (#2) and solve, I get pure imaginary, 0, pure real, respectively (hence my #4, and indeed, the three roots must be distinct to get that).

This was definitely one to use polar form.
 
  • #10
BvU
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Same thing. Still I get an imaginary number as sum, zero as a product and one real nonzero term in sum of dyadic products ... :oldconfused:
 
  • #11
PeroK
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Same thing. Still I get an imaginary number as sum, zero as a product and one real nonzero term in sum of dyadic products ... :oldconfused:

If ##z^2 = i \bar{z}## then, first, we get ##|z| = 1## and then ##z^3 = i##.
 
  • #12
BvU
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So ##\cos{\pi\over 6} + i\sin {\pi\over 6},\ \ 0,\ \ \cos{5\pi\over 6} + i\sin {5\pi\over 6}##.
Sum imaginary, product zero, and sum of dyadic products real, respectively ... :oldconfused:
 
  • #13
PeroK
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So ##\cos{\pi\over 6} + i\sin {\pi\over 6},\ \ 0,\ \ \cos{5\pi\over 6} + i\sin {5\pi\over 6}##.
Sum imaginary, product zero, and sum of dyadic products real, respectively ... :oldconfused:

I just read off the coefficients of ##z^3 - i = (z-z_1)(z-z_2)(z-z_3)##

##z = 0## is a fourth solution, but the problem states that the ##z_i## are non-zero. There's your problem!
 
  • #14
Ray Vickson
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So ##\cos{\pi\over 6} + i\sin {\pi\over 6},\ \ 0,\ \ \cos{5\pi\over 6} + i\sin {5\pi\over 6}##.
Sum imaginary, product zero, and sum of dyadic products real, respectively ... :oldconfused:

No: from ##z^2 = i \bar{z}## it follows that either ##z=0## or else ##|z| = 1## and then ##z^3 = i##. There are three nonzero roots of ##z^3 = i##.
 
Last edited:
  • #15
BvU
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Finally got it ..
o:) Thanks guys !
 

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