# Nonzero complex numbers

## 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.

## 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!

BvU
Homework Helper
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## ?

Buzz Bloom
Gold Member
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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BvU
Homework Helper
No typos in the problem statement ?

haruspex
Homework Helper
Gold Member
2020 Award
z1+z2+z3, z1z2z3, z1z2+z2z3+z3z1
Do those three expressions remind you of any standard formulas?

Yes @haruspex I got the answer .....the relation between coeffecients and roots....

PeroK
Homework Helper
Gold Member
2020 Award

## 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?

BvU
Homework Helper
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).

PeroK
Homework Helper
Gold Member
2020 Award
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.

BvU
Homework Helper
Same thing. Still I get an imaginary number as sum, zero as a product and one real nonzero term in sum of dyadic products ...

PeroK
Homework Helper
Gold Member
2020 Award
Same thing. Still I get an imaginary number as sum, zero as a product and one real nonzero term in sum of dyadic products ...

If ##z^2 = i \bar{z}## then, first, we get ##|z| = 1## and then ##z^3 = i##.

BvU
Homework Helper
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 ...

PeroK
Homework Helper
Gold Member
2020 Award
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 ...

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!

Ray Vickson
Homework Helper
Dearly Missed
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 ...

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:
BvU