# Nonzero complex numbers

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1. Sep 15, 2016

### cr7einstein

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

3. 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!

2. Sep 15, 2016

### BvU

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. Sep 15, 2016

### Buzz Bloom

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. Sep 15, 2016

### BvU

No typos in the problem statement ?

5. Sep 15, 2016

### haruspex

Do those three expressions remind you of any standard formulas?

6. Sep 16, 2016

### cr7einstein

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

7. Sep 16, 2016

### PeroK

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. Sep 20, 2016

### BvU

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. Sep 20, 2016

### PeroK

This was definitely one to use polar form.

10. Sep 20, 2016

### BvU

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

11. Sep 20, 2016

### PeroK

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

12. Sep 20, 2016

### BvU

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 ...

13. Sep 20, 2016

### PeroK

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. Sep 20, 2016

### Ray Vickson

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: Sep 20, 2016
15. Sep 20, 2016

### BvU

Finally got it ..
Thanks guys !