What are the properties of nonzero complex numbers satisfying z^2 = i\bar{z}?

In summary, the equation for z2 = i zbar can be solved for z in polar form and the three roots of z2 = i zbar are distinct.
  • #1
cr7einstein
87
2

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!
 
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  • #2
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
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
No typos in the problem statement ?
 
  • #5
cr7einstein said:
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
cr7einstein said:

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
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
BvU said:
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
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
BvU said:
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
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
BvU said:
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
BvU said:
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
Finally got it ..
o:) Thanks guys !
 
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1. What are nonzero complex numbers?

Nonzero complex numbers are numbers that contain both a real part and an imaginary part. They are written in the form a + bi, where a and b are real numbers and i is the imaginary unit equal to the square root of -1. Nonzero complex numbers are different from real numbers, which only contain a real part, and imaginary numbers, which only contain an imaginary part.

2. How are nonzero complex numbers represented on the complex plane?

Nonzero complex numbers are represented on the complex plane as points with coordinates (a,b), where a is the real part and b is the imaginary part. The horizontal axis represents the real numbers, while the vertical axis represents the imaginary numbers. The origin (0,0) on the complex plane represents the number 0, which is neither real nor imaginary.

3. What is the magnitude of a nonzero complex number?

The magnitude of a nonzero complex number is its distance from the origin on the complex plane. It is calculated using the Pythagorean theorem, by taking the square root of the sum of the squares of the real and imaginary parts. The magnitude of a nonzero complex number is always a positive real number.

4. How do you add and subtract nonzero complex numbers?

To add or subtract nonzero complex numbers, you simply combine the real parts and the imaginary parts separately. For example, to add (3+2i) and (5-4i), you would add 3 and 5 to get the real part, and add 2i and -4i to get the imaginary part, resulting in (8-2i). This follows the same rules as adding and subtracting polynomials.

5. What is the conjugate of a nonzero complex number?

The conjugate of a nonzero complex number a + bi is the complex number a - bi. In other words, the conjugate of a nonzero complex number simply changes the sign of the imaginary part. The product of a nonzero complex number and its conjugate always results in a real number, as the imaginary parts cancel out.

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