# Complex Solutions

1. Mar 12, 2012

### Rubik

1. The problem statement, all variables and given/known data
Find all complex solutions to $\bar{z}$ = z

2. Relevant equations

z = x + iy and $\bar{z}$ = x - iy

3. The attempt at a solution
What does it mean by find all complex solutions?

$\bar{z}$ = z
0 = x + iy - x + iy
0 = 2iy
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Mar 12, 2012

### aanandpatel

Two complex numbers are only equal if their real parts are equal and their imaginary parts are equal so you may have to equate real and imaginary parts to find the values of x and y.

3. Mar 12, 2012

### SammyS

Staff Emeritus
If 0 = 2iy, then ...
1. What must x be for this to be true?

2. What must y be for this to be true?​

4. Mar 13, 2012

### aanandpatel

2y = 0 and x=0

5. Mar 13, 2012

### HallsofIvy

Staff Emeritus
How do you arrive at "x= 0" from an equation that does not have an "x" in it??

6. Mar 13, 2012

### NewtonianAlch

Well the basic form is x + iy, so we know the x part of the complex number must be equal to zero if it's not there.

7. Mar 13, 2012

### SammyS

Staff Emeritus
Look at the equation (from post #1):
0 = x + iy - x + iy​
Is there any x that will not satisfy this, if y=0 ? If there is such an x, what is it ?

8. Mar 14, 2012

### aanandpatel

Yes when I say x=0 it means that the 'real part' of the solution is 0

9. Mar 14, 2012

### HallsofIvy

Staff Emeritus
Yes, and as you have been told repeatedly, that is wrong. The equation 2iy= 0 does NOT say "x= 0 because x isn't there". The fact that x is not in that equation means that the equation does not tell you anything about x. Suppose z= 4+ 0i. What is $\overline{z}$?

10. Mar 14, 2012

### HallsofIvy

Staff Emeritus
Good point. I really hadn't thought of that! Okay, how about a simple sequence of real numbers?