
#1
Mar1212, 09:31 PM

P: 97

1. The problem statement, all variables and given/known data
Find all complex solutions to [itex]\bar{z}[/itex] = z 2. Relevant equations z = x + iy and [itex]\bar{z}[/itex] = x  iy 3. The attempt at a solution What does it mean by find all complex solutions? [itex]\bar{z}[/itex] = 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
Mar1212, 09:36 PM

P: 16

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
Mar1212, 10:59 PM

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P: 7,418

1. What must x be for this to be true? 



#4
Mar1312, 06:45 AM

P: 16

Complex Solutions
2y = 0 and x=0




#5
Mar1312, 06:52 AM

Math
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PF Gold
P: 38,902

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




#6
Mar1312, 07:07 AM

P: 440

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
Mar1312, 08:47 AM

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0 = x + iy  x + iyIs there any x that will not satisfy this, if y=0 ? If there is such an x, what is it ? 



#8
Mar1412, 12:28 AM

P: 16

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




#9
Mar1412, 05:57 AM

Math
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PF Gold
P: 38,902

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 [itex]\overline{z}[/itex]?




#10
Mar1412, 06:08 AM

Math
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PF Gold
P: 38,902

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



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