Number of solutions to complex eq.

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Homework Help Overview

The discussion revolves around determining the number of solutions to the complex equation z^2 - iz(conjugate) = 1/4. Participants are exploring the nature of complex equations and the implications of their structure on the number of solutions.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants present differing views on the number of solutions, with some suggesting there are two based on the equation's power, while others claim three solutions. Questions arise regarding the conditions under which these solutions hold and the reasoning behind the counts.

Discussion Status

The discussion is active, with participants questioning each other's reasoning and attempting to clarify the conditions that affect the number of solutions. Some guidance is offered regarding the relationship between the equation's power and the number of solutions, but no consensus has been reached.

Contextual Notes

There are indications of confusion regarding the assumptions made about the equation, particularly in relation to the conditions under which the solutions are derived. Participants are also grappling with the implications of complex numbers in their reasoning.

kasse
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Homework Statement



How many solutions does

z^2-iz(conjugate)=1/4

have?

2. The attempt at a solution

z^2=(1/4)+y+ix

Since the RHS is a complex number, the eq. has two solutions.

Correct?
 
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I found 3 solutions. Why don't you solve the equation ?
 
dextercioby said:
I found 3 solutions. Why don't you solve the equation ?

I learned that the number of solutions equals the power of the eq., that is 2...
 
dextercioby said:
Under what conditions ?

w^n=z
 
dextercioby said:
Okay, but did you solve the equation ?

Can't make it further than this:

(x^2-y^2-y-(1/4)+i(2xy-x)=0
 
Last edited:
dextercioby said:
Okay then. What follows next ?

Ah, y=2x since 2xy-x has got to be 0?

But then...no idea.
 
  • #10
Well, if 2xy=x, then i see 2 solutions, either x=0, or y=1/2.
 

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