How Do You Solve the Complex Equation \( z^2 + 4\overline{z} + 4 = 0 \)?

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Discussion Overview

The discussion revolves around solving the complex equation \( z^2 + 4\overline{z} + 4 = 0 \). Participants explore different methods for finding solutions, including factoring and using the quadratic formula, while addressing the complexities introduced by the conjugate of the complex variable.

Discussion Character

  • Homework-related, Mathematical reasoning

Main Points Raised

  • One participant requests help with solving the equation, indicating difficulty with the problem.
  • Another participant suggests that the equation is a quadratic polynomial and questions the first participant's familiarity with factoring or the quadratic formula.
  • A participant speculates that the answer might be \( 2i^2 \), but does not provide further justification.
  • Another participant proposes that the answer could simply be \( 2 \), also without elaboration.
  • A later reply clarifies that the equation involves a conjugate and suggests substituting \( z = a + bi \) to separate real and imaginary parts, leading to two equations that can be solved simultaneously.

Areas of Agreement / Disagreement

There is no consensus on the solutions to the equation, as participants propose different answers and methods without agreement on a definitive approach or result.

Contextual Notes

Participants have not fully resolved the implications of using the conjugate in the equation, and there are assumptions regarding the methods of solving quadratic equations that remain unexamined.

ronho1234
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find all solutions to z^2+4z ̅+4=0 where z is a complex number. Please help with this qn I'm having difficulty. thanks
 
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It's a quadratic polynomial. Do you not know how to factor these/the quadratic formula?
 
so is the answer just 2i^2?
 
or is it 2?
 
ronho1234 said:
find all solutions to z^2+4z ̅+4=0 where z is a complex number. Please help with this qn I'm having difficulty. thanks

That's a conjugate isn't it?

Suppose so then you have:

[tex]z^2+4\overline{z}+4=0[/tex]

Then let [itex]z=a+bi[/itex]

so:

[tex](a+bi)^2+4(a-bi)+4=0[/tex]

equate real and imaginaries to zero. That's two equations in two unknowns. Solve them.
 

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