How do I Solve 4z - 2conj(z) + i = 0?

In summary, the conversation is about solving the equation 4z - 2conj(z) + i = 0, where the calculator provided an incorrect solution of z = -1/2*i. The person discussing the problem checked their own solution and confirmed it to be correct, showing that the calculator's solution is not accurate.
  • #1
squaremeplz
124
0

Homework Statement



hi

in trying to solve 4z - 2conj(z) + i = 0



Homework Equations





The Attempt at a Solution



the calculator spits out z = -1/2*i

i did 4(a+bi) - 2(a - bi) + i = 0

4a + 4bi - 2a + 2bi + i = 0

2a + 6bi + i = 0

i get z = -1/6*i

did i mess up somewhere? also, are there other solutions?

Thanks!
 
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  • #2
You are right. The calculator is wrong. It looks like it solved 4z-2z+i=0.
 
  • #3
Oftentimes it's difficult to get a solution, but simple to check. If z = -i/6, then
[tex]4z - 2\bar{z} + i[/tex]
[tex]4(-i/6) -2(i/6) + i = 0 [/tex]
This shows that z = -i/6 is the solution to your equation. Similarly, you can show that what the calculator produced is not the solution.
 

1. What is "z" in the equation 4z - 2conj(z) + i = 0?

"z" represents a complex number in the form of a + bi, where "a" and "b" are real numbers and "i" is the imaginary unit (i = √-1).

2. How do I solve for "z"?

To solve for "z", you can use the quadratic formula or factor the equation into two linear equations and solve for each separately. Remember to consider both the real and imaginary parts of the equation.

3. What does "conj(z)" mean?

"conj(z)" stands for the complex conjugate of the complex number "z". It is the same as "z" but with the sign of the imaginary part changed (e.g. conj(3+4i) = 3 - 4i).

4. How many solutions does the equation 4z - 2conj(z) + i = 0 have?

Since this is a quadratic equation, it will have two complex solutions. This is because the quadratic formula includes both a positive and negative solution.

5. Can the equation have real solutions?

Yes, it is possible for the equation to have real solutions. This would occur if the imaginary part of the complex solutions is equal to 0 (i.e. the solutions would be purely real numbers).

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