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

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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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You are right. The calculator is wrong. It looks like it solved 4z-2z+i=0.
 
Oftentimes it's difficult to get a solution, but simple to check. If z = -i/6, then
4z - 2\bar{z} + i
4(-i/6) -2(i/6) + i = 0
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.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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