SOLUTION: Find all complex numbers z such that |z|^2-2\bar{z}+iz=2i

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Question 1085367: Find all complex numbers z such that
|z|^2-2\bar{z}+iz=2i
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
I'm assuming the initial equation is

If so, then let z+=+a%2Bbi where (a,b are real numbers)

The complex conjugate of z is

Using the pythagorean theorem, and a visual representation of z, we can say

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Using those equations, we can make a bit of substitutions and rearrangements to get the following















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From that last equation, in the section above, we can equate the real and imaginary parts to form these two equations

a%5E2%2Bb%5E2-2a-b+=+0

2b%2Ba+=+2

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Solve 2b%2Ba+=+2 for 'a'

2b%2Ba+=+2

2b%2Ba-2b+=+2-2b

a+=+2-2b

Now plug this into a%5E2%2Bb%5E2-2a-b+=+0 and solve for b

a%5E2%2Bb%5E2-2a-b+=+0

%282-2b%29%5E2%2Bb%5E2-2%282-2b%29-b+=+0

4-8b%2B4b%5E2%2Bb%5E2-4%2B4b-b+=+0

%284b%5E2%2Bb%5E2%29%2B%28-8b%2B4b-b%29%2B%284-4%29+=+0

5b%5E2+-+5b+=+0

5b%28b+-+1%29+=+0

5b=0 or b+-+1+=+0

b=0 or b+=+1

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If b=0, then

a+=+2-2b

a+=+2-2%2A0

a+=+2

Therefore one solution is z+=+a%2Bbi+=+2%2B0i+=+2 which is purely a real number.

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If b+=+1, then

a+=+2-2b

a+=+2-2%2A1

a+=+0

The other solution is z+=+a%2Bbi+=+0%2B1i+=+i which is a purely imaginary number

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Summary:

The two solutions are z+=+2%2B0i (which is the same as z+=+2) and z+=+0%2B1i (which is the same as z+=+i)