Solving the Equation for a Complex Number

adam640
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Homework Statement



Solve the equation \overline{y} (y - 2) = 2\overline{y} + 15 - 8i for complex number y


Homework Equations


\overline{y}y = a2 + b2


The Attempt at a Solution


a2+b2+8i-4ib-15=0
a2+b(b-4i)+8i-15=0

Pretty clueless where to go from here? Or if I've even gone in the right direction. Any help would be greatly appreciated.
Thanks,
Adam
 
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Check your arithmetic and when you are sure it is correct set it equal to 0 + 0i and equate real and imaginary parts.
 
I have set the equation to = 0 but I don't understand how I can set to 0i though? I'm sure about the maths so far.
 
0 = 0 + 0i

Say you have a simple complex equation:
a + 3 + 2bi = 0

This is the same as saying
(a + 3) + 2bi = 0 + 0i

Equate the real and imaginary coefficients:
a + 3 = 0 => a = -3
2b = 0 => b = 0
 
Thanks for the help, from what you've said I've gathered...

From a2+b(b-4i)+8i-15=0

0=0+0i
Therefore
From a2+b(b-4i)+8i-15=0+0i
a2-15=0
=> a = [STRIKE]\sqrt{15}[/STRIKE]
b2-4b-8 = 0
b is solved with the quadratic?

Is this correct so far?
 
adam640 said:
Thanks for the help, from what you've said I've gathered...

From a2+b(b-4i)+8i-15=0

That equation isn't correct. Check your work

0=0+0i
Therefore
From a2+b(b-4i)+8i-15=0+0i
a2-15=0
=> a = [STRIKE]\sqrt{15}[/STRIKE]
b2-4b-8 = 0

And that last step is wrong too. b(b-4i) isn't (b2-4b)i.

b is solved with the quadratic?

Is this correct so far?
 
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